Lever Calculations

The operating principle of levers can be used to calculate the forces and distances when a lever is operated.

The basic lever equation is \( F_1 d_1  = F_2 d_2 \), where \( F_1 \) and \( d_1 \) are the input force and movements, and \( F_2 \) and \( d_2 \) are the output force and movement.

The mechanical advantage, \( M \), is defined as the ratio of output force to input force. Therefore, a class 3 lever is useful for increasing the speed of the load.

\( M = \frac{F_2}{F_1} \)

Levers of class 1 can have any value of mechanical advantage.

Levers of class 2 always have a mechanical advantage greater than 1, since the load is always closer to the fulcrum than the effort (i.e. \( d_2 < d_1 \)).

Levers of class 3 always have a mechanical advantage less than 1, since the load is always further from the fulcrum than the effort (i.e. \( d_2 > d_1 \)).

Do not confuse having a mechanical advantage greater than 1 with usefulness! Mechanical advantage is just the ratio of output and input forces.

By the law of the lever, \( M = \frac{d_1}{d_2} \). This means that having a mechanical advantage less than 1 means that the distance ratio must be greater than one.

Calculations

The law of the lever can be used to compute forces and distances. 

Example 1

A wheelbarrow with handles that extend 1.2 metres back from the fulcrum is used to carry a 60 kg load. The centre of mass of the load is 0.4 m from the fulcrum. What total lifting force is required on the handles to carry this load.

Solution: We make use of the law of the lever. First, we need to find the weight of the load. This will be F_2.

\( F_2 = 60 \cdot 9.81 \) = 588.6 N.

We also know that \( d_1 \) = 1.2 m, and \( d_2 \) = 0.4 m.

We rearrange the law of the lever to get...

\( F_1 = F_2 \cdot \frac{d_2}{d_1} = 588.6 \cdot \frac{0.4}{1.2} = 588.6 \cdot \frac{1}{3} \) = 196.2 N.

Or, the mass of 60 kg will be perceived by the person pushing the wheelbarrow as a mass of 20 kg.  

Example 2

A hammer is used to remove a nail. The pull-out force of the nail is 1500 N, the from the nail to the fulcrum is 50 mm, the and the handle is 350 mm long. How much force does the user need to exert at the end of the handle to pull out the nail?

Solution: We make use of the law of the lever. First, we need to convert all the measurements to metres. We get that \( F_2 \) = 1500 N, \( d_2 \) = 0.05 m, and \( d_1 \) = 0.35 m. We need to solve for \( F_1 \).

We rearrange the law of the lever to get...

\( F_1 = F_2 \cdot \frac{d_2}{d_1} = 1500 \cdot \frac{0.05}{0.35} = 1500 \cdot \frac{1}{7} \) = 214.34 N.

Or, the nail pulling force of 1200 N gets converted to a much more manageable force of 214.3 N (equivalent to lifting 21.8 kg against Earth's gravity).

A tennis ball launcher is frequently used to throw a tennis ball at a greater speed. If the person's arm length is 0.72 m, and the tennis ball launcher is another 0.5 m long, how many times faster can the person throw the ball?

Solution:  The tennis ball launcher is a class 3 lever, since the fulcrum (the shoulder joint) is located at one end of the lever. The effort is exerted between the shoulder joint and the end of the person's arm. In this case, we don't need to know the forces, as we can use the mechanical advantage directly to get the speed ratio. In this case, \( d_1 \) = 0.72 m, and \( d_2 \) = 1.22 m (0.72 + 0.5).

\( M = \frac{d_1}{d_2} = \frac{0.72}{1.22} \) = 0.5902

The mechanical advantage is less than one, as expected. However, this is for the forces. The "mechanical advantage" for distance (the ratio of output distance to input distance) is the reciprocal of \( M \). So the output distance will be \( \frac{1}{0.5902} \), or 1.694 times the input distance.

The same gain applies to speed. You can throw a tennis ball faster for a given arm movement, as the ball launcher effectively makes your throwing arm longer. The same principle applies to a spear thrower, an ancient implement used to impart greater speed to spears.