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A rope pulls the sled up the hill with a force of 60 N, and the sled is in equilibrium on the point of slipping up the hill.","difficulty":1,"difficultyBand":"medium","tier":null,"totalMarks":9,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"7a31a26c-0f30-4c0b-9223-91e0a864e514","label":"a","content":"Find the coefficient of friction between the sled and the hill.","markscheme":"- Resolves perpendicular: $R = 50 \\cos 45 = 25\\sqrt{2} \\approx 35.36$ M1\n- Resolves parallel (up plane): $60 = 50 \\sin 45 + \\mu R$ M1\n- Substitutes values: $60 = 35.36 + \\mu(35.36)$ M1\n- Rearranges: $24.64 = 35.36\\mu$ M1\n- Calculates $\\mu = \\frac{24.64}{35.36}$ M1\n- Final answer $\\mu = 0.697$ A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Correct resolution perpendicular to the plane.","code":"M"},{"id":"p2","marks":1,"description":"Correct equilibrium equation parallel to the plane.","code":"M"},{"id":"p3","marks":1,"description":"Correct substitution of numerical values into the parallel equation.","code":"M"},{"id":"p4","marks":1,"description":"Correct rearrangement of the equation.","code":"M"},{"id":"p5","marks":1,"description":"Solving for the ratio of $\\mu$.","code":"M"},{"id":"p6","marks":1,"description":"Correct final value for the coefficient of friction.","code":"A"}],"finalAnswers":["0.697"],"generatedAt":"2026-05-06T14:05:17.493Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"resolving_axes_plane","label":"Resolving Along the Plane","steps":[{"type":"text","order":0,"title":"Identify the forces","content":"To solve this problem using the **Resolving Along the Plane** approach, we first identify all the forces acting on the sled:\n\n1. **Weight ($W$):** $50\\text{ N}$ acting vertically downwards.\n2. **Normal Reaction Force ($R$):** acting perpendicular to and away from the inclined plane.\n3. **Tension/Pulling Force ($P$):** $60\\text{ N}$ acting parallel to and up the hill.\n4. **Frictional Force ($F$):** since the sled is on the point of slipping **up** the hill, the friction force acts in the opposite direction, which is **down** the hill.\n\nSince the sled is on the verge of moving, friction is at its limiting value: \n\n$$F = \\mu R$$","highlights":[{"text":"weight 50 N","location":"specification"},{"text":"inclined at an angle of 45°","location":"specification"},{"text":"pulls the sled up the hill with a force of 60 N","location":"specification"},{"text":"on the point of slipping up the hill","location":"specification"}]},{"type":"text","order":1,"title":"Resolve perpendicular to plane","content":"We resolve the forces in the direction perpendicular to the inclined hill. \n\nThe component of the weight perpendicular to the hill is $50 \\cos 45^\\circ$.\n\nSince there is no acceleration perpendicular to the hill, these forces must balance:\n\n$$R = 50 \\cos 45^\\circ$$\n\nUsing $\\cos 45^\\circ = \\frac{\\sqrt{2}}{2}$:\n\n$$R = 25\\sqrt{2} \\approx 35.36\\text{ N}$$"},{"type":"text","order":2,"title":"Resolve parallel to plane","content":"Next, we resolve forces parallel to the inclined plane (up and down the hill).\n\n* **Forces acting up the hill:** \n The pulling force of $60\\text{ N}$.\n* **Forces acting down the hill:** \n The component of weight acting down the hill, $50 \\sin 45^\\circ$, plus the friction force, $F = \\mu R$.\n\nSetting these forces equal for equilibrium:\n\n$$60 = 50 \\sin 45^\\circ + \\mu R$$"},{"type":"text","order":3,"title":"Substitute the values","content":"Now substitute the known values of $R \\approx 35.36\\text{ N}$ and $50 \\sin 45^\\circ = 50\\left(\\frac{\\sqrt{2}}{2}\\right) = 25\\sqrt{2} \\approx 35.36\\text{ N}$ into our parallel equilibrium equation:\n\n$$60 = 35.36 + \\mu(35.36)$$"},{"type":"text","order":4,"title":"Rearrange and solve","content":"Subtract $35.36$ from both sides to isolate the friction term:\n\n$$60 - 35.36 = 35.36\\mu$$\n\n$$24.64 = 35.36\\mu$$\n\nNow, solve for the coefficient of friction, $\\mu$:\n\n$$\\mu = \\frac{24.64}{35.36} \\approx 0.697$$"},{"type":"text","order":5,"title":"State the final answer","content":"Therefore, the coefficient of friction between the sled and the hill is:\n\n$$\\mu = 0.697\\text{ (to 3 significant figures)}$$"}]},{"id":"resolving_horizontal_vertical","label":"Resolving Horizontally and Vertically","steps":[{"type":"text","order":0,"title":"Identify forces and direction","content":"To solve this problem, we start by identifying all the forces acting on the sled:\n1. The **weight** ($W = 50\\text{ N}$) acting vertically downwards.\n2. The **normal reaction** ($R$) acting perpendicular to the inclined plane.\n3. The **tension/pulling force** ($P = 60\\text{ N}$) acting up the slope.\n4. The **friction force** ($F$) opposing the relative motion.\n\nSince the sled is on the point of slipping **up the hill**, friction acts in the opposite direction, which is **down the hill**.","highlights":[{"text":"A sled of weight 50 N","location":"specification"},{"text":"inclined at an angle of 45° to the horizontal","location":"specification"},{"text":"pulls the sled up the hill with a force of 60 N","location":"specification"},{"text":"on the point of slipping up the hill","location":"specification"}]},{"type":"text","order":1,"title":"Resolve perpendicular to plane","content":"First, let's resolve the forces perpendicular to the inclined plane to find the normal reaction, $R$.\n\nThe components perpendicular to the incline are the normal reaction force pushing away from the hill and the component of the weight pressing into the hill:\n\n$$R = 50\\cos(45^\\circ)$$\n\nUsing $\\cos(45^\\circ) = \\frac{\\sqrt{2}}{2}$:\n\n$$R = 25\\sqrt{2} \\approx 35.36\\text{ N}$$\n\nThis gives us our first mark **(M1)** for correctly finding the normal reaction."},{"type":"text","order":2,"title":"Set up parallel equation","content":"Next, we resolve forces parallel to the incline. \n\nSince the sled is in equilibrium, the total force pulling the sled up the hill is balanced by the forces opposing it down the hill (the component of gravity and the friction force $F$):\n\n$$P = W\\sin(45^\\circ) + F$$\n\nSubstituting our known values:\n\n$$60 = 50\\sin(45^\\circ) + F$$\n\nThis step earns the second mark **(M1)**."},{"type":"text","order":3,"title":"Incorporate friction formula","content":"Since the sled is on the point of slipping, friction is at its maximum limiting value, given by the formula:\n\n$$F = \\mu R$$\n\nNow, substitute $F = \\mu R$ and our calculated value of $R = 35.36\\text{ N}$ into our parallel resolution equation:\n\n$$60 = 50\\sin(45^\\circ) + \\mu(35.36)$$\n\nSince $50\\sin(45^\\circ) = 35.36$, we can write:\n\n$$60 = 35.36 + 35.36\\mu$$\n\nThis substitution and setup earn the third mark **(M1)**."},{"type":"text","order":4,"title":"Rearrange for mu","content":"To solve for the coefficient of friction, $\\mu$, we first isolate the term containing $\\mu$:\n\n$$60 - 35.36 = 35.36\\mu$$\n\n$$24.64 = 35.36\\mu$$\n\nThis rearrangement earns the fourth mark **(M1)**."},{"type":"text","order":5,"title":"Calculate final coefficient","content":"Finally, divide to isolate $\\mu$:\n\n$$\\mu = \\frac{24.64}{35.36}$$\n\nUsing a calculator to compute the value:\n\n$$\\mu \\approx 0.697$$\n\nThese final calculation steps earn the remaining marks: **(M1)** for the division setup, and **(A1)** for the final accurate value of $\\mu = 0.697$ (to 3 significant figures).","calculator":[{"gdc":{"expression":"24.64 / 35.36","instructions":"Type the expression into the main run screen and press EXE:\n24.64 / 35.36"},"ti84":{"expression":"24.64 / 35.36","instructions":"Enter the division calculation directly on the home screen:\nPress [2] [4] [.] [6] [4] [/] [3] [5] [.] [3] [6] [ENTER]"},"desmos":{"expression":"24.64 / 35.36","instructions":"Type the division into the calculation bar to see the decimal result:"},"description":"Calculate the coefficient of friction by dividing the remaining horizontal force by the normal force."}]}]}],"generatedAt":"2026-06-09T03:34:32.917Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"19d2d64a-e8dc-49bd-909c-f3e00601421c","label":"b","content":"The 60 N force is removed. Determine whether or not the sled moves.","markscheme":"- Calculates component of weight down hill: $50 \\sin 45 \\approx 35.36$ N M1\n- Calculates max friction: $F_{max} = \\mu R = 0.697 \\times 35.36 \\approx 24.65$ N M1\n- Compares forces: $35.36 > 24.65$, therefore the sled moves A1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Calculates the force attempting to pull the sled down (35.36 N).","code":"M"},{"id":"p2","marks":1,"description":"Calculates the limiting friction force (24.65 N).","code":"M"},{"id":"p3","marks":1,"description":"Makes correct comparison and concludes that the sled moves down the hill.","code":"A"}],"finalAnswers":["The sled moves"],"generatedAt":"2026-05-06T14:05:17.493Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"force_comparison","label":"Force Comparison Method","steps":[{"type":"text","order":0,"title":"Recall the parameters","content":"From the first part of the question, we have already found the coefficient of static friction between the sled and the hill:\n\n$$\\mu \\approx 0.697$$\n\nWe are also given the weight of the sled, $W = 50\\text{ N}$, and the angle of inclination, $\\theta = 45^\\circ$. \n\nWhen the $60\\text{ N}$ pulling force is removed, the only forces acting parallel to the hill are the component of gravity (weight) pulling the sled down, and the friction force acting up the hill to oppose this potential downward motion.","highlights":[{"text":"The 60 N force is removed. Determine whether or not the sled moves.","location":"specification"}]},{"type":"text","order":1,"title":"Calculate gravity component","content":"First, let's find the component of the sled's weight acting down the slope, which acts to pull the sled downwards:\n\n$$W_{\\text{parallel}} = W \\sin\\theta$$\n\nSubstituting the values:\n\n$$W_{\\text{parallel}} = 50 \\sin(45^\\circ) \\approx 35.36\\text{ N}$$\n\nThis is the downward force that friction must resist to keep the sled from sliding."},{"type":"text","order":2,"title":"Calculate normal reaction","content":"Next, we calculate the normal reaction force, $R$, acting perpendicular to the incline:\n\n$$R = W \\cos\\theta = 50 \\cos(45^\\circ) \\approx 35.36\\text{ N}$$"},{"type":"text","order":3,"title":"Find maximum static friction","content":"The maximum possible force that static friction can exert to prevent the sled from moving is given by:\n\n$$F_{\\text{max}} = \\mu R$$\n\nUsing $\\mu \\approx 0.697$ and $R \\approx 35.36\\text{ N}$:\n\n$$F_{\\text{max}} = 0.697 \\times 35.36 \\approx 24.65\\text{ N}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$W_{\\text{parallel}} = \\htmlData{anchor=w-val}{\\color{@red}{35.36\\text{ N}}}$"},{"latex":"$$F_{\\text{max}} = \\htmlData{anchor=f-val}{\\color{@sky}{24.65\\text{ N}}}$"},{"text":"Comparing the two values:"},{"latex":"$$\\htmlData{anchor=w-comp}{\\color{@red}{35.36\\text{ N}}} > \\htmlData{anchor=f-comp}{\\color{@sky}{24.65\\text{ N}}}$"}],"highlights":[{"color":"red","style":"text-color","anchor":"w-val"},{"color":"sky","style":"text-color","anchor":"f-val"}],"connections":[{"to":"w-comp","from":"w-val","color":"red","label":"Gravity force","style":"solid"},{"to":"f-comp","from":"f-val","color":"sky","label":"Max friction","style":"solid"}]},"order":4,"title":"Compare forces and conclude","description":"Since the component of the weight pulling the sled down the slope ($35.36\\text{ N}$) is greater than the maximum possible static friction force ($24.65\\text{ N}$), the static friction is insufficient to hold the sled in equilibrium. Therefore, **the sled moves** (it slides down the hill)."}]},{"id":"coefficient_angle_comparison","label":"Angle of Repose Method","steps":[{"type":"text","order":0,"title":"Understand angle of repose","content":"To determine whether the sled will slip down the hill when the pulling force is removed, we can use the **Angle of Repose** concept.\n\nThe angle of repose, $\theta_c$, is the steepest angle at which a material or object can remain resting on an inclined surface without sliding down.\n\nFrom mechanics theory, an object placed on an inclined plane will slide down under its own weight if and only if the tangent of the angle of inclination ($\theta$) is greater than the coefficient of static friction ($\\mu$):\n\n$$\\tan \\theta > \\mu$$"},{"type":"text","order":1,"title":"Find static friction coefficient","content":"From the first part of the question (where the sled is on the point of slipping *up* the hill with a $60\\text{ N}$ pulling force), the coefficient of static friction was calculated as:\n\n$$\\mu \\approx 0.697$$\n\nThis value represents the maximum ratio of frictional force to normal reaction force that the surface can support before sliding occurs."