Question

A lever is used to lift a heavy load. When a 52 N force pushes the left end of the lever down 1.28 m, the load rises 0.23m.
1) What is the ideal mechanical advantage of this lever?
2) What is the work input?
3) Calculate the weight of the load if the lever is 100% efficient

Answer 1

1. The ideal lever does not dissipate or store energy, which means there is no friction in the hinge or bending in the beam. In this case, the power into the lever equals the power out, and the ratio of output to input force is given by the ratio of the distances from the fulcrum to the points of application of these forces. This is known as the law of the lever.

Mathematically, this is expressed by

$$M_{in} = F_{in} d_{in} = F_{out} d_{out} = M_{out},$$

2. Work input = effort force $(F_{x}) * \text{effort distance } (D_{x})$

3.

$$F_{1} d_{1} = F_{2} d_{2} \rightarrow \frac{F_{1}}{F_{2}} = \frac{d_{2}}{d_{1}}, \frac{l_{1}}{l_{2}} = \frac{d_{1}}{d_{2}} \rightarrow \frac{F_{1}}{F_{2}} = \frac{l_{2}}{l_{1}} \rightarrow F_{2} = \frac{F_{1} \cdot l_{1}}{l_{2}} = \frac{52 * 1.28}{0.23} \approx 289.4 \, \text{N}$$

The weight of load is equal to $F_{2}$.