Question

A square, of 0.36 m on a side, is mounted so that it can rotate about an axis that passes through the center of the square. the axis is perpendicular to the plane of the square. a force of 15 n lies in this plane and is applied to the square. what is the magnitude of the maximum torque that such a force could produce?

A. 0.648 N·m
B. 2.7 N·m
C. 5.4 N·m
D. 1.35 N·m

Answer 1

The magnitude of the maximum torque that a force of 15 N could produce on a square mounted to rotate about its center is 2.7 N·m (Option B). This is calculated using the formula for torque, which is the product of the force and the lever arm.

Step-by-step explanation:

To find the magnitude of the maximum torque that a force could produce on a square that can rotate around its center, you need to consider the force applied at the farthest point from the axis of rotation (at the edge of the square) and perpendicular to the lever arm (the side of the square). In this case, the side of the square is 0.36 m, which means the maximum lever arm distance from the center to the edge is half of that, 0.18 m (since the axis passes through the center). The torque (τ) is calculated as the product of the force (F) and the lever arm (r), so τ = r × F. The maximum torque would occur when the force is applied perpendicularly, hence τ = 0.18 m × 15 N = 2.7 N·m.