A massless meter stick on a horizontal frictionless table top is pivoted at the 80-cm mark.A force is applied perpendicularly to the end of the stick at 0 cm, as shown.A second force (not shown)is applied perpendicularly at the 60-cm mark.The forces are in the plane of the table top.If the stick does not move, the force exerted by the pivot on the stick:
A)must be zero
B)must be in the same direction as and have magnitude
C)must be directed opposite to and have magnitude
D)must be in the same direction as and have magnitude
E)must be directed opposite to and have magnitude

Question

A massless meter stick on a horizontal frictionless table top is pivoted at the 80-cm mark.A force  is applied perpendicularly to the end of the stick at 0 cm, as shown.A second force  (not shown)is applied perpendicularly at the 60-cm mark.The forces are in the plane of the table top.If the stick does not move, the force exerted by the pivot on the stick: 
A)must be zero 
B)must be in the same direction as and have magnitude 
C)must be directed opposite to and have magnitude 
D)must be in the same direction as and have magnitude 
E)must be directed opposite to and have magnitude

Quizplus · Accepted Answer

Since the meter stick is not moving, the net torque acting on it is zero. Thus, the torque due to the force applied at 0cm must be equal and opposite to the torque due to the force applied at 60cm. The torque due to a force is given by the product of the force and the distance between the force and the pivot point. Let x be the force applied at 0cm and y be the force applied at 60cm. Then we have:

x(80cm) = y(20cm)
y = 4x

So the force applied at 60cm is four times the force applied at 0cm. Since the force applied at 0cm is perpendicular to the meter stick and passes through the pivot point, it does not produce any torque about the pivot point, and thus the pivot exerts no force on that end of the stick. However, the force applied at 60cm produces a clockwise torque about the pivot point, and thus the pivot exerts a counterclockwise force on that end of the stick. This force must be in the same direction as the force applied at 60cm, and its magnitude must be equal and opposite to the torque due to that force. Therefore, the answer is B – the pivot exerts a force in the same direction as the force applied at 60cm, with magnitude 4x (where x is the force applied at 0cm).

A Massless Meter Stick on a Horizontal Frictionless Table Top