A man opens a nut by applying a force of 250 N by using a lever handle of length 0.6 m. What shall be the length of the handle if he wants to open it by applying a force of 75 N?

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Solution

Torque $τ = r F$
Or $r_{1} F_{1} = r_{2} F_{2}$
Or $0.6 \times 250 = r_{2} \times 75$
$⟹ r_{2} = 2 m$

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A man opens a nut by applying a force of 250 N by using a lever handle of length 0.6 m. What shall be the length of the handle if he wants to open it by applying a force of 75 N?

Torque τ=rFOr r1F1=r2F2Or 0.6×250=r2×75⟹ r2=2 m

Torque $τ = r F$
Or $r_{1} F_{1} = r_{2} F_{2}$
Or $0.6 \times 250 = r_{2} \times 75$
$⟹ r_{2} = 2 m$