2.5 Moments, levers and gears — Practise

Work through these on your own. Keep your working neat: the method matters as much as the answer.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A force of 20 N acts at a perpendicular distance of 0.30 m from a pivot. Calculate the moment of the force.
A force of 250 N is applied to a spanner at a perpendicular distance of 0.14 m from the nut. Calculate the moment of the force about the nut.
The moment of a force is 45 N m. The force is 150 N. Calculate the perpendicular distance from the pivot to the line of action of the force.
A child of weight 400 N sits 1.5 m from the pivot of a see-saw. A second child sits 2.5 m from the pivot on the other side. The see-saw is balanced. Calculate the weight of the second child.
A crowbar is used to lift a heavy stone. The effort of 150 N is applied 1.0 m from the pivot. The load acts 0.20 m from the pivot. Calculate the force on the stone.
A wheelbarrow carries a load of weight 500 N. The load acts 0.40 m from the wheel, and the person lifts at 1.2 m from the wheel. Calculate the upward force the person must use.
A driving gear with 40 teeth meshes with a driven gear with 10 teeth. The driving gear has a moment of 6.0 N m. Calculate the moment of the driven gear.
Explain how a lever transmits the rotational effect of a force, and why it can make a job easier.
Explain how two gears transmit the rotational effect of a force, and what decides the moment of the driven gear.
A spanner is used to loosen a nut. The hand applies 180 N at 0.25 m from the nut. Calculate the moment. Then calculate the force the spanner must apply at 0.050 m from the nut to give the same moment.

Answers and concise solutions

1. Moment of the force

Answer: $M = Fd = 20 \times 0.30 = 6.0\,\text{N m}$ .

2. Moment about the nut

Answer: $M = Fd = 250 \times 0.14 = 35\,\text{N m}$ .

3. Distance from the pivot

Answer: $d = \dfrac{M}{F} = \dfrac{45}{150} = 0.30\,\text{m}$ .

4. Weight of the second child

Answer: The moments balance, so $F_1 d_1 = F_2 d_2$ . Then $F_2 = \dfrac{400 \times 1.5}{2.5} = 240\,\text{N}$ .

5. Force on the stone

Answer: The moments are equal: $F_1 d_1 = F_2 d_2$ . The force on the stone is $F_2 = \dfrac{150 \times 1.0}{0.20} = 750\,\text{N}$ . The crowbar multiplies the effort five times.

6. Upward force at the handles

Answer: The moments balance: $F_1 d_1 = F_2 d_2$ . The force is $F_2 = \dfrac{500 \times 0.40}{1.2} = 166.7\,\text{N}$ , so $170\,\text{N}$ to 2 significant figures.

7. Moment of the driven gear

Answer: The gear ratio is $\dfrac{40}{10} = 4$ , so the moment is multiplied by 4: $M_2 = 6.0 \times 4 = 24\,\text{N m}$ .

8. How a lever works

A lever turns about a pivot. A force applied to the lever makes the lever rotate about the pivot.
The turning effect is the moment of the force, $M = Fd$ , where $d$ is the perpendicular distance from the pivot.
If the effort acts further from the pivot than the load, a small effort gives a large moment. The lever then produces a larger force on the load than the effort put in.

9. How gears work

The teeth of the two gears mesh, so one gear turns the other.
The force on the teeth transfers the rotational effect from one gear to the other.
The moments are in the same ratio as the numbers of teeth, so a small gear turns a large gear with a larger moment and a slower turn.

10. Moment and force from the spanner

Answer: The moment is $M = Fd = 180 \times 0.25 = 45\,\text{N m}$ . At 0.050 m the force is $F = \dfrac{M}{d} = \dfrac{45}{0.050} = 900\,\text{N} = 9.0 \times 10^2\,\text{N}$ .