# A lever is used to lift a heavy load. When a 52 N force pushes the left end of the lever down 0.71 m, the load rises 0.3m. If the lever is only 78% efficient, what would be the weight of the load it could lift with this output?

Nov 28, 2015

$96 \text{N}$

#### Explanation:

Applying the conservation of energy we can say that the work done on the load is equal to the work done by the effort.

Since work done = force x distance moved in the direction of the force we can write:

$L \times {d}_{l} = E \times {d}_{e}$

$\therefore \frac{L \times {d}_{l}}{E \times {d}_{e}} = 1$

The term $\frac{L}{E}$ is called the mechanical advantage $\left(M A\right)$.

The term ${d}_{e} / {d}_{l}$ is called the velocity ratio $\left(V R\right)$.

So we can write:

$\frac{M A}{V R} = 1$

This assumes that the machine is 100% efficient and that there are no energy losses due to friction etc. However, we do not live in an ideal world and machines are rarely 100% efficient.

The ratio MA/VR is used as a measure of efficiency and is usually expressed as a percentage:

Efficiency = $\frac{M A}{V R} \times 100$

$\therefore$Efficiency$= \left(\frac{\frac{L}{E}}{{d}_{e} / {d}_{l}}\right) \times 100$

$\therefore$Efficiency$= \frac{L \times {d}_{l}}{E \times {d}_{e}} \times 100$

Now we can apply this to this problem:

$78 = \frac{L \times 0.3}{52 \times 0.71} \times 100$

$\therefore L = \frac{78 \times 52 \times 0.71}{0.3 \times 100}$

$L = 96 \text{N}$

You don't need to go through this derivation every time you get a problem like this.

Just remember these 3 points:

1.

$M A = \frac{L}{E}$

2.

$V R = {d}_{e} / {d}_{l}$

3.

Efficiency =$\frac{M A}{V R} \times 100$