},{"type":"text","order":2,"title":"Compare incline to friction","content":"Now, we find the tangent of the incline angle, $\\theta = 45^\\circ$:\n\n$$\\tan 45^\\circ = 1.0$$\n\nNext, we compare $\\tan \\theta$ to the coefficient of static friction $\\mu$:\n\n$$\\tan 45^\\circ = 1.0 > 0.697 = \\mu$$"},{"type":"text","order":3,"title":"Conclude if sled moves","content":"Since the tangent of the incline angle is greater than the coefficient of static friction:\n\n$$\\tan \\theta > \\mu$$\n\nthe slope of the hill ($45^\\circ$) is steeper than the angle of repose for this surface interaction (which is $\\theta_c = \\arctan(0.697) \\approx 34.9^\\circ$). \n\nTherefore, the static friction force is not strong enough to hold the sled against gravity, and the **sled will move (slide down the hill)**."}]}],"generatedAt":"2026-06-09T03:35:40.841Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"7a31a26c-0f30-4c0b-9223-91e0a864e514:resolving_axes_plane","label":"(a) Resolving Along the Plane","steps":[{"type":"text","order":0,"title":"Identify the forces","content":"To solve this problem using the **Resolving Along the Plane** approach, we first identify all the forces acting on the sled:\n\n1. **Weight ($W$):** $50\\text{ N}$ acting vertically downwards.\n2. **Normal Reaction Force ($R$):** acting perpendicular to and away from the inclined plane.\n3. **Tension/Pulling Force ($P$):** $60\\text{ N}$ acting parallel to and up the hill.\n4. **Frictional Force ($F$):** since the sled is on the point of slipping **up** the hill, the friction force acts in the opposite direction, which is **down** the hill.\n\nSince the sled is on the verge of moving, friction is at its limiting value: \n\n$$F = \\mu R$$","highlights":[{"text":"weight 50 N","location":"specification"},{"text":"inclined at an angle of 45°","location":"specification"},{"text":"pulls the sled up the hill with a force of 60 N","location":"specification"},{"text":"on the point of slipping up the hill","location":"specification"}]},{"type":"text","order":1,"title":"Resolve perpendicular to plane","content":"We resolve the forces in the direction perpendicular to the inclined hill. \n\nThe component of the weight perpendicular to the hill is $50 \\cos 45^\\circ$.\n\nSince there is no acceleration perpendicular to the hill, these forces must balance:\n\n$$R = 50 \\cos 45^\\circ$$\n\nUsing $\\cos 45^\\circ = \\frac{\\sqrt{2}}{2}$:\n\n$$R = 25\\sqrt{2} \\approx 35.36\\text{ N}$$"},{"type":"text","order":2,"title":"Resolve parallel to plane","content":"Next, we resolve forces parallel to the inclined plane (up and down the hill).\n\n* **Forces acting up the hill:** \n The pulling force of $60\\text{ N}$.\n* **Forces acting down the hill:** \n The component of weight acting down the hill, $50 \\sin 45^\\circ$, plus the friction force, $F = \\mu R$.\n\nSetting these forces equal for equilibrium:\n\n$$60 = 50 \\sin 45^\\circ + \\mu R$$"},{"type":"text","order":3,"title":"Substitute the values","content":"Now substitute the known values of $R \\approx 35.36\\text{ N}$ and $50 \\sin 45^\\circ = 50\\left(\\frac{\\sqrt{2}}{2}\\right) = 25\\sqrt{2} \\approx 35.36\\text{ N}$ into our parallel equilibrium equation:\n\n$$60 = 35.36 + \\mu(35.36)$$"},{"type":"text","order":4,"title":"Rearrange and solve","content":"Subtract $35.36$ from both sides to isolate the friction term:\n\n$$60 - 35.36 = 35.36\\mu$$\n\n$$24.64 = 35.36\\mu$$\n\nNow, solve for the coefficient of friction, $\\mu$:\n\n$$\\mu = \\frac{24.64}{35.36} \\approx 0.697$$"},{"type":"text","order":5,"title":"State the final answer","content":"Therefore, the coefficient of friction between the sled and the hill is:\n\n$$\\mu = 0.697\\text{ (to 3 significant figures)}$$"}]},{"id":"7a31a26c-0f30-4c0b-9223-91e0a864e514:resolving_horizontal_vertical","label":"(a) Resolving Horizontally and Vertically","steps":[{"type":"text","order":0,"title":"Identify forces and direction","content":"To solve this problem, we start by identifying all the forces acting on the sled:\n1. The **weight** ($W = 50\\text{ N}$) acting vertically downwards.\n2. The **normal reaction** ($R$) acting perpendicular to the inclined plane.\n3. The **tension/pulling force** ($P = 60\\text{ N}$) acting up the slope.\n4. The **friction force** ($F$) opposing the relative motion.\n\nSince the sled is on the point of slipping **up the hill**, friction acts in the opposite direction, which is **down the hill**.","highlights":[{"text":"A sled of weight 50 N","location":"specification"},{"text":"inclined at an angle of 45° to the horizontal","location":"specification"},{"text":"pulls the sled up the hill with a force of 60 N","location":"specification"},{"text":"on the point of slipping up the hill","location":"specification"}]},{"type":"text","order":1,"title":"Resolve perpendicular to plane","content":"First, let's resolve the forces perpendicular to the inclined plane to find the normal reaction, $R$.\n\nThe components perpendicular to the incline are the normal reaction force pushing away from the hill and the component of the weight pressing into the hill:\n\n$$R = 50\\cos(45^\\circ)$$\n\nUsing $\\cos(45^\\circ) = \\frac{\\sqrt{2}}{2}$:\n\n$$R = 25\\sqrt{2} \\approx 35.36\\text{ N}$$\n\nThis gives us our first mark **(M1)** for correctly finding the normal reaction."},{"type":"text","order":2,"title":"Set up parallel equation","content":"Next, we resolve forces parallel to the incline. \n\nSince the sled is in equilibrium, the total force pulling the sled up the hill is balanced by the forces opposing it down the hill (the component of gravity and the friction force $F$):\n\n$$P = W\\sin(45^\\circ) + F$$\n\nSubstituting our known values:\n\n$$60 = 50\\sin(45^\\circ) + F$$\n\nThis step earns the second mark **(M1)**."},{"type":"text","order":3,"title":"Incorporate friction formula","content":"Since the sled is on the point of slipping, friction is at its maximum limiting value, given by the formula:\n\n$$F = \\mu R$$\n\nNow, substitute $F = \\mu R$ and our calculated value of $R = 35.36\\text{ N}$ into our parallel resolution equation:\n\n$$60 = 50\\sin(45^\\circ) + \\mu(35.36)$$\n\nSince $50\\sin(45^\\circ) = 35.36$, we can write:\n\n$$60 = 35.36 + 35.36\\mu$$\n\nThis substitution and setup earn the third mark **(M1)**."},{"type":"text","order":4,"title":"Rearrange for mu","content":"To solve for the coefficient of friction, $\\mu$, we first isolate the term containing $\\mu$:\n\n$$60 - 35.36 = 35.36\\mu$$\n\n$$24.64 = 35.36\\mu$$\n\nThis rearrangement earns the fourth mark **(M1)**."},{"type":"text","order":5,"title":"Calculate final coefficient","content":"Finally, divide to isolate $\\mu$:\n\n$$\\mu = \\frac{24.64}{35.36}$$\n\nUsing a calculator to compute the value:\n\n$$\\mu \\approx 0.697$$\n\nThese final calculation steps earn the remaining marks: **(M1)** for the division setup, and **(A1)** for the final accurate value of $\\mu = 0.697$ (to 3 significant figures).","calculator":[{"gdc":{"expression":"24.64 / 35.36","instructions":"Type the expression into the main run screen and press EXE:\n24.64 / 35.36"},"ti84":{"expression":"24.64 / 35.36","instructions":"Enter the division calculation directly on the home screen:\nPress [2] [4] [.] [6] [4] [/] [3] [5] [.] [3] [6] [ENTER]"},"desmos":{"expression":"24.64 / 35.36","instructions":"Type the division into the calculation bar to see the decimal result:"},"description":"Calculate the coefficient of friction by dividing the remaining horizontal force by the normal force."}]}]},{"id":"19d2d64a-e8dc-49bd-909c-f3e00601421c:force_comparison","label":"(b) Force Comparison Method","steps":[{"type":"text","order":0,"title":"Recall the parameters","content":"From the first part of the question, we have already found the coefficient of static friction between the sled and the hill:\n\n$$\\mu \\approx 0.697$$\n\nWe are also given the weight of the sled, $W = 50\\text{ N}$, and the angle of inclination, $\\theta = 45^\\circ$. \n\nWhen the $60\\text{ N}$ pulling force is removed, the only forces acting parallel to the hill are the component of gravity (weight) pulling the sled down, and the friction force acting up the hill to oppose this potential downward motion.","highlights":[{"text":"The 60 N force is removed. Determine whether or not the sled moves.","location":"specification"}]},{"type":"text","order":1,"title":"Calculate gravity component","content":"First, let's find the component of the sled's weight acting down the slope, which acts to pull the sled downwards:\n\n$$W_{\\text{parallel}} = W \\sin\\theta$$\n\nSubstituting the values:\n\n$$W_{\\text{parallel}} = 50 \\sin(45^\\circ) \\approx 35.36\\text{ N}$$\n\nThis is the downward force that friction must resist to keep the sled from sliding."},{"type":"text","order":2,"title":"Calculate normal reaction","content":"Next, we calculate the normal reaction force, $R$, acting perpendicular to the incline:\n\n$$R = W \\cos\\theta = 50 \\cos(45^\\circ) \\approx 35.36\\text{ N}$$"},{"type":"text","order":3,"title":"Find maximum static friction","content":"The maximum possible force that static friction can exert to prevent the sled from moving is given by:\n\n$$F_{\\text{max}} = \\mu R$$\n\nUsing $\\mu \\approx 0.697$ and $R \\approx 35.36\\text{ N}$:\n\n$$F_{\\text{max}} = 0.697 \\times 35.36 \\approx 24.65\\text{ N}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$W_{\\text{parallel}} = \\htmlData{anchor=w-val}{\\color{@red}{35.36\\text{ N}}}$"},{"latex":"$$F_{\\text{max}} = \\htmlData{anchor=f-val}{\\color{@sky}{24.65\\text{ N}}}$"},{"text":"Comparing the two values:"},{"latex":"$$\\htmlData{anchor=w-comp}{\\color{@red}{35.36\\text{ N}}} > \\htmlData{anchor=f-comp}{\\color{@sky}{24.65\\text{ N}}}$"}],"highlights":[{"color":"red","style":"text-color","anchor":"w-val"},{"color":"sky","style":"text-color","anchor":"f-val"}],"connections":[{"to":"w-comp","from":"w-val","color":"red","label":"Gravity force","style":"solid"},{"to":"f-comp","from":"f-val","color":"sky","label":"Max friction","style":"solid"}]},"order":4,"title":"Compare forces and conclude","description":"Since the component of the weight pulling the sled down the slope ($35.36\\text{ N}$) is greater than the maximum possible static friction force ($24.65\\text{ N}$), the static friction is insufficient to hold the sled in equilibrium. Therefore, **the sled moves** (it slides down the hill)."}]},{"id":"19d2d64a-e8dc-49bd-909c-f3e00601421c:coefficient_angle_comparison","label":"(b) Angle of Repose Method","steps":[{"type":"text","order":0,"title":"Understand angle of repose","content":"To determine whether the sled will slip down the hill when the pulling force is removed, we can use the **Angle of Repose** concept.\n\nThe angle of repose, $\\theta_c$, is the steepest angle at which a material or object can remain resting on an inclined surface without sliding down.\n\nFrom mechanics theory, an object placed on an inclined plane will slide down under its own weight if and only if the tangent of the angle of inclination ($\\theta$) is greater than the coefficient of static friction ($\\mu$):\n\n$$\\tan \\theta > \\mu$$"},{"type":"text","order":1,"title":"Find static friction coefficient","content":"From the first part of the question (where the sled is on the point of slipping *up* the hill with a $60\\text{ N}$ pulling force), the coefficient of static friction was calculated as:\n\n$$\\mu \\approx 0.697$$\n\nThis value represents the maximum ratio of frictional force to normal reaction force that the surface can support before sliding occurs."},{"type":"text","order":2,"title":"Compare incline to friction","content":"Now, we find the tangent of the incline angle, $\\theta = 45^\\circ$:\n\n$$\\tan 45^\\circ = 1.0$$\n\nNext, we compare $\\tan \\theta$ to the coefficient of static friction $\\mu$:\n\n$$\\tan 45^\\circ = 1.0 > 0.697 = \\mu$$"},{"type":"text","order":3,"title":"Conclude if sled moves","content":"Since the tangent of the incline angle is greater than the coefficient of static friction:\n\n$$\\tan \\theta > \\mu$$\n\nthe slope of the hill ($45^\\circ$) is steeper than the angle of repose for this surface interaction (which is $\\theta_c = \\arctan(0.697) \\approx 34.9^\\circ$). \n\nTherefore, the static friction force is not strong enough to hold the sled against gravity, and the **sled will move (slide down the hill)**."