At what angle does the magnitude of the torque have a value?

Question

The figure shows a lower leg being exercised. It has a weight attached to the footand is extended at and angle with respect to the vertical. Consider a rotational axis at the knee

The figure shows a lower leg being exercised. It has a 49-N weight attached to the foot and is exteded at an angle (theta) with respect to the vertical. Consider a rotational axis at the knee, as indicated. a) When (theta)=78(degrees), find the manitude of the torque that the weight creates. b) At what angle (theta) does the magnitude of the torque have a value of 15 N*m?

Answer 1

There is no figure, can you add a link to show one?

Answer 2

A force applied at the end of a lever generates a torque, except at certain angles.

Torque is r x F, or "r cross F", indicating the cross product (vector product). The cross product in this case means

T = |r| |F| sin(theta)

or torque equals magnitude of r * magnitude of F * sin(angle between the two vectors). This is zero when the angle between the lever arm vector and force vector is 90 degrees or 270 degrees. At any other angle, the sin() term is not zero, so there will be a torque.