Need help to find the magnitude of net torque

Question

Four forces in the x−y plane act on an irregularly shaped object that is free to rotate about the z axis. Note that the z axis, which passes through the origin, is perpendicular to the x−y plane and points out of the screen.

Force F 1, with a magnitude of F1=1.36N, is applied at a point on the y axis a distance r1=2.63m from the origin at an angle α=29.6∘ below the horizontal, as shown.

Force F 2, with a magnitude of F2=3.72N, is applied at a point that is a distance r2=4.14m from the origin and is perpendicular to a line connecting its point of application to the origin.

Force F 3, with a magnitude of F3=3.72N, is applied at a point on the x axis a distance r3=2.63m from the origin at an angle of β=18.7∘ with the vertical, as shown.

Force F 4, with a magnitude of F4=1.36N, is applied at the origin.

What is the magnitude, in newton-meters, of the net torque about the z

axis?

Answer 1

First, you can eliminate the force that acts at the origin, as forces that act at the axis of rotation do not produce a torque. Next, you will apply the torque equation t = rFsinθ where r is the distance from the axis of rotation, F is the magnitude of the force applied, and θ is the angle between the r vector and the force vector. Remember that θ is measured with the tails of the vectors touching. For t1, θ is 119.6 degrees, and assuming that a counterclockwise rotation is positive, the torque from this segment is t1 = (1.36N)(2.63)sin(119.6) = 3.11 Nm. For t2, θ is 90 degrees as the force acts perpendicular to the r vector. Therefore, sin(90) = 1 and t2 = -(3.72N)(4.14m) = -15.40 Nm (negative because rotation is clockwise). Finally, t3 is also applied clockwise and is thus negative and θ will be 108.7 degrees. Thus, t3 = -(3.72)(2.63)sin(108.7) = -9.26 Nm. Adding all of these torque together you get a net torque of -21.55 Nm or 21.55 Nm in the clockwise direction.