Question

A mechanic can open a nut by applying a force of 120N while using a lever handle of length 40cm. What is the length of the handle required if he wants to apply the force of only 30N.​

Answer 1

Answer:

Solution :

In the 1st case, the force applied,

F = 120 N and the magnitude of the perpendicular distance (called radius vector),

r=40cm=40100m

Torque (τ)=120N×40100=48Nm.

In the 2nd case, applied force F = 40 N and the length of the radius vector is to be found.

r=?

τ=F×r is the same in both the cases.

Hence,

48=40×r

∴r=1.2m. Hence, to apply a force of 40 N for the same torque, the mechanic should use a spanner of length 1.2 m.

Answer 2

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