Forces, Density and Pressure: Question 2

Worked solution

Part (a): The principle of moments

The principle of moments states that, for a body in rotational equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.

Part (b): Finding the force F

Take moments about the pivot P.

Distances from P:

• The load of $40\text{ N}$ hangs at A, which is $2.0\text{ m}$ from P.
• The beam’s weight ($100\text{ N}$) acts at its centre of gravity, the midpoint of the beam, $3.0\text{ m}$ from A. Since P is $2.0\text{ m}$ from A, the centre of gravity is $3.0 - 2.0 = 1.0\text{ m}$ from P, on the B side.
• The force $F$ acts at B, which is $6.0 - 2.0 = 4.0\text{ m}$ from P.

Setting up the equation:

The beam’s weight (on the B side) tends to rotate the beam so that B dips down, call this the clockwise sense. The load at A (on the A side) and the upward force $F$ at B both tend to rotate the beam anticlockwise about P: the load by making A dip down, and $F$ by lifting B up. Both oppose the weight’s tendency, so both are on the anticlockwise side of the equation.

By the principle of moments: $\text{clockwise moments} = \text{anticlockwise moments}$ $100 \times 1.0 = (40 \times 2.0) + (F \times 4.0)$ $100 = 80 + 4F$ $4F = 20$ $F = 5.0\text{ N}$

Part (c): Torque of a couple

A couple consists of two equal, opposite, parallel forces of $6.0\text{ N}$ each, whose lines of action are separated by a perpendicular distance of $0.50\text{ m}$. The torque of a couple is given by: $\text{torque} = \text{one force} \times \text{perpendicular distance between the forces}$ $\text{torque} = 6.0 \times 0.50$ $\text{torque} = 3.0\text{ N m}$

Final answers

• (a) Principle of moments: sum of clockwise moments $=$ sum of anticlockwise moments, about any point, for a body in equilibrium.
• (b) $F = \boxed{5.0}\text{ N}$
• (c) Torque of the couple $= \boxed{3.0}\text{ N m}$