A 5 kg particle rests on a rough plane inclined at angle $\theta$, where $\tan \theta =\frac{3}{4}$. Find the normal reaction and the frictional force.

1. Draw the forces: weight $5g$ vertically downwards, normal reaction $R$ perpendicular to the plane, and friction $F$ up the plane because the particle would tend to slide down.

2. Use the 3-4-5 triangle from $\tan \theta =\frac{3}{4}$:

   $$\sin \theta =\frac{3}{5},\cos \theta =\frac{4}{5}$$

3. Resolve perpendicular to the plane to find the normal reaction, which is 39.2 N:

   $$R=5g\cos \theta =5g\cdot \frac{4}{5}=4g$$

4. Resolve parallel to the plane to find the frictional force, which is 29.4 N up the plane:

   $$F=5g\sin \theta =5g\cdot \frac{3}{5}=3g$$

A uniform rod of length 2 m and mass 4 kg is hinged to a vertical wall at $A$. A string joins the free end $B$ to a point $C$ on the wall, where $AC=1.5$ m. The rod is horizontal. Find the tension and the force at the hinge.

1. The string forms a triangle with horizontal side 2 m, vertical side 1.5 m and hypotenuse 2.5 m. So the tension components at $B$ are $\frac{4T}{5}$ horizontally left and $\frac{3T}{5}$ vertically up.

2. Take moments about $A$. The hinge force has no moment about $A$:

   $$2\left(\frac{3T}{5}\right)=4g\Rightarrow T=\frac{10g}{3}$$

3. So the tension is 32.7 N. Let the hinge reaction have components $X$ to the right and $Y$ upwards:

   $$X=\frac{4T}{5},Y+\frac{3T}{5}=4g$$

4. Substitute $T=\frac{10g}{3}$:

   $$X=\frac{8g}{3},Y=2g$$

5. The resultant hinge force has magnitude in newtons and direction $\varphi$ above the horizontal:

   $$\begin{array}{rl}magnitude& =\sqrt{{\left(\frac{8g}{3}\right)}^2+(2g{)}^2}=\frac{10g}{3}\approx 32.7,\\ \tan \varphi & =\frac{2g}{8g/3}=\frac{3}{4},\\ \varphi & \approx 36.9^{\circ }\end{array}$$

A uniform 6 m ladder of mass 16 kg rests with its foot on rough horizontal ground and its top against a smooth vertical wall. The coefficient of friction at the ground is 0.25. The ladder is on the point of slipping and makes angle $\theta$ with the ground. Find the friction and $\theta$.

1. Forces: weight $16g$ at the midpoint, ground reaction $R$ upwards, friction $F$ at the ground, and wall reaction $H$ horizontally at the top. The wall is smooth, so there is no friction at the wall.

2. Resolve vertically and use limiting friction. The friction is 39.2 N:

   $$R=16g,F=0.25R=4g$$

3. Resolve horizontally:

   $$H=F=4g$$

4. Take moments about the foot of the ladder. The wall reaction has perpendicular distance $6\sin \theta$; the weight has perpendicular distance $3\cos \theta$:

   $$H(6\sin \theta )=16g(3\cos \theta )$$

5. Substitute $H=4g$ and solve:

   $$4g(6\sin \theta )=48g\cos \theta \Rightarrow \tan \theta =2\Rightarrow \theta \approx 63.4^{\circ }$$

A 2 kg particle on a rough 30° plane is connected over a smooth pulley to a hanging 3 kg particle. The coefficient of friction on the plane is 0.25. The system is released, with the 3 kg particle moving downwards. Find the acceleration, tension, speed after the hanging particle descends 1.6 m, and the force on the pulley.

1. For the 2 kg particle, take up the plane as positive. Since it moves up the plane, friction acts down the plane. The normal reaction is $R=2g\cos 30^{\circ }$.

2. Write Newton’s second law for both particles:

   $$\begin{array}{rl}T-2g\sin 30^{\circ }-0.25(2g\cos 30^{\circ })& =2a,\\ 3g-T& =3a\end{array}$$

3. Add the equations to eliminate $T$. The acceleration is about 3.07 m s⁻²:

   $$a=\frac{3g-2g\sin 30^{\circ }-0.25(2g\cos 30^{\circ })}{5}\approx 3.07$$

4. Use the hanging particle equation to find the tension, about 20.2 N:

   $$T=3g-3a\approx 20.2$$

5. Use $v^2=u^2+2as$ with $u=0$ and $s=1.6$. The speed is about 3.14 m s⁻¹:

   $$v^2=0^2+2(3.07)(1.6),v\approx 3.14$$

6. The force on the pulley is the resultant of two tensions. The angle between the string sections is 60°, so the magnitude is about 35.0 N, along the angle bisector:

$$
P=\sqrt{T^2+T^2+2T^2\cos60^\circ}=T\sqrt{3}\approx35.0
$$

A 2 kg particle is on a rough plane with $\tan \theta =\frac{3}{4}$ and coefficient of friction $\frac{5}{8}$. It has just come to rest after moving up the plane, and the string is now slack. Determine whether it remains at rest.

1. With the string slack, $T=0$. For the particle to remain at rest, friction must be able to balance the downhill component of weight.

2. Use $\sin \theta =\frac{3}{5}$ and $\cos \theta =\frac{4}{5}$:

   $$F_{needed}=2g\cdot \frac{3}{5}=\frac{6g}{5},R=2g\cdot \frac{4}{5}=\frac{8g}{5}$$

3. Find the maximum possible friction:

   $$F_{\max }=\mu R=\frac{5}{8}\cdot \frac{8g}{5}=g$$

4. Since $F_{\max }<F_{needed}$, friction is too small. The particle will not remain at rest; it will slide down the plane.