}]}],"generatedAt":"2026-06-09T03:35:50.877Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"38782f6d-5927-4b8a-9ac7-0d9e2aae0f3f","revisionId":"e7c4a70f-f072-4fe3-b340-d99e0ee408ab","questionNumber":null,"questionType":"LA","specification":"A small ornament $O$ is hung from a tree branch by a light inextensible thread. A horizontal breeze exerts a constant force of $12\\text{ N}$ on the ornament. The ornament is in equilibrium with the thread making an angle of $40^\\circ$ with the vertical.","difficulty":1,"difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"eb45d4d1-6afa-4dda-8f2d-d69b7cdd6b84","label":"a","content":"Find the tension in the thread.","markscheme":"- Resolving horizontally: $T \\sin 40 = 12$ M1\n- $T = 18.7\\text{ N}$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Correct resolution of forces in the horizontal direction.","code":"M"},{"id":"p2","marks":1,"description":"Accurate tension result rounded to 3sf.","code":"A"}],"finalAnswers":["18.7\\text{ N}","18.7"],"generatedAt":"2026-05-06T14:03:47.127Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"horizontal_resolution","label":"Resolving Forces Horizontally","steps":[{"type":"text","order":0,"title":"Identify the forces","content":"Let's first identify all the forces acting on the ornament $O$:\n\n1. The weight of the ornament acting vertically downwards.\n2. A horizontal force from the breeze of magnitude $12\\text{ N}$ acting horizontally.\n3. The tension $T$ in the thread acting along the thread at an angle of $40^\\circ$ to the vertical.\n\nSince the ornament is in equilibrium, the forces must balance in both the horizontal and vertical directions.","highlights":[{"text":"horizontal breeze exerts a constant force of 12 N on the ornament","location":"specification"},{"text":"thread making an angle of 40° with the vertical","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Horizontal component of tension} = \\htmlData{anchor=tension-comp}{\\color{@sky}{T \\sin 40^\\circ}}$"},{"latex":"$$\\text{Horizontal force from breeze} = \\htmlData{anchor=breeze-force}{\\color{@amber}{12\\text{ N}}}$"},{"latex":"$$\\htmlData{anchor=eq}{\\color{@sky}{T \\sin 40^\\circ}} = \\htmlData{anchor=eq-val}{\\color{@amber}{12}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"tension-comp"},{"color":"amber","style":"text-color","anchor":"breeze-force"}],"connections":[{"to":"eq","from":"tension-comp","color":"sky","label":null,"style":"solid"},{"to":"eq-val","from":"breeze-force","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Resolve horizontally","description":"We resolve the forces acting on the ornament horizontally. Because the angle is measured relative to the vertical, the horizontal component of the tension is opposite to the angle, which means we use the sine function: $T \\sin 40^\\circ$.\n\nSetting the horizontal forces equal to each other gives the equilibrium equation."},{"type":"text","order":2,"title":"Solve for tension","content":"To find the tension $T$, we rearrange our horizontal equilibrium equation:\n\n$$T = \\frac{12}{\\sin 40^\\circ}$$\n\nUsing a calculator, we evaluate this expression:\n\n$$T \\approx 18.7\\text{ N}\\text{ (to 3 significant figures)}$$\n\n*Note: In the exam, setting up the horizontal equilibrium equation correctly earns you the first mark (M1), and obtaining the final correct value of $18.7\\text{ N}$ earns the accuracy mark (A1).*","calculator":[{"gdc":{"expression":"12 / sin(40)","instructions":"1. Set angle settings to degrees.\n2. Calculate: 12 / sin(40)."},"ti84":{"expression":"12 / sin(40)","instructions":"1. Press [MODE] and ensure DEGREE is highlighted.\n2. Press [2nd] [MODE] to return to the home screen.\n3. Input: 12 / sin(40) and press [ENTER]."},"desmos":{"expression":"12 / \\sin(40)","instructions":"1. Click the wrench icon (Graph Settings) and select Degrees.\n2. Type: 12 / sin(40) into the input line."},"description":"Calculate the value of $T = \\frac{12}{\\sin 40^\\circ}$ using your scientific or graphing calculator. Ensure your calculator is set to **Degree** mode."}]}]},{"id":"force_triangle","label":"Triangle of Forces","steps":[{"type":"text","order":0,"title":"Identify the forces acting on the ornament","content":"To solve this using the triangle of forces method, let's first list the three forces acting on the ornament in equilibrium:\n\n1. **Tension ($T$):** Acts along the thread, at an angle of $40^\\circ$ to the vertical.\n2. **Horizontal Breeze ($F = 12\\text{ N}$):** Acts horizontally to the right (or left).\n3. **Weight ($W$):** Acts vertically downwards.\n\nSince the ornament is in static equilibrium, these three forces must form a closed triangle of forces.","highlights":[{"text":"horizontal breeze exerts a constant force of 12 N","location":"specification"},{"text":"thread making an angle of 40° with the vertical","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{Force vectors form a closed right-angled triangle:}"},{"latex":"\\htmlData{anchor=opp}{\\color{@sky}{12\\text{ N}}} \\quad \\text{(Opposite side, horizontal breeze)}"},{"latex":"\\htmlData{anchor=hyp}{\\color{@purple}{T}} \\quad \\text{(Hypotenuse, tension along the thread)}"},{"latex":"\\htmlData{anchor=theta}{\\color{@amber}{40^\\circ}} \\quad \\text{(Angle between tension and vertical weight vector)}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"opp"},{"color":"purple","style":"text-color","anchor":"hyp"},{"color":"amber","style":"text-color","anchor":"theta"}],"connections":[{"to":"hyp","from":"opp","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Construct the triangle of forces","description":"Since the breeze is horizontal and the weight is vertical, they are perpendicular. This gives us a right-angled triangle where the tension $T$ is the hypotenuse, the horizontal force of $12\\text{ N}$ is the side opposite the angle of $40^\\circ$, and the vertical weight $W$ is adjacent to the angle."},{"type":"text","order":2,"title":"Set up the trigonometric ratio","content":"Using the trigonometric definition of sine in a right-angled triangle:\n\n$$\\sin(\\theta) = \\frac{\\text{Opposite}}{\\text{Hypotenuse}}$$\n\nSubstituting our values into the formula gives:\n\n$$\\sin(40^\\circ) = \\frac{12}{T}$$\n\nRearranging to solve for the tension $T$:\n\n$$T = \\frac{12}{\\sin(40^\\circ)}$$"},{"type":"text","order":3,"title":"Calculate the tension value","content":"Now we compute the final value of the tension $T$:\n\n$$T = \\frac{12}{0.64278...} \\approx 18.7\\text{ N} \\text{ (to 3 sig. figs.)}$$\n\n**Markscheme Guidance:**\n- Setting up the correct ratio, e.g. $T \\sin(40^\\circ) = 12$, earns the method mark (**M1**).\n- Finding the correct value of $18.7\\text{ N}$ earns the final accuracy mark (**A1**).","calculator":[{"gdc":{"expression":"12 / sin(40)","instructions":"Set angle mode to Degrees. Enter the division of 12 by the sine of 40."},"ti84":{"expression":"12 / sin(40)","instructions":"Ensure your calculator is in DEGREE mode. Press [MODE], select 'Degree' and press [ENTER]. Then press [CLEAR] and enter the calculation:"},"desmos":{"expression":"12 / \\sin(40)","instructions":"Ensure the toggle in the wrench menu is set to 'Degrees' and type:"},"description":"Calculate the value of the tension using trigonometry."}]}]},{"id":"lamis_theorem","label":"Lami's Theorem","steps":[{"type":"text","order":0,"title":"Identify the forces","content":"The ornament $O$ is in equilibrium under the action of three coplanar forces:\n1. The **Tension** ($T$) acting along the thread, at $40^\\circ$ to the vertical.\n2. The **horizontal breeze force** ($F = 12\\text{ N}$).\n3. The **Weight** ($W$) of the ornament acting vertically downwards.","highlights":[{"text":"horizontal breeze exerts a constant force of 12 N","location":"specification"},{"text":"thread making an angle of 40° with the vertical","location":"specification"}]},{"type":"text","order":1,"title":"Find angles between forces","content":"To apply Lami's Theorem, we need the angles between each pair of forces:\n\n* **Angle between Weight ($W$) and Breeze ($F$):** Since weight is vertical and the breeze is horizontal, the angle between them is:\n $$\\theta_{W, F} = 90^\\circ$$\n\n* **Angle between Tension ($T$) and Weight ($W$):** The thread is pulled to the side, making an angle of $40^\\circ$ with the upward vertical. The angle to the downward vertical (the direction of weight) is:\n $$\\theta_{T, W} = 180^\\circ - 40^\\circ = 140^\\circ$$\n\n* **Angle between Tension ($T$) and Breeze ($F$):** Since the upward vertical and horizontal breeze are perpendicular ($90^\\circ$), and the tension is deflected $40^\\circ$ away from the breeze side, the angle is:\n $$\\theta_{T, F} = 90^\\circ + 40^\\circ = 130^\\circ$$\n\nLet's check that they sum to $360^\\circ$:\n$$90^\\circ + 140^\\circ + 130^\\circ = 360^\\circ$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{\\htmlData{anchor=force-t}{T}}{\\sin(\\htmlData{anchor=ang-wf}{90^\\circ})} = \\frac{\\htmlData{anchor=force-f}{12}}{\\sin(\\htmlData{anchor=ang-tw}{140^\\circ})} = \\frac{\\htmlData{anchor=force-w}{W}}{\\sin(\\htmlData{anchor=ang-tf}{130^\\circ})}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"force-t"},{"color":"sky","style":"text-color","anchor":"ang-wf"},{"color":"amber","style":"text-color","anchor":"force-f"},{"color":"amber","style":"text-color","anchor":"ang-tw"}],"connections":[{"to":"ang-wf","from":"force-t","color":"sky","label":"opposite angle","style":"solid"},{"to":"ang-tw","from":"force-f","color":"amber","label":"opposite angle","style":"solid"}]},"order":2,"title":"Apply Lami's Theorem","description":"Lami's Theorem states that for three coplanar forces in equilibrium, each force is proportional to the sine of the angle between the other two forces."},{"type":"text","order":3,"title":"Solve for Tension","content":"Using the relation between $T$ and the horizontal breeze force:\n\n$$\\frac{T}{\\sin 90^\\circ} = \\frac{12}{\\sin 140^\\circ}$$\n\nSince $\\sin 90^\\circ = 1$, we can simplify this directly to find $T$:\n\n$$T = \\frac{12}{\\sin 140^\\circ}$$"},{"type":"text","order":4,"title":"Calculate the final value","content":"Using a calculator to find the value of $T$:\n\n$$T = \\frac{12}{0.6428} \\approx 18.7\\text{ N}$$\n\nThis matches the markscheme requirement where resolving horizontally also yields $T \\sin 40^\\circ = 12 \\implies T = 18.7\\text{ N}$ (noting that $\\sin 140^\\circ = \\sin 40^\\circ$).","calculator":[{"gdc":{"expression":"12 / sin(40)","instructions":"1. Ensure your calculator's angle setting is set to 'Degree'.\n2. Enter the division expression.\n3. Press [EXE] or [ENTER]."},"ti84":{"expression":"12 / sin(40)","instructions":"1. Press [MODE] and ensure 'DEGREE' is highlighted.\n2. Press [2nd] [QUIT] to return to the home screen.\n3. Input the fraction expression.\n4. Press [ENTER]."},"desmos":{"expression":"12 / \\sin(40)","instructions":"1. Click the wrench icon (Graph Settings) in the upper-right corner and select 'Degrees'.\n2. Type the expression into an active cell to see the result."},"description":"Calculate the tension value with your graphing calculator in Degree mode."}]}]}],"generatedAt":"2026-06-09T04:22:00.176Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"78f632e3-852a-4a04-9c77-5e0e74cfb300","label":"b","content":"Find the weight of the ornament.","markscheme":"- Resolving vertically: $T \\cos 40 = W$ M1\n- $18.7 \\cos 40 = W$ M1\n- $W = 14.3\\text{ N}$ A1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Correct resolution of forces in the vertical direction.","code":"M"},{"id":"p2","marks":1,"description":"Substituting tension from part (a) into the vertical equation.","code":"M"},{"id":"p3","marks":1,"description":"Accurate weight result rounded to 3sf.","code":"A"}],"finalAnswers":["14.3\\text{ N}","14.3"],"generatedAt":"2026-05-06T14:03:47.127Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"resolving_components","label":"Resolving Components","steps":[{"type":"text","order":0,"title":"Analyze the forces","content":"To find the weight of the ornament, let's first identify all the forces acting on the ornament $O$:\n\n1. The weight of the ornament, $W$, acting vertically downwards.\n2. The force of the breeze, $F = 12\\text{ N}$, acting horizontally.\n3. The tension in the thread, $T$, acting along the thread at an angle of $40^\\circ$ to the vertical.\n\nSince the ornament is in equilibrium, the resultant force acting on it is zero. This means the horizontal forces must balance, and the vertical forces must balance.","highlights":[{"text":"horizontal breeze exerts a constant force of 12 N","location":"specification"},{"text":"thread making an angle of 40° with the vertical","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$T \\sin 40^\\circ = \\htmlData{anchor=force-val}{12}$"},{"latex":"$$T = \\frac{\\htmlData{anchor=force-num}{12}}{\\sin 40^\\circ}$"},{"latex":"$$T \\approx \\htmlData{anchor=tension-val}{18.67\\text{ N}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"force-val"},{"color":"amber","style":"text-color","anchor":"tension-val"}],"connections":[{"to":"force-num","from":"force-val","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Resolve forces horizontally","description":"By resolving the forces horizontally, we balance the horizontal component of the tension with the force of the breeze. This allows us to calculate the magnitude of the tension, $T$."