Forces, Density and Pressure: Question 7

Worked solution

Part (a): Why the reactions sum to the weight

The rod is in vertical equilibrium, so the resultant force on it is zero. The only vertical forces acting on the rod are the two upward reactions from the spring balances at P and Q, and the single downward weight of the rod. For the resultant vertical force to be zero, the two upward reactions must together equal the downward weight: $R_P + R_Q = W$ $11 + 7 = 18\text{ N} \checkmark$

Part (b): Locating the centre of gravity

Take moments about P. The reaction at P has zero perpendicular distance from P, so it contributes no moment there. The weight $W$ acts downward at the (unknown) centre of gravity, a distance $x$ from P; the reaction at Q acts upward at the far end, a distance equal to the full length of the rod, $2.4\text{ m}$, from P.

The weight tends to rotate the rod so that the loaded side dips down; the upward reaction at Q, acting on the opposite side, opposes this rotation. By the principle of moments (taking the weight’s moment and the reaction at Q’s moment as equal and opposite senses about P): $R_Q \times 2.4 = W \times x$ $7 \times 2.4 = 18 \times x$ $16.8 = 18x$ $x = \frac{16.8}{18} = 0.9\overline{3}\text{ m} \approx 0.933\text{ m (3 s.f.)}$

So the centre of gravity is about $0.933\text{ m}$ from P, i.e. $2.4 - 0.933 = 1.467\text{ m} \approx 1.47\text{ m}$ from Q. Since $0.933\text{ m}$ is less than half the rod’s length ($1.2\text{ m}$), the centre of gravity lies closer to P than to Q, consistent with the spring balance at P reading the larger force ($11\text{ N} > 7\text{ N}$), since the support nearer the centre of gravity carries the greater share of the weight.

Part (c): Conditions for equilibrium

A rigid body is in equilibrium only when both of the following hold simultaneously:

1. The resultant force on the body is zero (equivalently, the forces resolved in any two perpendicular directions each sum to zero).
2. The resultant moment about any point is zero (the sum of clockwise moments equals the sum of anticlockwise moments about every point, not just one particular point).

Final answers

• (a) Resultant vertical force is zero, so $R_P + R_Q = W$.
• (b) Centre of gravity is $\boxed{0.933}\text{ m}$ from P.
• (c) Resultant force zero and resultant moment (about any point) zero.