Question

A man can open a nut by applying a force of 150 N by using a lever handle of length 0.4 m. What should be the length of the handle. If he is able to open it by applying a force of 60 N?

Answer 1

Step 1: Given,

Initially the amount of force, $F_1=150N$

The initial length of the lever handle. $L_1=0.4m$

The final amount of force, $F_2=60N$

Let the length of the handle be L₂

Step 2: Calculation for the length of the handle

$$\begin{gathered} \mathrm{Moment}\mathrm{of}\mathrm{force}=\mathrm{Force}\times \mathrm{distance} \\ \mathrm{Moment}\mathrm{of}\mathrm{force}=\mathrm{constant} \\ {\mathrm{F}}_1\times {\mathrm{L}}_1={\mathrm{F}}_2\times {\mathrm{L}}_2 \\ 150N\times 0.4m=60N\times {\mathrm{L}}_2 \\ {\mathrm{L}}_2=\frac{150N\times 0.4m}{60N} \\ {\mathrm{L}}_2=\frac{60}{60} \\ {\mathrm{L}}_2=1\mathrm{m} \end{gathered}$$

Therefore, the length of the handle will be 1 m.