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Physics

A mechanic can open a nut by applying a force of 150 N while using a lever handle of length 40 cm. How long a handle is required if he wants to open it by applying a force of only 50 N ?

Force

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Answer

Given,

F1 = 150 N

I1 = 40 cm = 0.4 m

F2 = 50 N

I2 = ?

∵ F1 x I1 = F2 x I2

150 x 0.4 = 50 x I2

I2 = 150×0.450\dfrac{150 \times 0.4}{50}

I2 = 1.2 m = 120 cm

∴ Length of handle required for 50 N force = 120 cm.

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