Question

A machinist is using a wrench to loosen a nut. The wrench is 25.0 {\rm cm} long, and he exerts a 17.0-{\rm N} force at the end of the handle at 37 ^\circ with the handle

What is the magnitude of the torque does the machinist exert about the center of the nut?

What is the direction of the torque in part (A).

What is the maximum torque he could exert with this force?
How should the force mentioned in part (C) be oriented ?

Answer 1

The magnitude of the torque exerted by the machinist is given by multiplying the perpendicular distance from the point of application to the center of the nut by the magnitude of the force. The direction of the torque is clockwise in this case. The maximum torque occurs when the force is applied perpendicular to the handle of the wrench.

Step-by-step explanation:

To calculate the torque exerted by the machinist, we need to find the perpendicular distance from the point where the force is applied to the center of the nut. Using trigonometry, we can find this distance by multiplying the length of the wrench (25.0 cm) by the sine of the angle between the force and the handle (37°).

The torque is then calculated by multiplying this perpendicular distance by the magnitude of the force. So, the magnitude of the torque exerted by the machinist is:

Torque = (25.0 cm x sin(37°)) x 17.0 N

The direction of the torque in this case is clockwise, since the force is being exerted at an angle of 37° with the handle.

The maximum torque the machinist could exert is obtained when the force is applied perpendicular to the handle (at 90°). So, the maximum torque is:

Torque = (25.0 cm x sin(90°)) x 17.0 N

The force mentioned in part (C) should be oriented perpendicular to the handle of the wrench.

Answer 2

torque = force* perpendicular distance from the axis of rotation
τ=force∗r where r is the perpendicular distance from the axis of rotation


direction is upwards


maximum torque can be applied if the direction of force is perpendicular to the axis of rotation