Question

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 torqu?

Answer 1

In first case,

Mass = 50 kg

Initial length = R = 1 m

F₁ = mg = 50 × 9.8 = 490 N

Torque 1 = R × F₁ = 1 × 490 = 490 Nm

Now in second case,

New length = r = R − 20% × R = R − 0.2 R = 0.8 R = 0.8 × 1 = 0.8 m

Let required force = F₂

So,

Torque 2 = r × F₂ = 0.8 × F₂

According to the question,

Torque 1 = Torque 2

490 = 0.8 × F₂

F₂ = `490/0.8` = 612.5 N

Here, F₂-F₁ = 612.5 − 490 = 122.5 N

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