A uniform rod of length 2.4 m and mass 5 kg is held horizontally by two identical vertical strings, one at each end. Find the tension in each string.

1. Since the rod is uniform, its weight acts at the midpoint of the rod.

2. The rod is symmetrical and the strings are identical, so the two tensions are equal. Call each tension $T$.

3. Use vertical equilibrium: total upward force equals total downward force.

   $$2T=5g$$

4. Solve for $T$.

   $$T=\frac{5g}{2}$$

5. Using $g=9.8$, each string has tension 24.5 N.

A uniform seesaw of length 4 m is supported at its centre. A child of mass 32 kg sits at the left end. A child of mass 40 kg sits on the right, at distance $x$ m from the centre. The seesaw is horizontal and in equilibrium. Find $x$.

1. The pivot is at the centre, so the left end is 2 m from the pivot.

2. The seesaw is uniform, so its own weight acts at the centre. Its line of action passes through the pivot, so it has zero moment about the pivot.

3. The 32 kg child produces an anticlockwise moment:

   $$32g\times 2$$

4. The 40 kg child produces a clockwise moment:

   $$40g\times x$$

5. For equilibrium, clockwise moment equals anticlockwise moment.

   $$40gx=32g\times 2$$

6. Cancel $g$ and solve.

   $$x=\frac{64}{40}=1.6$$

7. The child should sit 1.6 m from the centre.

A uniform rod $AB$ has length 3 m and mass 6 kg. It rests horizontally on two supports at $C$ and $D$, where $AC=0.6$ m and $AD=2.1$ m. Find the reactions at the two supports.

1. Let the reactions at $C$ and $D$ be $R_C$ and $R_D$.

2. The rod is uniform, so its weight $6g$ acts at the midpoint, 1.5 m from $A$.

3. Work out the distances from $C$: the distance from $C$ to $D$ is 1.5 m, and the distance from $C$ to the centre of mass is 0.9 m.

4. Take moments about $C$. The reaction at $C$ has no moment.

   $$R_D\times 1.5=6g\times 0.9$$

5. Solve for $R_D$.

   $$R_D=\frac{6g\times 0.9}{1.5}=3.6g$$

6. So the reaction at $D$ is 35.3 N, using $g=9.8$.

7. Now use vertical equilibrium.

   $$R_C+R_D=6g$$

8. Substitute $R_D=3.6g$.

   $$R_C=6g-3.6g=2.4g$$

9. So the reaction at $C$ is 23.5 N.

The same uniform 3 m rod of mass 6 kg rests on supports at $C$ and $D$, where $AC=0.6$ m and $AD=2.1$ m. A block is placed at $B$. The rod is on the point of tilting about $D$. Find the weight of the block and the reaction at $D$.

1. Since the rod is about to tilt about $D$, the reaction at $C$ is zero.

2. Let the weight of the block be $W$. The block is at $B$, which is 0.9 m to the right of $D$.

3. The rod’s weight $6g$ acts at its midpoint, 1.5 m from $A$. This is 0.6 m to the left of $D$.

4. Take moments about $D$.

   $$W\times 0.9=6g\times 0.6$$

5. Solve for $W$.

   $$W=\frac{6g\times 0.6}{0.9}=4g$$

6. So the block has weight 39.2 N.

7. Use vertical equilibrium. The only upward reaction is now at $D$.

   $$R_D=6g+4g=10g$$

8. Therefore the reaction at $D$ is 98 N.

A non-uniform rod $AB$ has length 5 m and mass 15 kg. It is held horizontally by vertical strings at $A$ and at a point $C$, where $AC=4$ m. The tension at $A$ is twice the tension at $C$. Find the tension at $C$ and the distance of the rod’s centre of mass from $A$.

1. Let the tension at $C$ be $T$. Then the tension at $A$ is $2T$.

2. Use vertical equilibrium.

   $$2T+T=15g$$

3. Solve for $T$.

   $$T=5g$$

4. So the tension at $C$ is 49 N.

5. Let the centre of mass be $x$ m from $A$.

6. Take moments about $A$. The tension at $A$ has zero moment.

   $$T\times 4=15g\times x$$

7. Substitute $T=5g$.

   $$5g\times 4=15gx$$

8. Cancel $g$ and solve.

   $$x=\frac{20}{15}=\frac{4}{3}$$

9. The centre of mass is $\frac{4}{3}$ m from $A$.

A non-uniform rod $AB$ is 180 cm long and has mass 12 kg. An 8 kg block is placed at a point 30 cm from $A$. The rod is held horizontally by vertical strings at $A$ and $B$. The tension at $A$ is 20 N greater than the tension at $B$. Find the distance of the rod’s centre of mass from $A$, to the nearest cm.

1. Convert the distances to metres: the rod is 1.8 m long, and the block is 0.30 m from $A$.

2. Let the tensions be $T_A$ and $T_B$, where $T_A=T_B+20$.

3. Use vertical equilibrium.

   $$T_A+T_B=20g$$

4. Substitute $g=9.8$ and $T_A=T_B+20$.

   $$\begin{array}{rl}{T}_B+20+T_B& =196\\ 2T_B& =176\\ T_B& =88\end{array}$$

5. Let the rod’s centre of mass be $x$ m from $A$.

6. Take moments about $A$. The tension at $A$ has zero moment.

   $$88\times 1.8=12g\times x+8g\times 0.30$$

7. Substitute $g=9.8$.

   $$158.4=117.6x+23.52$$

8. Solve for $x$.

   $$x=\frac{158.4-23.52}{117.6}\approx 1.148$$

9. The centre of mass is about 1.15 m from $A$, so the answer is 115 cm to the nearest cm.