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

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

2- What is the maximum torque he could exert with this force?

3- How should the force mentioned in part (2) be oriented as?

The force is directed into the page.

The force is directed out of the page.

The force is perpendicular to the wrench.

The force is parallel to the wrench.

Three forces are applied to a wheel of radius 0.350 m, as shown in the figure . One force is perpendicular to the rim , one is tangent to it, and the other one makes a 40.0 degree angle with the radius.(The force perpendicular to the rim = 11.9 N; The force making a 40 degree angle = 14.6N; The force that is tanget to it = 8.50 N)

– What is the magnitude of the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and passing through its center?

10 pts for best answer

### 1 Answer

1) Torque = force . perpendicular distance = 17 . 0.25 . sin 37˚ = 2.56 Nm

2) Max torque would be 17 . 0.25 = 4.25 Nm

3) The force should be perpendicular to the wrench.

Without the diagram, I can’t do the next bit… too complex.

Sorry.

Source(s): Old teacher