Answer to Question #266902 in Mechanics | Relativity for Thel

Question

A 50-N force is applied at the end of the wrench handle that is 24-cm long. The force is applied in a direction perpendicular to the handle as shown.

A. What is the torque applied to the nut by the wrench?

B. What would the torque be if the force were applied half-way up the handle instead of at the end?

Answer 1

a) Express the relationship between torque and force.

$r=F\times L$

Here, r is the torque applied to the nut, F is the force applied by the wrench and l is the length of the wrench.

Substitute 50N for F, and 24 cm for l to r find.

$r=50N\times 24cm \\ =50N(24cm\times \frac{10^2m}{1cm}) \\ =12 Nm$

Therefore, the torque applied to the nut by the wrench is 12 Nm

(b)

Express the relationship between torque and force when the handle is at half the way.

$r=F\frac{L}{2}$

Substitute 50N for F, and 24 cm for l to find r.

$r=50N\times 24cm \\ =50N(\frac{24cm}{2}\times \frac{10^2m}{1cm}) \\ =6 Nm$

Therefore, torque applied to the nut by the wrench when the handle is half way is 6 Nm