Question

If the torque required to loosen a nut on the wheel of a car has a magnitude of 60.0 N m, what minimum perpendicular force must be exerted by a mechanic at the end of a 45.0 cm wrench to loosen the nut?

Answer 1

Explanation:

Considering that the force

F

was applied at the end of the wrench,we can have minimum force to serve the purpose

So,

τ

=

r

×

F

Or,

F

=

40

0.30

=

133.32

N

Answer 2

133.32N is the minimum perpendicular force must be exerted by a mechanic at the end.

Explanation:

Given: torque required to loosen a nut on the wheel of a car has a magnitude = τ = 60.0 Nm

 Radius of nut = 45.0 cm

by converting the radius into meter = 45.0 / 100 = 0.45 m

∴ r =  0.45 m

To find: Minimum force =?

By using formula: torque = distance × Force

  τ = r × F

∴ F =  τ / r

     = 60.0 Nm / 0.45 m

     = 133.3333 N

     ≈ 133.32 N_(approximately)

∴ 133.32N is the minimum perpendicular force must be exerted by a mechanic at the end.