Types of lever
Levers
A A simple machine consisting of a pivot, effort and load. consists of:
- a A point around which something can rotate or turn.
- an Force used to move a load over a distance.
- a The overall force that is exerted, usually by a mass or object, on a surface.
The table shows some examples of the different types of lever:
| Arrangement of components | Examples |
| effort – pivot – load | see-saw, crowbar, scissors |
| pivot – load – effort | wheelbarrow, nutcracker |
| pivot – effort – load | tweezers, cooking tongs |
| Arrangement of components | effort – pivot – load |
|---|---|
| Examples | see-saw, crowbar, scissors |
| Arrangement of components | pivot – load – effort |
|---|---|
| Examples | wheelbarrow, nutcracker |
| Arrangement of components | pivot – effort – load |
|---|---|
| Examples | tweezers, cooking tongs |
Simple levers and rotation
A simple lever could be a solid beam laid across a pivot. As effort is applied to rotate one end about the pivot, the opposite end is also rotated about the pivot in the same direction. This has the effect of rotating or lifting the load.
Levers such as this one make use of moments to act as a Something that increases the effect of a force.. They allow a larger force to act upon the load than is supplied by the effort, so it is easier to move large or heavy objects.
Example
A solid beam 0.5 m long is laid across a pivot to form a simple lever. The pivot is 0.1 m from the end of the beam. Calculate the heaviest load that could be lifted using a force of 500 N.
First calculate the moment due to the 500 N force. To do this, distance will also need to be calculated. To lift the greatest load, the effort must be applied furthest from the pivot.
Calculate the greatest distance from the pivot:
\(0.5 - 0.1 = 0.4~m\)
Then use the values to calculate the moment:
\(M = F~d\)
\(M = 500 \times 0.4\)
\(M = 200~Nm\)
Use the answer above to calculate the maximum force 0.1 m from the pivot.
Rearrange \(M = F d\) to find \(F\):
\(F = \frac{M}{d}\)
\(F = \frac{200}{0.1}\)
\(F = 2,000~N\)
The heaviest load that could be lifted by this arrangement is 2,000 N. The lever has the effect of multiplying the force by 5 times – it is a 5× force multiplier.
Example
A student is lifting a ball weighing 50 N. The distance from his elbow to the ball is 40 cm and the distance from his elbow to the muscle responsible for lifting is 4 cm. Calculate the effort exerted by the muscle.
First calculate the moment due to the 40 N force.
\(M = F~d\)
\(M = 50 \times 0.4\)
\(M = 20~Nm\)
Use the answer above to calculate the force exerted 0.04 m from the pivot.
Rearrange \(M = F~d\) to find F:
\(F = \frac{M}{d}\)
\(F = \frac{20}{0.04}\)
\(F = 500~N\)
In this case, the effort is greater than the load.
Gears
A toothed wheel used with other gears to turn axles at different speeds. are wheels with toothed edges that rotate on an A bar, rod or shaft which passes through a wheel or gear. The wheel or gear will rotate around the axle. or shaft. The teeth of one gear fit into the teeth of another gear. This lets one gear turn the other, meaning one axle or shaft can be used to turn another shaft.
Rotation and transmission of forces by gears
As one gear turns, the other gear must also turn. Where the gears meet, the teeth must both move in the same direction. In the diagram, the teeth of both gears move upwards. This means that the gears rotate in opposite directions.
The forces acting on the teeth are identical for both gears, but their moments are different:
- If a larger gear is driven by a smaller gear, the large gear will rotate more slowly but will have a greater moment. For example, a low gear on a bike or car.
- If a smaller gear is driven by a larger gear, the small gear will rotate more quickly but will have a smaller moment. For example, a high gear on a bike or car.
Example
A gear with a radius of 0.1 m is turned by a gear with a radius of 0.05 m. The moment of the smaller gear is 20 Nm. Calculate the moment of the larger gear.
First calculate the force on the teeth of the smaller gear.
Rearrange \(M = F~d\) to find F:
\(F = \frac{M}{d}\)
\(F = \frac{20}{0.05}\)
\(F = 400~N\)
Use the answer above to calculate the moment of the larger gear:
\(M = F~d\)
\(M = 400 \times 0.1\)
\(M = 40~Nm\)
Turning a gear that has double the radius doubles the turning effect – it is a 2× force multiplier.