},{"type":"text","order":2,"title":"Calculate tension value","content":"Let's perform the calculation to find the tension $T$ using our calculator:\n\n$$T = \\frac{12}{\\sin 40^\\circ} \\approx 18.67\\text{ N}$$","calculator":[{"gdc":{"expression":"12 / sin(40)","instructions":"Set angle units to degrees in system settings. In the run-matrix menu, input: 12 / sin(40)"},"ti84":{"expression":"12 / sin(40)","instructions":"Ensure your calculator is in DEGREE mode. Press [MODE], select 'Degree', then press [ENTER]. Then enter the calculation: 12 / sin(40)"},"desmos":{"expression":"12 / \\sin(40)","instructions":"Ensure Desmos is in 'Degrees' mode (click the wrench icon in the top right). Type: 12 / sin(40)"},"description":"Calculate the tension $T$ using a calculator in degree mode."}]},{"type":"hyper_latex","block":{"nodes":[{"text":"By resolving vertically, the vertical component of the tension balances the weight:"},{"latex":"$$\\htmlData{anchor=t-var}{T} \\cos 40^\\circ = W$"},{"latex":"$$\\htmlData{anchor=t-sub}{18.67} \\cos 40^\\circ = W$"},{"latex":"$$W \\approx \\htmlData{anchor=w-val}{14.3\\text{ N}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"t-var"},{"color":"amber","style":"text-color","anchor":"t-sub"},{"color":"green","style":"box","anchor":"w-val"}],"connections":[{"to":"t-sub","from":"t-var","color":"amber","label":"substitute T","style":"solid"}]},"order":3,"title":"Resolve forces vertically","description":"Now, we resolve the forces vertically to set up our equation for the weight $W$, and substitute our value of tension $T$ into it."},{"type":"text","order":4,"title":"Calculate final weight","content":"Let's perform the final multiplication to find the weight $W$:\n\n$$W = 18.67 \\times \\cos 40^\\circ \\approx 14.3\\text{ N}$$\n\nThis matches the target answer from the markscheme. Under exam conditions, the marks are awarded as follows:\n- **M1**: Correctly resolving vertically to get the equation $T \\cos 40^\\circ = W$.\n- **M1**: Substituting the tension value from Part 1 ($18.7\\text{ N}$ or $18.67\\text{ N}$) into the vertical equation.\n- **A1**: Finding the final weight $W = 14.3\\text{ N}$ (to 3 significant figures).","calculator":[{"gdc":{"expression":"18.67 * cos(40)","instructions":"Use Ans * cos(40) to keep precision, or input: 18.67 * cos(40)"},"ti84":{"expression":"18.67 * cos(40)","instructions":"With the previous answer on screen, press [*] then [cos] [40] [ENTER]. Or type: 18.67 * cos(40)"},"desmos":{"expression":"18.67 * \\cos(40)","instructions":"Type: 18.67 * cos(40)"},"description":"Calculate the weight $W$ using the unrounded tension value from your calculator history for maximum precision."}]}]},{"id":"force_triangle","label":"Force Triangle Method","steps":[{"type":"text","order":0,"title":"Identify the forces","content":"The ornament is in equilibrium under the action of three coplanar forces:\n1. The vertical **weight** ($W$) acting downwards.\n2. The constant **horizontal force** ($F = 12\\text{ N}$) acting horizontally due to the breeze.\n3. The **tension** ($T$) in the thread acting along the thread at an angle of $40^\\circ$ to the vertical.\n\nSince these three forces are in equilibrium, they must form a closed triangle of forces.","highlights":[{"text":"horizontal breeze exerts a constant force of 12 N","location":"specification"},{"text":"thread making an angle of 40° with the vertical","location":"specification"}]},{"type":"text","order":1,"title":"Construct the force triangle","content":"Because the weight ($W$) is vertical and the breeze force ($F = 12\\text{ N}$) is horizontal, they are perpendicular ($90^\\circ$ to each other).\n\nThis allows us to construct a right-angled force triangle:\n- The horizontal side representing the breeze force: $12\\text{ N}$\n- The vertical side representing the weight: $W$\n- The hypotenuse representing the tension: $T$\n\nThe angle between the tension ($T$) and the vertical ($W$) is $40^\\circ$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\tan(40^\\circ) = \\frac{\\text{Opposite}}{\\text{Adjacent}} = \\frac{\\htmlData{anchor=force}{\\color{@sky}{12}}}{\\htmlData{anchor=weight}{\\color{@amber}{W}}}$$"},{"latex":"$$$\\htmlData{anchor=weight-rearranged}{\\color{@amber}{W}} = \\frac{\\htmlData{anchor=force-rearranged}{\\color{@sky}{12}}}{\\tan(40^\\circ)}$$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"weight"},{"color":"sky","style":"text-color","anchor":"force"}],"connections":[{"to":"weight-rearranged","from":"weight","color":"amber","label":"rearrange for W","style":"solid"},{"to":"force-rearranged","from":"force","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Relate forces using trigonometry","description":"In our right-angled triangle, the side opposite the $40^\\circ$ angle is the horizontal breeze force of $12\\text{ N}$, and the adjacent side is the vertical weight $W$."},{"type":"text","order":3,"title":"Calculate the weight","content":"Now we evaluate this expression using a scientific calculator to find $W$:\n\n$$W = \\frac{12}{\\tan(40^\\circ)} \\approx 14.3\\text{ N}\\text{ (3 s.f.)}$$\n\nThis method bypasses finding the tension $T$ first, giving us the weight directly and efficiently!","calculator":[{"gdc":{"expression":"12 / tan(40)","instructions":"1. Set angle settings to degrees.\n2. Input: 12 / tan(40)\n3. Press [EXE] / [ENTER]."},"ti84":{"expression":"12 / tan(40)","instructions":"1. Press [MODE] and ensure 'DEGREE' is highlighted, then press [CLEAR].\n2. Enter: 12 / tan(40)\n3. Press [ENTER]."},"desmos":{"expression":"12 / \\tan(40^{\\circ})","instructions":"1. Click the wrench icon (Settings) and select 'Degrees'.\n2. Type: 12 / tan(40)"},"description":"Calculate the value of $12 / \\tan(40^\\circ)$ in degree mode."}]}]},{"id":"lamis_theorem","label":"Lami's Theorem","steps":[{"type":"text","order":0,"title":"Identify the forces","content":"To solve this problem using **Lami's Theorem**, we first identify the three coplanar forces acting on the ornament $O$ in equilibrium:\n1. The tension, $T$, acting along the thread at an angle of $40^\\circ$ to the vertical.\n2. The constant horizontal breeze of $12\\text{ N}$ acting horizontally.\n3. The weight, $W$, acting vertically downwards.","highlights":[{"text":"horizontal breeze exerts a constant force of 12 N","location":"specification"},{"text":"making an angle of 40° with the vertical","location":"specification"}]},{"type":"text","order":1,"title":"Find angles between forces","content":"Next, we find the angle between each pair of forces to set up Lami's Theorem:\n- **Angle between $W$ and the breeze ($12\\text{ N}$):** Since $W$ is vertical and the breeze is horizontal, the angle between them is:\n $$\\theta_{W, 12} = 90^\\circ$$\n- **Angle between $T$ and $W$:** The thread makes an angle of $40^\\circ$ with the upward vertical. The force $W$ acts straight downwards, so the angle between $T$ and $W$ is:\n $$\\theta_{T, W} = 180^\\circ - 40^\\circ = 140^\\circ$$\n- **Angle between $T$ and the breeze ($12\\text{ N}$):** The angle between the upward vertical and the horizontal force is $90^\\circ$. Since the thread is deflected away from the breeze by $40^\\circ$, the angle between $T$ and the breeze is:\n $$\\theta_{T, 12} = 90^\\circ + 40^\\circ = 130^\\circ$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Lami's Theorem states that for three forces in equilibrium, each force is proportional to the sine of the angle between the other two forces:"},{"latex":"$$$\\frac{\\htmlData{anchor=force-w}{\\color{@sky}{W}}}{\\sin(\\htmlData{anchor=angle-w}{\\color{@sky}{130^\\circ}})} = \\frac{\\htmlData{anchor=force-f}{\\color{@amber}{12}}}{\\sin(\\htmlData{anchor=angle-f}{\\color{@amber}{140^\\circ}})}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"force-w"},{"color":"sky","style":"text-color","anchor":"angle-w"},{"color":"amber","style":"text-color","anchor":"force-f"},{"color":"amber","style":"text-color","anchor":"angle-f"}],"connections":[{"to":"angle-w","from":"force-w","color":"sky","label":"opposite angle","style":"solid"},{"to":"angle-f","from":"force-f","color":"amber","label":"opposite angle","style":"solid"}]},"order":2,"title":"Apply Lami's Theorem","description":"We choose the ratios involving the known horizontal force ($12\\text{ N}$) and the unknown weight ($W$)."},{"type":"text","order":3,"title":"Calculate the weight","content":"Now, we rearrange the equation to solve for $W$:\n\n$$W = \\frac{12 \\sin(130^\\circ)}{\\sin(140^\\circ)}$$\n\nLet's calculate this value using a calculator:\n\n$$W \\approx 14.301\\text{ N}$$","calculator":[{"gdc":{"expression":"12 * sin(130) / sin(140)","instructions":"Set your calculator to Degree mode and enter the expression in the run-matrix menu."},"ti84":{"expression":"12 * sin(130) / sin(140)","instructions":"Ensure the calculator is in Degree mode: press [MODE], select [DEGREE], and press [ENTER]. Then enter the calculation."},"desmos":{"expression":"12 * \\sin(130) / \\sin(140)","instructions":"Make sure Desmos is set to Degrees (click the wrench icon in the top-right corner to change settings), then type the expression."},"description":"Evaluate the expression to find the weight in Newtons."}]},{"type":"text","order":4,"title":"State final answer","content":"Rounding our result to $3$ significant figures:\n\n$$W = 14.3\\text{ N}$$"}]}],"generatedAt":"2026-06-09T04:22:52.900Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"eb45d4d1-6afa-4dda-8f2d-d69b7cdd6b84:horizontal_resolution","label":"(a) Resolving Forces Horizontally","steps":[{"type":"text","order":0,"title":"Identify the forces","content":"Let's first identify all the forces acting on the ornament $O$:\n\n1. The weight of the ornament acting vertically downwards.\n2. A horizontal force from the breeze of magnitude $12\\text{ N}$ acting horizontally.\n3. The tension $T$ in the thread acting along the thread at an angle of $40^\\circ$ to the vertical.\n\nSince the ornament is in equilibrium, the forces must balance in both the horizontal and vertical directions.","highlights":[{"text":"horizontal breeze exerts a constant force of 12 N on the ornament","location":"specification"},{"text":"thread making an angle of 40° with the vertical","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Horizontal component of tension} = \\htmlData{anchor=tension-comp}{\\color{@sky}{T \\sin 40^\\circ}}$"},{"latex":"$$\\text{Horizontal force from breeze} = \\htmlData{anchor=breeze-force}{\\color{@amber}{12\\text{ N}}}$"},{"latex":"$$\\htmlData{anchor=eq}{\\color{@sky}{T \\sin 40^\\circ}} = \\htmlData{anchor=eq-val}{\\color{@amber}{12}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"tension-comp"},{"color":"amber","style":"text-color","anchor":"breeze-force"}],"connections":[{"to":"eq","from":"tension-comp","color":"sky","label":null,"style":"solid"},{"to":"eq-val","from":"breeze-force","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Resolve horizontally","description":"We resolve the forces acting on the ornament horizontally. Because the angle is measured relative to the vertical, the horizontal component of the tension is opposite to the angle, which means we use the sine function: $T \\sin 40^\\circ$.\n\nSetting the horizontal forces equal to each other gives the equilibrium equation."},{"type":"text","order":2,"title":"Solve for tension","content":"To find the tension $T$, we rearrange our horizontal equilibrium equation:\n\n$$T = \\frac{12}{\\sin 40^\\circ}$$\n\nUsing a calculator, we evaluate this expression:\n\n$$T \\approx 18.7\\text{ N}\\text{ (to 3 significant figures)}$$\n\n*Note: In the exam, setting up the horizontal equilibrium equation correctly earns you the first mark (M1), and obtaining the final correct value of $18.7\\text{ N}$ earns the accuracy mark (A1).*","calculator":[{"gdc":{"expression":"12 / sin(40)","instructions":"1. Set angle settings to degrees.\n2. Calculate: 12 / sin(40)."},"ti84":{"expression":"12 / sin(40)","instructions":"1. Press [MODE] and ensure DEGREE is highlighted.\n2. Press [2nd] [MODE] to return to the home screen.\n3. Input: 12 / sin(40) and press [ENTER]."},"desmos":{"expression":"12 / \\sin(40)","instructions":"1. Click the wrench icon (Graph Settings) and select Degrees.\n2. Type: 12 / sin(40) into the input line."},"description":"Calculate the value of $T = \\frac{12}{\\sin 40^\\circ}$ using your scientific or graphing calculator. Ensure your calculator is set to **Degree** mode."