Arod is lying on the top of a table. one end of the rod is hinged to the table so that the rod can rotate freely on the tabletop. two forces, both parallel to the tabletop, act on the rod at the same place. one force is directed perpendicular to the rod and has a magnitude of 37.8 n. the second force has a magnitude of 56.5 n and is directed at an angle θ with respect to the rod. if the sum of the torques due to the two forces is zero, what must be the angle θ?

The solution for this problem is:

For the torques to be equivalent the constituent of the forces vertical to the rod must be equal and in the opposite direction 

So 37.8 = 56.5 * sin (θ) 

So θ = arc sin (37.8/56.5)

= arcsin 0.670212766 = sin⁻¹ 0.670212766 = 42° 5' 0.558"

= 42.08348831° + k×360°  (k=..-1,0,1,..)

= -317.91651169°, 42.08348831°, 402.08348831°, ..

= 0.73449543rad + k×2π  (k=..-1,0,1,..)

= -1.76620284π, 0.23379716π, 2.23379716π,..

= 42.1 degrees

d is the correct answer

force tends to be low and long diustance masses are high