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Increasing Force with a Lever

by Ron Kurtus (revised 26 September 2016)

You can use a Class 1 or Class 2 lever to increase the force pushing on the load, according to where the fulcrum is located. To increase the force on the load, the length of the effort arm of the lever must be greater than the length of the load arm.

The equation for the forces relates to the work done and the distance relationships for a lever. From the equation, you can determine an unknown force or length.

Questions you may have include:

This lesson will answer those questions. Useful tool: Units Conversion



Using Class 1 or Class 2 lever

You can use a lever to increase the applied force in order to lift a heavy load. Class 1 or Class 2 levers are used to increase the output force.

Lifting heavy load with Class 1 lever

Lifting heavy load with Class 1 lever

An application of using a Class 1 lever is to lift a large rock.

Lifting heavy load with Class 2 lever

Lifting heavy load with Class 2 lever

A wheelbarrow is an example of using a Class 2 lever to lift and transport a heavy load. With a wheelbarrow, the wheel acts as the fulcrum.

Lever force equation

Assuming the resistive friction force at the fulcrum is negligible, the relationship between the input or effort and force on the load is dependent on the ratio of the arms of the lever, according to the equation:

FO/FI = dI/dO

where

Note: FO/FI is also the force mechanical advantage of the lever.

(See Force Mechanical Advantage for more information.)

Derivation

The derivation of this equation starts with the fact that work is a product of the force times the distance moved or displacement:

WI = FIDI

WO = FODO

where

and

According to the Law of Conservation of Energy, the output energy or work equals the input work:

WO = WI

Thus:

FODO = FIDI

DO/DI = FO/FI

Applying this relationship to the distance equation (DO/DI = dO/dI) for levers, you get:

FO/FI = dI/dO

(See Increasing Distance Moved with a Lever for more information.)

Application

How much effort is required to lift a load of 20 kilograms when the effort arm is 10 meters and the load arm is 1 meter?

Start by solving FOdO= FIdI for FI:

FI = FOdO/dI

Substitute in values:

FO = 20 kg

dO = 0.5 m

dI = 2 m

FI = 20*(0.5)/2 = 5 kg push required to lift the 20 kg weight.

Summary

You can use a Class 1 or Class 2 lever to increase the force pushing on the load, according to where the fulcrum is located. To increase the force on the load, the length of the effort arm of the lever must be greater than the length of the load arm.

The equation for the forces relates to the force mechanical advantage of the lever. The force equation is:

FO/FI = dI/dO

From the equation, you can determine an unknown force or length.


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Resources and references

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