}]}]},{"id":"eb45d4d1-6afa-4dda-8f2d-d69b7cdd6b84:force_triangle","label":"(a) Triangle of Forces","steps":[{"type":"text","order":0,"title":"Identify the forces acting on the ornament","content":"To solve this using the triangle of forces method, let's first list the three forces acting on the ornament in equilibrium:\n\n1. **Tension ($T$):** Acts along the thread, at an angle of $40^\\circ$ to the vertical.\n2. **Horizontal Breeze ($F = 12\\text{ N}$):** Acts horizontally to the right (or left).\n3. **Weight ($W$):** Acts vertically downwards.\n\nSince the ornament is in static equilibrium, these three forces must form a closed triangle of forces.","highlights":[{"text":"horizontal breeze exerts a constant force of 12 N","location":"specification"},{"text":"thread making an angle of 40° with the vertical","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{Force vectors form a closed right-angled triangle:}"},{"latex":"\\htmlData{anchor=opp}{\\color{@sky}{12\\text{ N}}} \\quad \\text{(Opposite side, horizontal breeze)}"},{"latex":"\\htmlData{anchor=hyp}{\\color{@purple}{T}} \\quad \\text{(Hypotenuse, tension along the thread)}"},{"latex":"\\htmlData{anchor=theta}{\\color{@amber}{40^\\circ}} \\quad \\text{(Angle between the tension and vertical sides of the force triangle)}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"opp"},{"color":"purple","style":"text-color","anchor":"hyp"},{"color":"amber","style":"text-color","anchor":"theta"}],"connections":[{"to":"hyp","from":"opp","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Construct the triangle of forces","description":"Since the breeze is horizontal and the weight is vertical, they are perpendicular. This gives us a right-angled triangle where the tension $T$ is the hypotenuse, the horizontal force of $12\\text{ N}$ is the side opposite the angle of $40^\\circ$, and the vertical weight $W$ is adjacent to the angle."},{"type":"text","order":2,"title":"Set up the trigonometric ratio","content":"Using the trigonometric definition of sine in a right-angled triangle:\n\n$$\\sin(\\theta) = \\frac{\\text{Opposite}}{\\text{Hypotenuse}}$$\n\nSubstituting our values into the formula gives:\n\n$$\\sin(40^\\circ) = \\frac{12}{T}$$\n\nRearranging to solve for the tension $T$:\n\n$$T = \\frac{12}{\\sin(40^\\circ)}$$"},{"type":"text","order":3,"title":"Calculate the tension value","content":"Now we compute the final value of the tension $T$:\n\n$$T = \\frac{12}{0.64278...} \\approx 18.7\\text{ N} \\text{ (to 3 sig. figs.)}$$\n\n**Markscheme Guidance:**\n- Setting up the correct ratio, e.g. $T \\sin(40^\\circ) = 12$, earns the method mark (**M1**).\n- Finding the correct value of $18.7\\text{ N}$ earns the final accuracy mark (**A1**).","calculator":[{"gdc":{"expression":"12 / sin(40)","instructions":"Set angle mode to Degrees. Enter the division of 12 by the sine of 40."},"ti84":{"expression":"12 / sin(40)","instructions":"Ensure your calculator is in DEGREE mode. Press [MODE], select 'Degree' and press [ENTER]. Then press [CLEAR] and enter the calculation:"},"desmos":{"expression":"12 / \\sin(40)","instructions":"Ensure the toggle in the wrench menu is set to 'Degrees' and type:"},"description":"Calculate the value of the tension using trigonometry."}]}]},{"id":"eb45d4d1-6afa-4dda-8f2d-d69b7cdd6b84:lamis_theorem","label":"(a) Lami's Theorem","steps":[{"type":"text","order":0,"title":"Identify the forces","content":"The ornament $O$ is in equilibrium under the action of three coplanar forces:\n1. The **Tension** ($T$) acting along the thread, at $40^\\circ$ to the vertical.\n2. The **horizontal breeze force** ($F = 12\\text{ N}$).\n3. The **Weight** ($W$) of the ornament acting vertically downwards.","highlights":[{"text":"horizontal breeze exerts a constant force of 12 N","location":"specification"},{"text":"thread making an angle of 40° with the vertical","location":"specification"}]},{"type":"text","order":1,"title":"Find angles between forces","content":"To apply Lami's Theorem, we need the angles between each pair of forces:\n\n* **Angle between Weight ($W$) and Breeze ($F$):** Since weight is vertical and the breeze is horizontal, the angle between them is:\n $$\\theta_{W, F} = 90^\\circ$$\n\n* **Angle between Tension ($T$) and Weight ($W$):** The thread is pulled to the side, making an angle of $40^\\circ$ with the upward vertical. The angle to the downward vertical (the direction of weight) is:\n $$\\theta_{T, W} = 180^\\circ - 40^\\circ = 140^\\circ$$\n\n* **Angle between Tension ($T$) and Breeze ($F$):** Since the upward vertical and horizontal breeze are perpendicular ($90^\\circ$), and the tension is deflected $40^\\circ$ away from the breeze side, the angle is:\n $$\\theta_{T, F} = 90^\\circ + 40^\\circ = 130^\\circ$$\n\nLet's check that they sum to $360^\\circ$:\n$$90^\\circ + 140^\\circ + 130^\\circ = 360^\\circ$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{\\htmlData{anchor=force-t}{T}}{\\sin(\\htmlData{anchor=ang-wf}{90^\\circ})} = \\frac{\\htmlData{anchor=force-f}{12}}{\\sin(\\htmlData{anchor=ang-tw}{140^\\circ})} = \\frac{\\htmlData{anchor=force-w}{W}}{\\sin(\\htmlData{anchor=ang-tf}{130^\\circ})}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"force-t"},{"color":"sky","style":"text-color","anchor":"ang-wf"},{"color":"amber","style":"text-color","anchor":"force-f"},{"color":"amber","style":"text-color","anchor":"ang-tw"}],"connections":[{"to":"ang-wf","from":"force-t","color":"sky","label":"opposite angle","style":"solid"},{"to":"ang-tw","from":"force-f","color":"amber","label":"opposite angle","style":"solid"}]},"order":2,"title":"Apply Lami's Theorem","description":"Lami's Theorem states that for three coplanar forces in equilibrium, each force is proportional to the sine of the angle between the other two forces."},{"type":"text","order":3,"title":"Solve for Tension","content":"Using the relation between $T$ and the horizontal breeze force:\n\n$$\\frac{T}{\\sin 90^\\circ} = \\frac{12}{\\sin 140^\\circ}$$\n\nSince $\\sin 90^\\circ = 1$, we can simplify this directly to find $T$:\n\n$$T = \\frac{12}{\\sin 140^\\circ}$$"},{"type":"text","order":4,"title":"Calculate the final value","content":"Using a calculator to find the value of $T$:\n\n$$T = \\frac{12}{0.6428} \\approx 18.7\\text{ N}$$\n\nThis matches the markscheme requirement where resolving horizontally also yields $T \\sin 40^\\circ = 12 \\implies T = 18.7\\text{ N}$ (noting that $\\sin 140^\\circ = \\sin 40^\\circ$).","calculator":[{"gdc":{"expression":"12 / sin(40)","instructions":"1. Ensure your calculator's angle setting is set to 'Degree'.\n2. Enter the division expression.\n3. Press [EXE] or [ENTER]."},"ti84":{"expression":"12 / sin(40)","instructions":"1. Press [MODE] and ensure 'DEGREE' is highlighted.\n2. Press [2nd] [QUIT] to return to the home screen.\n3. Input the fraction expression.\n4. Press [ENTER]."},"desmos":{"expression":"12 / \\sin(40)","instructions":"1. Click the wrench icon (Graph Settings) in the upper-right corner and select 'Degrees'.\n2. Type the expression into an active cell to see the result."},"description":"Calculate the tension value with your graphing calculator in Degree mode."}]}]},{"id":"78f632e3-852a-4a04-9c77-5e0e74cfb300:resolving_components","label":"(b) Resolving Components","steps":[{"type":"text","order":0,"title":"Analyze the forces","content":"To find the weight of the ornament, let's first identify all the forces acting on the ornament $O$:\n\n1. The weight of the ornament, $W$, acting vertically downwards.\n2. The force of the breeze, $F = 12\\text{ N}$, acting horizontally.\n3. The tension in the thread, $T$, acting along the thread at an angle of $40^\\circ$ to the vertical.\n\nSince the ornament is in equilibrium, the resultant force acting on it is zero. This means the horizontal forces must balance, and the vertical forces must balance.","highlights":[{"text":"horizontal breeze exerts a constant force of 12 N","location":"specification"},{"text":"thread making an angle of 40° with the vertical","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$T \\sin 40^\\circ = \\htmlData{anchor=force-val}{12}$"},{"latex":"$$T = \\frac{\\htmlData{anchor=force-num}{12}}{\\sin 40^\\circ}$"},{"latex":"$$T \\approx \\htmlData{anchor=tension-val}{18.67\\text{ N}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"force-val"},{"color":"amber","style":"text-color","anchor":"tension-val"}],"connections":[{"to":"force-num","from":"force-val","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Resolve forces horizontally","description":"By resolving the forces horizontally, we balance the horizontal component of the tension with the force of the breeze. This allows us to calculate the magnitude of the tension, $T$."},{"type":"text","order":2,"title":"Calculate tension value","content":"Let's perform the calculation to find the tension $T$ using our calculator:\n\n$$T = \\frac{12}{\\sin 40^\\circ} \\approx 18.67\\text{ N}$$","calculator":[{"gdc":{"expression":"12 / sin(40)","instructions":"Set angle units to degrees in system settings. In the run-matrix menu, input: 12 / sin(40)"},"ti84":{"expression":"12 / sin(40)","instructions":"Ensure your calculator is in DEGREE mode. Press [MODE], select 'Degree', then press [ENTER]. Then enter the calculation: 12 / sin(40)"},"desmos":{"expression":"12 / \\sin(40)","instructions":"Ensure Desmos is in 'Degrees' mode (click the wrench icon in the top right). Type: 12 / sin(40)"},"description":"Calculate the tension $T$ using a calculator in degree mode."}]},{"type":"hyper_latex","block":{"nodes":[{"text":"By resolving vertically, the vertical component of the tension balances the weight:"},{"latex":"$$\\htmlData{anchor=t-var}{T} \\cos 40^\\circ = W$"},{"latex":"$$\\htmlData{anchor=t-sub}{18.67} \\cos 40^\\circ = W$"},{"latex":"$$W \\approx \\htmlData{anchor=w-val}{14.3\\text{ N}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"t-var"},{"color":"amber","style":"text-color","anchor":"t-sub"},{"color":"green","style":"box","anchor":"w-val"}],"connections":[{"to":"t-sub","from":"t-var","color":"amber","label":"substitute T","style":"solid"}]},"order":3,"title":"Resolve forces vertically","description":"Now, we resolve the forces vertically to set up our equation for the weight $W$, and substitute our value of tension $T$ into it."},{"type":"text","order":4,"title":"Calculate final weight","content":"Let's perform the final multiplication to find the weight $W$:\n\n$$W = 18.67 \\times \\cos 40^\\circ \\approx 14.3\\text{ N}$$\n\nThis matches the target answer from the markscheme. Under exam conditions, the marks are awarded as follows:\n- **M1**: Correctly resolving vertically to get the equation $T \\cos 40^\\circ = W$.\n- **M1**: Substituting the tension value from Part 1 ($18.7\\text{ N}$ or $18.67\\text{ N}$) into the vertical equation.\n- **A1**: Finding the final weight $W = 14.3\\text{ N}$ (to 3 significant figures).","calculator":[{"gdc":{"expression":"18.67 * cos(40)","instructions":"Use Ans * cos(40) to keep precision, or input: 18.67 * cos(40)"},"ti84":{"expression":"18.67 * cos(40)","instructions":"With the previous answer on screen, press [*] then [cos] [40] [ENTER]. Or type: 18.67 * cos(40)"},"desmos":{"expression":"18.67 * \\cos(40)","instructions":"Type: 18.67 * cos(40)"},"description":"Calculate the weight $W$ using the unrounded tension value from your calculator history for maximum precision."}]}]},{"id":"78f632e3-852a-4a04-9c77-5e0e74cfb300:force_triangle","label":"(b) Force Triangle Method","steps":[{"type":"text","order":0,"title":"Identify the forces","content":"The ornament is in equilibrium under the action of three coplanar forces:\n1. The vertical **weight** ($W$) acting downwards.\n2. The constant **horizontal force** ($F = 12\\text{ N}$) acting horizontally due to the breeze.\n3. The **tension** ($T$) in the thread acting along the thread at an angle of $40^\\circ$ to the vertical.\n\nSince these three forces are in equilibrium, they must form a closed triangle of forces.","highlights":[{"text":"horizontal breeze exerts a constant force of 12 N","location":"specification"},{"text":"thread making an angle of 40° with the vertical","location":"specification"}]},{"type":"text","order":1,"title":"Construct the force triangle","content":"Because the weight ($W$) is vertical and the breeze force ($F = 12\\text{ N}$) is horizontal, they are perpendicular ($90^\\circ$ to each other).