Physics

A 50 kg mass is being lifted by using a 1 metre long lever. If the length of the lever is reduced by 20%, how much more or less force is to be applied to achieve the same torque?

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In first case,

mass = 50 kg

Initial length = R = 1m

F₁ = mg = 50 x 9.8 = 490 N

Torque 1 = R x F₁ = 1 x 490 = 490 Nm

Now in second case,

New length = r = R - 20% x R = R - 0.2 R = 0.8 R = 0.8 x 1 = 0.8m

Let required force = F₂

So,

Torque 2 = r x F₂ = 0.8 x F₂

According to the question,

Torque 1 = Torque 2

490 = 0.8 x F₂

F₂ = $\dfrac {49}{0.8}$ = 612.5N

Here, F₂-F₁ = 612.5 - 490 = 122.5 N

∴ Force required to produce the same torque will increase by 122.5 N

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