A square, 0.52 m on a side, is mounted so that it can rotate about an axis that passes through the center of the square. This axis is perpendicular to the plane of the square. A force of 15.0 N lies in this plane and is applied to the square. What is the magnitude of the maximum torque (in N*m) such that a force could produce?

didn't I just do this? see your other post.

No I don't think I posted this question :(

sorry wrong post!!!

To find the magnitude of the maximum torque that a force could produce, we need to know the lever arm (distance between the axis of rotation and the line of action of the force) and the magnitude of the force itself.

First, let's find the lever arm. Since the square is rotating about an axis that passes through its center, the lever arm is simply half the length of one of its sides.

Given that the square has a side length of 0.52 m, the lever arm will be:

Lever arm = 0.52 m / 2 = 0.26 m

Next, we can calculate the torque by using the formula:

Torque = force x lever arm

Using the given force of 15.0 N and the calculated lever arm of 0.26 m:

Torque = 15.0 N x 0.26 m = 3.9 N*m

So, the maximum torque that could be produced by the applied force is 3.9 N*m.