A man drags a 71.5-kg crate across the floor at a constant velocity by pulling on a strap attached to the bottom of the crate. The crate is tilted 22.5° above the horizontal, and the strap is inclined 64.0° above the horizontal. The center of gravity of the crate coincides with its geometrical center, as indicated in the drawing. Find the magnitude of the tension in the strap. (Image attached)
torque = F * l
The Attempt at a Solution
Since the crate is not rotating, the torques must balance. From what I can
see, there are 3 total torques - two clockwise and one counterclockwise.
I set up my equations like this:
t1-t3 are the 3 torques.
T is the tension on the rope.
theta1 is the angle between the box and horizontal.
theta2 is the angle between the rope and horizontal.
t1 = mg * .9/2 * cos(theta1) <--- Torque due to gravity
t2 = T * cos(theta2) * .9 * sin(theta1) <- Torque due to horiz. comp. of tension
t3 = T * sin(theta2) * .9 * cos(theta1) <- Torque due to vert. comp of tension
t1 + t2 = t3 ; Solved for T.
I have a feeling I'm setting up the torque due to gravity wrong.
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