Question

A mechanic can open a nut by applying 120 N force using a lever of 50 cm length. How long should the handle be, if he wishes to open, it by applying a force of only 40 N?

• A. 1 m
• B. 2 m
• C. 1.5 m
• D. 2.5 m

Answer 1

The correct option is C

1.5 m

Step 1: Given data and assumptions made

Initial force, F₁ = 120 N

Initial length, L₁= 50 cm =0.5 m

Final force, F₂ = 40 N

Let L₂ be the length of the handle

Step 2: Formula & solution

$$\begin{gathered} \mathrm{Moment}\mathrm{of}\mathrm{force}=\mathrm{Force}\times \mathrm{distance} \\ {\mathrm{F}}_1{\mathrm{L}}_1={\mathrm{F}}_2{\mathrm{L}}_2 \\ 120\mathrm{N}\times 0.5\mathrm{m}=40\mathrm{N}\times {\mathrm{L}}_2 \\ {\mathrm{L}}_2=\frac{120N\times 0.5m}{40N} \\ {\mathrm{L}}_2=1.5\mathrm{m} \end{gathered}$$

So, the length of the handle will be 1.5 m. Hence, C is the correct option.