A uniform stick 1 m long is placed horizontally on the ground along an east-west axis. A force of 1.0 N is applied to the center of the stick in a direction 30^o west of north. The torque exerted by the force relative to the east end of the stick is A) zero. B) 0.25 m, clockwise. C) 0.43 m, clockwise. D) 0.25 m, counterclockwise. E) 0.43 m, counterclockwise.

Question

Quizplus · Accepted Answer

To find the torque, we need to find the force's moment arm, which is the perpendicular distance from the line of action of the force to the pivot point (in this case, the east end of the stick). We can draw a diagram to help us visualize the problem:

30°
         |-------|
         |       |
      N  |   *   |  S
         |       |
         |-------|

The force is applied at an angle of 30° west of north, which means it has both northward and westward components. We can use trigonometry to find the magnitudes of these components:

cos 30° = adjacent/hypotenuse
adjacent = hypotenuse * cos 30°
adjacent = 1.0 N * cos 30°
adjacent = 0.87 N

sin 30° = opposite/hypotenuse
opposite = hypotenuse * sin 30°
opposite = 1.0 N * sin 30°
opposite = 0.50 N

The westward component is the one that creates the torque, so we need to find the moment arm corresponding to that component. Since the force is applied at the center of the stick, the moment arm is half the length of the stick:

moment arm = 1 m / 2
moment arm = 0.50 m

The torque is then given by:

torque = force * moment arm * sin θ (where θ is the angle between the force and moment arm)

torque = 0.87 N * 0.50 m * sin 30°
torque = 0.43 Nm (clockwise)

So the answer is (C) 0.43 m, clockwise.

A Uniform Stick 1 M Long Is Placed Horizontally on the Ground