\n\nThis allows us to construct a right-angled force triangle:\n- The horizontal side representing the breeze force: $12\\text{ N}$\n- The vertical side representing the weight: $W$\n- The hypotenuse representing the tension: $T$\n\nThe angle between the tension ($T$) and the vertical ($W$) is $40^\\circ$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\tan(40^\\circ) = \\frac{\\text{Opposite}}{\\text{Adjacent}} = \\frac{\\htmlData{anchor=force}{\\color{@sky}{12}}}{\\htmlData{anchor=weight}{\\color{@amber}{W}}}$$"},{"latex":"$$$\\htmlData{anchor=weight-rearranged}{\\color{@amber}{W}} = \\frac{\\htmlData{anchor=force-rearranged}{\\color{@sky}{12}}}{\\tan(40^\\circ)}$$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"weight"},{"color":"sky","style":"text-color","anchor":"force"}],"connections":[{"to":"weight-rearranged","from":"weight","color":"amber","label":"rearrange for W","style":"solid"},{"to":"force-rearranged","from":"force","color":"sky","label":null,"style":"solid"}]},"order":2,"title":"Relate forces using trigonometry","description":"In our right-angled triangle, the side opposite the $40^\\circ$ angle is the horizontal breeze force of $12\\text{ N}$, and the adjacent side is the vertical weight $W$."},{"type":"text","order":3,"title":"Calculate the weight","content":"Now we evaluate this expression using a scientific calculator to find $W$:\n\n$$W = \\frac{12}{\\tan(40^\\circ)} \\approx 14.3\\text{ N}\\text{ (3 s.f.)}$$\n\nThis method bypasses finding the tension $T$ first, giving us the weight directly and efficiently!","calculator":[{"gdc":{"expression":"12 / tan(40)","instructions":"1. Set angle settings to degrees.\n2. Input: 12 / tan(40)\n3. Press [EXE] / [ENTER]."},"ti84":{"expression":"12 / tan(40)","instructions":"1. Press [MODE] and ensure 'DEGREE' is highlighted, then press [CLEAR].\n2. Enter: 12 / tan(40)\n3. Press [ENTER]."},"desmos":{"expression":"12 / \\tan(40^{\\circ})","instructions":"1. Click the wrench icon (Settings) and select 'Degrees'.\n2. Type: 12 / tan(40)"},"description":"Calculate the value of $12 / \\tan(40^\\circ)$ in degree mode."}]}]},{"id":"78f632e3-852a-4a04-9c77-5e0e74cfb300:lamis_theorem","label":"(b) Lami's Theorem","steps":[{"type":"text","order":0,"title":"Identify the forces","content":"To solve this problem using **Lami's Theorem**, we first identify the three coplanar forces acting on the ornament $O$ in equilibrium:\n1. The tension, $T$, acting along the thread at an angle of $40^\\circ$ to the vertical.\n2. The constant horizontal breeze of $12\\text{ N}$ acting horizontally.\n3. The weight, $W$, acting vertically downwards.","highlights":[{"text":"horizontal breeze exerts a constant force of 12 N","location":"specification"},{"text":"making an angle of 40° with the vertical","location":"specification"}]},{"type":"text","order":1,"title":"Find angles between forces","content":"Next, we find the angle between each pair of forces to set up Lami's Theorem:\n- **Angle between $W$ and the breeze ($12\\text{ N}$):** Since $W$ is vertical and the breeze is horizontal, the angle between them is:\n $$\\theta_{W, 12} = 90^\\circ$$\n- **Angle between $T$ and $W$:** The thread makes an angle of $40^\\circ$ with the upward vertical. The force $W$ acts straight downwards, so the angle between $T$ and $W$ is:\n $$\\theta_{T, W} = 180^\\circ - 40^\\circ = 140^\\circ$$\n- **Angle between $T$ and the breeze ($12\\text{ N}$):** The angle between the upward vertical and the horizontal force is $90^\\circ$. Since the thread is deflected away from the breeze by $40^\\circ$, the angle between $T$ and the breeze is:\n $$\\theta_{T, 12} = 90^\\circ + 40^\\circ = 130^\\circ$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Lami's Theorem states that for three forces in equilibrium, each force is proportional to the sine of the angle between the other two forces:"},{"latex":"$$$\\frac{\\htmlData{anchor=force-w}{\\color{@sky}{W}}}{\\sin(\\htmlData{anchor=angle-w}{\\color{@sky}{130^\\circ}})} = \\frac{\\htmlData{anchor=force-f}{\\color{@amber}{12}}}{\\sin(\\htmlData{anchor=angle-f}{\\color{@amber}{140^\\circ}})}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"force-w"},{"color":"sky","style":"text-color","anchor":"angle-w"},{"color":"amber","style":"text-color","anchor":"force-f"},{"color":"amber","style":"text-color","anchor":"angle-f"}],"connections":[{"to":"angle-w","from":"force-w","color":"sky","label":"opposite angle","style":"solid"},{"to":"angle-f","from":"force-f","color":"amber","label":"opposite angle","style":"solid"}]},"order":2,"title":"Apply Lami's Theorem","description":"We choose the ratios involving the known horizontal force ($12\\text{ N}$) and the unknown weight ($W$)."},{"type":"text","order":3,"title":"Calculate the weight","content":"Now, we rearrange the equation to solve for $W$:\n\n$$W = \\frac{12 \\sin(130^\\circ)}{\\sin(140^\\circ)}$$\n\nLet's calculate this value using a calculator:\n\n$$W \\approx 14.301\\text{ N}$$","calculator":[{"gdc":{"expression":"12 * sin(130) / sin(140)","instructions":"Set your calculator to Degree mode and enter the expression in the run-matrix menu."},"ti84":{"expression":"12 * sin(130) / sin(140)","instructions":"Ensure the calculator is in Degree mode: press [MODE], select [DEGREE], and press [ENTER]. Then enter the calculation."},"desmos":{"expression":"12 * \\sin(130) / \\sin(140)","instructions":"Make sure Desmos is set to Degrees (click the wrench icon in the top-right corner to change settings), then type the expression."},"description":"Evaluate the expression to find the weight in Newtons."}]},{"type":"text","order":4,"title":"State final answer","content":"Rounding our result to $3$ significant figures:\n\n$$W = 14.3\\text{ N}$$"}]}],"generatedAt":"2026-06-09T04:23:13.747Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"4cd7c753-d44c-4ed6-b81b-9f4da12259e7","revisionId":"280915ba-6946-4d21-bb40-4ebacfe1fac2","questionNumber":null,"questionType":"LA","specification":"A rough slope is inclined to the horizontal at an angle $\\phi$, where $\\tan \\phi = \\frac{7}{24}$. A package of mass $m$ kg is placed on the slope and released. The coefficient of friction between the package and the slope is $\\frac{1}{4}$.","difficulty":1,"difficultyBand":"medium","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"66640675-41a8-4b71-b157-a035ba62844c","label":"a","content":"Find the acceleration of the package down the slope.","markscheme":"- Trig values: $\\tan \\phi = 7/24 \\Rightarrow \\sin \\phi = 7/25, \\cos \\phi = 24/25$ B1\n- Resolves perpendicular: $R = mg \\cos \\phi = \\frac{24}{25}mg$ M1\n- Applies $F = ma$ parallel: $mg \\sin \\phi - \\mu R = ma$ M1\n- Substitutes: $mg(\\frac{7}{25}) - \\frac{1}{4}(\\frac{24}{25}mg) = ma$ M1\n- Simplifies: $(\\frac{7}{25} - \\frac{6}{25})g = a$ M1\n- Final answer: $a = \\frac{1}{25}g$ ms$^{-2}$ (or $0.04g$ or $0.392$ ms$^{-2}$) A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Determine $\\sin \\phi = 7/25$ and $\\cos \\phi = 24/25$.","code":"B"},{"id":"p2","marks":1,"description":"Calculate normal reaction $R$ in terms of $m, g$ and $\\cos \\phi$.","code":"M"},{"id":"p3","marks":1,"description":"Formulate $F=ma$ equation along the slope.","code":"M"},{"id":"p4","marks":1,"description":"Substitute numerical and algebraic values into the $F=ma$ equation.","code":"M"},{"id":"p5","marks":1,"description":"Cancel mass $m$ and perform arithmetic subtraction of terms.","code":"M"},{"id":"p6","marks":1,"description":"Obtain final acceleration $1/25 g$.","code":"A"}],"finalAnswers":["\\frac{1}{25}g","0.04g","0.392"],"generatedAt":"2026-05-06T14:04:53.885Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"exact_trig","label":"Exact Trigonometric Ratios","steps":[{"type":"text","order":0,"title":"Find exact trigonometric ratios","content":"We are given $\\tan \\phi = \\frac{7}{24}$. Instead of finding the angle $\\phi$ in degrees and introducing rounding errors, we can find the exact values for $\\sin \\phi$ and $\\cos \\phi$ using the properties of right-angled triangles.\n\nSince $\\tan \\phi = \\frac{\\text{Opposite}}{\\text{Adjacent}} = \\frac{7}{24}$:\n- $\\text{Opposite} = 7$\n- $\\text{Adjacent} = 24$\n\nUsing the Pythagorean theorem (or recognizing the 7-24-25 Pythagorean triple), the hypotenuse is:\n\n$$\\text{Hypotenuse} = \\sqrt{7^2 + 24^2} = \\sqrt{49 + 576} = \\sqrt{625} = 25$$\n\nNow we can write down the exact values of the trigonometric ratios:\n\n$$\\sin \\phi = \\frac{7}{25}$$\n$$\\cos \\phi = \\frac{24}{25}$$\n\nThis earns you the first mark **[B1]**."},{"type":"text","order":1,"title":"Resolve forces perpendicular to slope","content":"Next, we resolve forces perpendicular to the incline. \n\nThe forces acting perpendicular to the slope are:\n- The normal reaction force $R$ acting perpendicularly away from the slope.\n- The perpendicular component of the weight acting into the slope, which is $mg \\cos \\phi$.\n\nSince there is no motion perpendicular to the slope, the forces must balance:\n\n$$R = mg \\cos \\phi$$\n\nSubstituting our exact value for $\\cos \\phi$:\n\n$$R = mg \\left(\\frac{24}{25}\\right) = \\frac{24}{25}mg$$\n\nThis earns the second mark **[M1]**."},{"type":"text","order":2,"title":"Calculate friction force","content":"Since the package is released and moves down the slope, friction is at its maximum value (limiting friction). We use the friction formula:\n\n$$F = \\mu R$$\n\nWe are given the coefficient of friction $\\mu = \\frac{1}{4}$. Substituting our values:\n\n$$F = \\frac{1}{4} \\left( \\frac{24}{25}mg \\right)$$\n$$F = \\frac{6}{25}mg$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$mg \\htmlData{anchor=sin-ratio}{\\color{@sky}{\\sin \\phi}} - \\htmlData{anchor=friction-force}{\\color{@amber}{F}} = ma$"},{"latex":"$$mg \\left( \\htmlData{anchor=sin-sub}{\\color{@sky}{\\frac{7}{25}}} \\right) - \\htmlData{anchor=friction-sub}{\\color{@amber}{\\frac{6}{25}mg}} = ma$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-ratio"},{"color":"amber","style":"text-color","anchor":"friction-force"}],"connections":[{"to":"sin-sub","from":"sin-ratio","color":"sky","label":"Substitute","style":"solid"},{"to":"friction-sub","from":"friction-force","color":"amber","label":"Substitute","style":"solid"}]},"order":3,"title":"Set up equation of motion","description":"We apply Newton's second law ($F = ma$) parallel to the slope. The component of weight pulling the package down is $mg \\sin \\phi$, and the opposing force is friction $F$."},{"type":"text","order":4,"title":"Solve for acceleration","content":"We can now simplify the algebraic equation and solve for the acceleration $a$:\n\n$$\\frac{7}{25}mg - \\frac{6}{25}mg = ma$$\n\nFactor out $mg$ on the left side to simplify:\n\n$$\\left( \\frac{7}{25} - \\frac{6}{25} \\right)mg = ma$$\n\n$$\\frac{1}{25}mg = ma$$\n\nDivide both sides by the mass $m$ to isolate $a$:\n\n$$a = \\frac{1}{25}g \\text{ m s}^{-2}$$\n\nUsing the standard approximation for acceleration due to gravity $g = 9.8 \\text{ m s}^{-2}$, we can also express this in decimal form:\n\n$$a = 0.04g = 0.392 \\text{ m s}^{-2}$$\n\nThis earns the final marks **[M1]** for simplifying and **[A1]** for the correct final answer.","calculator":[{"gdc":{"expression":"0.04 * 9.8","instructions":"In the Run-Matrix menu, enter 0.04 * 9.8."},"ti84":{"expression":"0.04 * 9.8","instructions":"Type 0.04 * 9.8 and press ENTER."},"desmos":{"expression":"0.04 * 9.8","instructions":"Type 0.04 * 9.8 into the input line."},"description":"Calculate the numerical value of acceleration using $g = 9.8\\text{ m s}^{-2}$."}]}]},{"id":"decimal_approximation","label":"Decimal Approximation","steps":[{"type":"text","order":0,"title":"Find the angle of inclination","content":"We are given that the slope is inclined at an angle $\\phi$ where:\n\n$$\\tan \\phi = \\frac{7}{24}$$\n\nTo use the decimal approximation method, we first find the angle $\\phi$ using the inverse tangent function on our calculator:\n\n$$\\phi = \\arctan\\left(\\frac{7}{24}\\right) \\approx 16.2602^\\circ$$"},{"type":"text","order":1,"title":"Calculate decimal trig values","content":"Now we compute the decimal approximations for both $\\sin \\phi$ and $\\cos \\phi$ using the angle we just found:\n\n$$\\sin(16.2602^\\circ) \\approx 0.2800$$\n$$\\cos(16.2602^\\circ) \\approx 0.9600$$\n\n*(Note: These correspond to the exact values $\\frac{7}{25}$ and $\\frac{24}{25}$, but we will carry forward these decimal approximations to solve our system of equations.)*","calculator":[{"gdc":{"expression":"sin(16.2602)","instructions":"Calculate sine and cosine of the stored angle value."},"ti84":{"expression":"sin(16.2602)","instructions":"Press [SIN] (16.2602) [ENTER] and [COS] (16.2602) [ENTER]."},"desmos":{"expression":"\\sin(16.2602^\\circ)","instructions":"Type sin(16.2602) and cos(16.2602) with settings in Degrees."},"description":"Find sine and cosine of the angle"}]},{"type":"text","order":2,"title":"Resolve forces perpendicular","content":"The forces acting perpendicular to the inclined plane must balance because there is no motion in this direction.