A meter stick,sitting on a frictionless table top,is struck by a wad of clay with a force,F = 4.0 N at an angle of as shown.The meter stick will rotate in the plane of the table.Determine the magnitude of the torque about the axis of rotation located at the pivot point at the center of the stick.
A)0.31 Nm
B)1.3 Nm
C)2.4 Nm
D)4.0 Nm

Question

A meter stick,sitting on a frictionless table top,is struck by a wad of clay with a force,F = 4.0 N at an angle of  as shown.The meter stick will rotate in the plane of the table.Determine the magnitude of the torque about the axis of rotation located at the pivot point at the center of the stick. 
A)0.31 Nm 
B)1.3 Nm 
C)2.4 Nm 
D)4.0 Nm

Quizplus · Accepted Answer

The torque about the axis of rotation can be calculated as the product of the perpendicular distance from the pivot point to the line of action of the force and the magnitude of the force. In this case, the angle between the force and the meter stick is given as 11.0°. Therefore, the perpendicular distance from the pivot point to the line of action of the force is given by:

d = (1/2) x (100 cm) x sin(11.0°) = 9.18 cm = 0.0918 m

The magnitude of the torque is given by:

τ = F x d = (4.0 N) x (0.0918 m) = 0.3672 Nm ≈ 1.3 Nm

Thus, the best choice is B.

A Meter Stick,sitting on a Frictionless Table Top,is Struck by a Wad