\n\nLet $R$ be the normal reaction force. Resolving perpendicular to the slope:\n\n$$R = mg \\cos \\phi$$\n\nSubstituting our decimal approximation for $\\cos \\phi$:\n\n$$R \\approx 0.9600mg$$"},{"type":"text","order":3,"title":"Calculate the friction force","content":"Since the package is moving down the slope, the friction force $F$ is at its maximum limiting value:\n\n$$F = \\mu R$$\n\nUsing the given coefficient of friction $\\mu = \\frac{1}{4} = 0.25$, we calculate:\n\n$$F \\approx 0.25 \\times (0.9600mg) = 0.2400mg$$"},{"type":"text","order":4,"title":"Apply Newton's second law","content":"Now we resolve parallel to the slope (taking down the slope as the positive direction of motion):\n\n$$mg \\sin \\phi - F = ma$$\n\nSubstitute our decimal values for $\\sin \\phi$ and $F$:\n\n$$0.2800mg - 0.2400mg = ma$$\n\nFactor out the mass $m$ on both sides:\n\n$$m(0.2800g - 0.2400g) = ma$$\n\n$$0.2800g - 0.2400g = a$$"},{"type":"text","order":5,"title":"Find the final acceleration","content":"Subtracting the decimal values gives:\n\n$$a \\approx 0.0400g \\text{ m s}^{-2}$$\n\nUsing the standard approximation for gravity $g \\approx 9.8 \\text{ m s}^{-2}$:\n\n$$a \\approx 0.0400 \\times 9.8 = 0.392 \\text{ m s}^{-2}$$\n\n*(This matches the exact analytical fraction of $a = \\frac{1}{25}g$ m s$-2$)*","calculator":[{"gdc":{"expression":"0.04 * 9.8","instructions":"Multiply 0.04 by 9.8."},"ti84":{"expression":"0.04 * 9.8","instructions":"Press 0.04 * 9.8 [ENTER]."},"desmos":{"expression":"0.04 \\times 9.8","instructions":"Type 0.04 * 9.8."},"description":"Compute the final numerical acceleration"}]}]}],"generatedAt":"2026-06-09T05:22:56.005Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"a475dab9-5f28-48f2-9d76-4c84ddb3f2dc","label":"b","content":"A different package with mass $2m$ kg is placed on the same slope. Explain how the acceleration of this package would compare to your answer in part (a).","markscheme":"- States that the acceleration is identical B1\n- Explains that the force due to gravity and the friction force both scale linearly with mass, meaning mass cancels out of $F=ma$ B1","marks":2,"order":1,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"States acceleration is the same.","code":"B"},{"id":"p2","marks":1,"description":"Logic: Mass $m$ cancels from the governing equation.","code":"B"}],"finalAnswers":[],"generatedAt":"2026-05-06T14:04:53.885Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"algebraic","label":"Algebraic Derivation","steps":[{"type":"text","order":0,"title":"Set up general forces","content":"To see how the mass of the package affects its acceleration, let's write down the expressions for the forces acting on a general package of mass $M$ sliding down the slope.\n\nThe component of gravity acting down the slope is:\n$$F_{\\text{slope}} = M g \\sin \\phi$$\n\nThe normal reaction force $R$ perpendicular to the slope is:\n$$R = M g \\cos \\phi$$\n\nSince the package is moving, the friction force $F_f$ opposing the motion is at its maximum value:\n$$F_f = \\mu R = \\mu M g \\cos \\phi$$\nwhere $\\mu$ is the coefficient of friction."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$F_{\\text{net}} = \\htmlData{anchor=fnet}{\\color{@sky}{M a}}$"},{"latex":"$$F_{\\text{net}} = \\htmlData{anchor=fgrav}{\\color{@amber}{M g \\sin \\phi}} - \\htmlData{anchor=fric}{\\color{@purple}{\\mu M g \\cos \\phi}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"fnet"},{"color":"amber","style":"text-color","anchor":"fgrav"},{"color":"purple","style":"text-color","anchor":"fric"}],"connections":[{"to":"fgrav","from":"fnet","color":"sky","label":"equate to Ma","style":"solid"}]},"order":1,"title":"Write the equation of motion","description":"Using Newton's Second Law, $F_{\\text{net}} = Ma$, we can equate the net force down the slope to the product of mass and acceleration."},{"type":"text","order":2,"title":"Cancel the mass term","content":"By equating the two expressions for the net force, we get:\n\n$$M g \\sin \\phi - \\mu M g \\cos \\phi = M a$$\n\nNotice that every single term in the equation contains a factor of the mass $M$. We can divide the entire equation by $M$:\n\n$$a = g \\sin \\phi - \\mu g \\cos \\phi$$\n\nThis algebraic step shows us that the mass $M$ completely cancels out of the equation of motion."},{"type":"text","order":3,"title":"Compare the accelerations","content":"Because the acceleration formula\n\n$$a = g \\sin \\phi - \\mu g \\cos \\phi$$\n\ncontains no term for mass, the acceleration is entirely independent of the mass of the package.\n\nTherefore, a package of mass $2m$ kg will have the **identical acceleration** to the package of mass $m$ kg.\n\n* **Mark 1 (B1):** Correctly stating that the acceleration is identical.\n* **Mark 2 (B1):** Explaining that both the gravitational force component and the friction force scale linearly with mass, causing mass to cancel out of $F=ma$.","highlights":[{"text":"different package with mass 2m kg","location":"specification"},{"text":"Explain how the acceleration of this package would compare to your answer in part (a).","location":"specification"}]}]},{"id":"proportionality","label":"Proportionality Argument","steps":[{"type":"text","order":0,"title":"State the final conclusion","content":"The acceleration of the $2m$ kg package is **identical** to the acceleration of the $m$ kg package.\n\nIn exams, questions of this type can often be solved quickly using a **proportionality argument**. Let's break down why changing the mass has absolutely no effect on the acceleration.","highlights":[{"text":"Explain how the acceleration of this package would compare to your answer in part (a).","location":"specification"}]},{"type":"text","order":1,"title":"Analyze the driving force","content":"The driving force pulling the package down the slope is the component of its weight parallel to the inclined plane:\n\n$$F_{\\text{driving}} = mg \\sin \\phi$$\n\nSince gravity $g$ and the angle of inclination $\\phi$ are constants, the driving force is **directly proportional** to the mass:\n\n$$F_{\\text{driving}} \\propto m$$\n\nWhen the mass is doubled to $2m$, this pulling force also exactly doubles."},{"type":"text","order":2,"title":"Analyze the friction force","content":"The resistive force opposing the motion is friction, which depends on the normal reaction force $R = mg \\cos \\phi$:\n\n$$F_{\\text{friction}} = \\mu R = \\mu mg \\cos \\phi$$\n\nSince the coefficient of friction $\\mu$, the gravitational field strength $g$, and the slope angle $\\phi$ are all constant, the friction force is also **directly proportional** to the mass:\n\n$$F_{\\text{friction}} \\propto m$$\n\nDoubling the mass to $2m$ exactly doubles the frictional resistance force."},{"type":"hyper_latex","block":{"nodes":[{"text":"Since both the driving force and friction force scale linearly with mass, the net force does too:"},{"latex":"$$\\htmlData{anchor=fnet-1}{\\color{@orange}{F_{\\text{net}}}} = \\htmlData{anchor=fdrive-1}{\\color{@sky}{F_{\\text{driving}}}} - \\htmlData{anchor=ffric-1}{\\color{@purple}{F_{\\text{friction}}}}$"},{"latex":"$$\\htmlData{anchor=fnet-2}{\\color{@orange}{F_{\\text{net}}}} \\propto \\htmlData{anchor=m-1}{\\color{@green}{m}}$"},{"text":"To find the acceleration, we divide the net force by the mass:"},{"latex":"$$\\text{acceleration } a = \\frac{\\htmlData{anchor=fnet-3}{\\color{@orange}{F_{\\text{net}}}}}{\\htmlData{anchor=m-2}{\\color{@green}{m}}}$"}],"highlights":[{"color":"orange","style":"text-color","anchor":"fnet-1"},{"color":"orange","style":"text-color","anchor":"fnet-2"},{"color":"orange","style":"text-color","anchor":"fnet-3"},{"color":"green","style":"text-color","anchor":"m-1"},{"color":"green","style":"text-color","anchor":"m-2"}],"connections":[{"to":"fnet-2","from":"fnet-1","color":"orange","label":null,"style":"solid"},{"to":"fnet-3","from":"fnet-2","color":"orange","label":"scales with m","style":"dashed"},{"to":"m-2","from":"m-1","color":"green","label":"cancels out","style":"solid"}]},"order":3,"title":"Apply the proportionality argument","description":"Using Newton's second law ($F_{\\text{net}} = ma$), we can see how these proportionalities combine to determine the acceleration."},{"type":"text","order":4,"title":"Exam mark breakdown","content":"Here is how you would secure both marks in the exam:\n\n* **1 Mark (B1)**: State clearly that the acceleration remains **identical** (or is the same).\n* **1 Mark (B1)**: Explain that both the force due to gravity and the frictional force are proportional to mass, so mass cancels out when applying $$F = ma$$."}]}],"generatedAt":"2026-06-09T05:25:56.317Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"66640675-41a8-4b71-b157-a035ba62844c:exact_trig","label":"(a) Exact Trigonometric Ratios","steps":[{"type":"text","order":0,"title":"Find exact trigonometric ratios","content":"We are given $\\tan \\phi = \\frac{7}{24}$. Instead of finding the angle $\\phi$ in degrees and introducing rounding errors, we can find the exact values for $\\sin \\phi$ and $\\cos \\phi$ using the properties of right-angled triangles.\n\nSince $\\tan \\phi = \\frac{\\text{Opposite}}{\\text{Adjacent}} = \\frac{7}{24}$:\n- $\\text{Opposite} = 7$\n- $\\text{Adjacent} = 24$\n\nUsing the Pythagorean theorem (or recognizing the 7-24-25 Pythagorean triple), the hypotenuse is:\n\n$$\\text{Hypotenuse} = \\sqrt{7^2 + 24^2} = \\sqrt{49 + 576} = \\sqrt{625} = 25$$\n\nNow we can write down the exact values of the trigonometric ratios:\n\n$$\\sin \\phi = \\frac{7}{25}$$\n$$\\cos \\phi = \\frac{24}{25}$$\n\nThis earns you the first mark **[B1]**."},{"type":"text","order":1,"title":"Resolve forces perpendicular to slope","content":"Next, we resolve forces perpendicular to the incline. \n\nThe forces acting perpendicular to the slope are:\n- The normal reaction force $R$ acting perpendicularly away from the slope.\n- The perpendicular component of the weight acting into the slope, which is $mg \\cos \\phi$.\n\nSince there is no motion perpendicular to the slope, the forces must balance:\n\n$$R = mg \\cos \\phi$$\n\nSubstituting our exact value for $\\cos \\phi$:\n\n$$R = mg \\left(\\frac{24}{25}\\right) = \\frac{24}{25}mg$$\n\nThis earns the second mark **[M1]**."},{"type":"text","order":2,"title":"Calculate friction force","content":"Since the package is released and moves down the slope, friction is at its maximum value (limiting friction). We use the friction formula:\n\n$$F = \\mu R$$\n\nWe are given the coefficient of friction $\\mu = \\frac{1}{4}$. Substituting our values:\n\n$$F = \\frac{1}{4} \\left( \\frac{24}{25}mg \\right)$$\n$$F = \\frac{6}{25}mg$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$mg \\htmlData{anchor=sin-ratio}{\\color{@sky}{\\sin \\phi}} - \\htmlData{anchor=friction-force}{\\color{@amber}{F}} = ma$"},{"latex":"$$mg \\left( \\htmlData{anchor=sin-sub}{\\color{@sky}{\\frac{7}{25}}} \\right) - \\htmlData{anchor=friction-sub}{\\color{@amber}{\\frac{6}{25}mg}} = ma$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sin-ratio"},{"color":"amber","style":"text-color","anchor":"friction-force"}],"connections":[{"to":"sin-sub","from":"sin-ratio","color":"sky","label":"Substitute","style":"solid"},{"to":"friction-sub","from":"friction-force","color":"amber","label":"Substitute","style":"solid"}]},"order":3,"title":"Set up equation of motion","description":"We apply Newton's second law ($F = ma$) parallel to the slope. The component of weight pulling the package down is $mg \\sin \\phi$, and the opposing force is friction $F$."},{"type":"text","order":4,"title":"Solve for acceleration","content":"We can now simplify the algebraic equation and solve for the acceleration $a$:\n\n$$\\frac{7}{25}mg - \\frac{6}{25}mg = ma$$\n\nFactor out $mg$ on the left side to simplify:\n\n$$\\left( \\frac{7}{25} - \\frac{6}{25} \\right)mg = ma$$\n\n$$\\frac{1}{25}mg = ma$$\n\nDivide both sides by the mass $m$ to isolate $a$:\n\n$$a = \\frac{1}{25}g \\text{ m s}^{-2}$$\n\nUsing the standard approximation for acceleration due to gravity $g = 9.8 \\text{ m s}^{-2}$, we can also express this in decimal form:\n\n$$a = 0.04g = 0.392 \\text{ m s}^{-2}$$\n\nThis earns the final marks **[M1]** for simplifying and **[A1]** for the correct final answer.","calculator":[{"gdc":{"expression":"0.04 * 9.8","instructions":"In the Run-Matrix menu, enter 0.04 * 9.8."},"ti84":{"expression":"0.04 * 9.8","instructions":"Type 0.04 * 9.8 and press ENTER."},"desmos":{"expression":"0.04 * 9.8","instructions":"Type 0.04 * 9.8 into the input line."},"description":"Calculate the numerical value of acceleration using $g = 9.8\\text{ m s}^{-2}$."}]}]},{"id":"66640675-41a8-4b71-b157-a035ba62844c:decimal_approximation","label":"(a) Decimal Approximation","steps":[{"type":"text","order":0,"title":"Find the angle of inclination","content":"We are given that the slope is inclined at an angle $\\phi$ where:\n\n$$\\tan \\phi = \\frac{7}{24}$$\n\nTo use the decimal approximation method, we first find the angle $\\phi$ using the inverse tangent function on our calculator:\n\n$$\\phi = \\arctan\\left(\\frac{7}{24}\\right) \\approx 16.2602^\\circ$$"},{"type":"text","order":1,"title":"Calculate decimal trig values","content":"Now we compute the decimal approximations for both $\\sin \\phi$ and $\\cos \\phi$ using the angle we just found:\n\n$$\\sin(16.2602^\\circ) \\approx 0.2800$$\n$$\\cos(16.2602^\\circ) \\approx 0.9600$$\n\n*(Note: These correspond to the exact values $\\frac{7}{25}$ and $\\frac{24}{25}$, but we will carry forward these decimal approximations to solve our system of equations.)*","calculator":[{"gdc":{"expression":"sin(16.2602)","instructions":"Calculate sine and cosine of the stored angle value."},"ti84":{"expression":"sin(16.2602)","instructions":"Press [SIN] (16.2602) [ENTER] and [COS] (16.2602) [ENTER]."},"desmos":{"expression":"\\sin(16.2602^\\circ)","instructions":"Type sin(16.2602) and cos(16.2602) with settings in Degrees."},"description":"Find sine and cosine of the angle"}]},{"type":"text","order":2,"title":"Resolve forces perpendicular","content":"The forces acting perpendicular to the inclined plane must balance because there is no motion in this direction.\n\nLet $R$ be the normal reaction force. Resolving perpendicular to the slope:\n\n$$R = mg \\cos \\phi$$\n\nSubstituting our decimal approximation for $\\cos \\phi$:\n\n$$R \\approx 0.9600mg$$"},{"type":"text","order":3,"title":"Calculate the friction force","content":"Since the package is moving down the slope, the friction force $F$ is at its maximum limiting value:\n\n$$F = \\mu R$$\n\nUsing the given coefficient of friction $\\mu = \\frac{1}{4} = 0.25$, we calculate:\n\n$$F \\approx 0.25 \\times (0.9600mg) = 0.2400mg$$"},{"type":"text","order":4,"title":"Apply Newton's second law","content":"Now we resolve parallel to the slope (taking down the slope as the positive direction of motion):\n\n$$mg \\sin \\phi - F = ma$$\n\nSubstitute our decimal values for $\\sin \\phi$ and $F$:\n\n$$0.2800mg - 0.2400mg = ma$$\n\nFactor out the mass $m$ on both sides:\n\n$$m(0.2800g - 0.2400g) = ma$$\n\n$$0.2800g - 0.2400g = a$$"},{"type":"text","order":5,"title":"Find the final acceleration","content":"Subtracting the decimal values gives:\n\n$$a \\approx 0.0400g \\text{ m s}^{-2}$$\n\nUsing the standard approximation for gravity $g \\approx 9.8 \\text{ m s}^{-2}$:\n\n$$a \\approx 0.0400 \\times 9.8 = 0.392 \\text{ m s}^{-2}$$\n\n*(This matches the exact analytical fraction of $a = \\frac{1}{25}g$ m s$-2$)*","calculator":[{"gdc":{"expression":"0.04 * 9.8","instructions":"Multiply 0.04 by 9.8."},"ti84":{"expression":"0.04 * 9.8","instructions":"Press 0.04 * 9.8 [ENTER]."},"desmos":{"expression":"0.04 \\times 9.8","instructions":"Type 0.04 * 9.8."},"description":"Compute the final numerical acceleration"}]}]},{"id":"a475dab9-5f28-48f2-9d76-4c84ddb3f2dc:algebraic","label":"(b) Algebraic Derivation","steps":[{"type":"text","order":0,"title":"Set up general forces","content":"To see how the mass of the package affects its acceleration, let's write down the expressions for the forces acting on a general package of mass $M$ sliding down the slope.\n\nThe component of gravity acting down the slope is:\n$$F_{\\text{slope}} = M g \\sin \\phi$$\n\nThe normal reaction force $R$ perpendicular to the slope is:\n$$R = M g \\cos \\phi$$\n\nSince the package is moving, the friction force $F_f$ opposing the motion is at its maximum value:\n$$F_f = \\mu R = \\mu M g \\cos \\phi$$\nwhere $\\mu$ is the coefficient of friction."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$F_{\\text{net}} = \\htmlData{anchor=fnet}{\\color{@sky}{M a}}$"},{"latex":"$$F_{\\text{net}} = \\htmlData{anchor=fgrav}{\\color{@amber}{M g \\sin \\phi}} - \\htmlData{anchor=fric}{\\color{@purple}{\\mu M g \\cos \\phi}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"fnet"},{"color":"amber","style":"text-color","anchor":"fgrav"},{"color":"purple","style":"text-color","anchor":"fric"}],"connections":[{"to":"fgrav","from":"fnet","color":"sky","label":"equate to Ma","style":"solid"}]},"order":1,"title":"Write the equation of motion","description":"Using Newton's Second Law, $F_{\\text{net}} = Ma$, we can equate the net force down the slope to the product of mass and acceleration."},{"type":"text","order":2,"title":"Cancel the mass term","content":"By equating the two expressions for the net force, we get:\n\n$$M g \\sin \\phi - \\mu M g \\cos \\phi = M a$$\n\nNotice that every single term in the equation contains a factor of the mass $M$. We can divide the entire equation by $M$:\n\n$$a = g \\sin \\phi - \\mu g \\cos \\phi$$\n\nThis algebraic step shows us that the mass $M$ completely cancels out of the equation of motion."},{"type":"text","order":3,"title":"Compare the accelerations","content":"Because the acceleration formula\n\n$$a = g \\sin \\phi - \\mu g \\cos \\phi$$\n\ncontains no term for mass, the acceleration is entirely independent of the mass of the package.\n\nTherefore, a package of mass $2m$ kg will have the **identical acceleration** to the package of mass $m$ kg.\n\n* **Mark 1 (B1):** Correctly stating that the acceleration is identical.\n* **Mark 2 (B1):** Explaining that both the gravitational force component and the friction force scale linearly with mass, causing mass to cancel out of $F=ma$.","highlights":[{"text":"different package with mass 2m kg","location":"specification"},{"text":"Explain how the acceleration of this package would compare to your answer in part (a).","location":"specification"}]}]},{"id":"a475dab9-5f28-48f2-9d76-4c84ddb3f2dc:proportionality","label":"(b) Proportionality Argument","steps":[{"type":"text","order":0,"title":"State the final conclusion","content":"The acceleration of the $2m$ kg package is **identical** to the acceleration of the $m$ kg package.\n\nIn exams, questions of this type can often be solved quickly using a **proportionality argument**. Let's break down why changing the mass has absolutely no effect on the acceleration.","highlights":[{"text":"Explain how the acceleration of this package would compare to your answer in part (a).","location":"specification"}]},{"type":"text","order":1,"title":"Analyze the driving force","content":"The driving force pulling the package down the slope is the component of its weight parallel to the inclined plane:\n\n$$F_{\\text{driving}} = mg \\sin \\phi$$\n\nSince gravity $g$ and the angle of inclination $\\phi$ are constants, the driving force is **directly proportional** to the mass:\n\n$$F_{\\text{driving}} \\propto m$$\n\nWhen the mass is doubled to $2m$, this pulling force also exactly doubles."},{"type":"text","order":2,"title":"Analyze the friction force","content":"The resistive force opposing the motion is friction, which depends on the normal reaction force $R = mg \\cos \\phi$:\n\n$$F_{\\text{friction}} = \\mu R = \\mu mg \\cos \\phi$$\n\nSince the coefficient of friction $\\mu$, the gravitational field strength $g$, and the slope angle $\\phi$ are all constant, the friction force is also **directly proportional** to the mass:\n\n$$F_{\\text{friction}} \\propto m$$\n\nDoubling the mass to $2m$ exactly doubles the frictional resistance force."},{"type":"hyper_latex","block":{"nodes":[{"text":"Since both the driving force and friction force scale linearly with mass, the net force does too:"},{"latex":"$$\\htmlData{anchor=fnet-1}{\\color{@orange}{F_{\\text{net}}}} = \\htmlData{anchor=fdrive-1}{\\color{@sky}{F_{\\text{driving}}}} - \\htmlData{anchor=ffric-1}{\\color{@purple}{F_{\\text{friction}}}}$"},{"latex":"$$\\htmlData{anchor=fnet-2}{\\color{@orange}{F_{\\text{net}}}} \\propto \\htmlData{anchor=m-1}{\\color{@green}{m}}$"},{"text":"To find the acceleration, we divide the net force by the mass:"},{"latex":"$$\\text{acceleration } a = \\frac{\\htmlData{anchor=fnet-3}{\\color{@orange}{F_{\\text{net}}}}}{\\htmlData{anchor=m-2}{\\color{@green}{m}}}$"}],"highlights":[{"color":"orange","style":"text-color","anchor":"fnet-1"},{"color":"orange","style":"text-color","anchor":"fnet-2"},{"color":"orange","style":"text-color","anchor":"fnet-3"},{"color":"green","style":"text-color","anchor":"m-1"},{"color":"green","style":"text-color","anchor":"m-2"}],"connections":[{"to":"fnet-2","from":"fnet-1","color":"orange","label":null,"style":"solid"},{"to":"fnet-3","from":"fnet-2","color":"orange","label":"scales with m","style":"dashed"},{"to":"m-2","from":"m-1","color":"green","label":"cancels out","style":"solid"}]},"order":3,"title":"Apply the proportionality argument","description":"Using Newton's second law ($F_{\\text{net}} = ma$), we can see how these proportionalities combine to determine the acceleration."},{"type":"text","order":4,"title":"Exam mark breakdown","content":"Here is how you would secure both marks in the exam:\n\n* **1 Mark (B1)**: State clearly that the acceleration remains **identical** (or is the same).\n* **1 Mark (B1)**: Explain that both the force due to gravity and the frictional force are proportional to mass, so mass cancels out when applying $$F = ma$$."}]}],"generatedAt":"2026-06-09T05:26:10.019Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__ocr_mei","summaries":[{"id":"89bcd644-7a8c-49c0-a3d3-7043f3ccfa8e","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"9f7ba419-e7e3-48d5-bfa5-7e582c9665b3","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"a21345de-7186-4255-b838-1d05e18f1ca5","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"b08f341a-00c3-4518-b159-9a5e8ef5d4e9","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"e79530d9-1948-4601-ac69-a4a9c1e9c0b0","difficultyBand":"hard","calculatorUse":null,"totalMarks":11,"questionType":"LA","isCalculation":true},{"id":"e844cec2-c7d0-4a57-aada-a316046143e6","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"f915f046-4539-497e-a424-4e48792691a1","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"06b054a9-6014-4c43-bc75-9cab1a7e2a09","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"3da41302-3120-47e3-bc16-ec736d879adb","difficultyBand":"hard","calculatorUse":null,"totalMarks":11,"questionType":"LA","isCalculation":true},{"id":"42f724f6-33dd-4a34-a772-a5bdd624ced0","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"4433a7ec-3cbc-44b9-ae7a-06be6f956981","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"a002aeda-0103-48b7-be9f-7cf946c859ab","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"c2740c44-d525-4be0-a5b6-5a0acea80450","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"1a1a2422-2ea5-4956-983a-0c47da428a0f","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"38782f6d-5927-4b8a-9ac7-0d9e2aae0f3f","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"4cd7c753-d44c-4ed6-b81b-9f4da12259e7","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"71da9fb8-691e-41bf-b8bd-15f926cde0e2","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"803bd710-da2b-4d71-888e-1560713e30ae","difficultyBand":"hard","calculatorUse":null,"totalMarks":11,"questionType":"LA","isCalculation":true},{"id":"a16ebb2e-4507-4a9b-91aa-0bd38356c398","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"c18b79a7-1b92-43db-a335-b2cdb0a00142","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"0166a340-6fcc-43fe-90ce-5c0e692d7861","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"03a4d945-1887-4606-984d-e74f8c548119","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"0a62b5a9-db80-4384-95f6-5852d79c364c","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"1eb89cd6-980a-48ea-86d9-32df5839eac4","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"25714283-7fb3-4753-9280-a32dedd244a2","difficultyBand":"hard","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"296542b1-c93b-4d61-ae0b-3120c9efa3c6","difficultyBand":"hard","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"2b99c8c7-3c4e-4e02-801c-17df917d4324","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"2fb29051-1860-4f54-900f-f62c62c814cb","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"3124d710-1ba9-4380-a591-6ae916d074a4","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"32d46706-74ea-41a4-a23a-6ff736a1488e","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"332f2401-8c8e-4d98-b9f2-60ed3c99cec8","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"33f3e563-3547-43b7-ba80-4f4a2b72e02e","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"3676b702-5721-4dac-b873-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Forces","topicId":"d95ba0a5-dd02-4efb-91aa-eb2b64e806fe","topicSidebar":"$L3c","topicSlug":"forces","topicTitle":"3.4 Forces"}]}],"$L3d","$L3e"]}]

3.4 Forces

Question 13

3.4 Forces Questions

3.4 Forces

Question 13

3.4 Forces Questions