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When you arrive at Duke's Dude Ranch, you are greeted by the large wooden sign shown. The left end of the sign is held in place by a bolt and the right end is tied to a rope that makes an angle of 20.0° with the horizontal. The sign is 3.20 m long and has a mass of 16.0 kg.
a) Draw a FBD showing the three forces acting on the sign.
b) Write down the torques due to all three forces, choosing the axis of rotation to be at the bolt. (Note that the torque due to F is zero because r is zero at the point of contact.)
c) Write down the torque equilibrium equation and solve for the magnitude of the tension T.
d) Write down the force equilibrium equation in the x direction. (Answer: Fx - Tcos20.0° = 0) Solve this equation for Fx.
e) Write down the force equilibrium equation in the y direction. (Answer: Fy - mg + Tsin20.0° = 0) Solve this equation for Fy.
In the drawing, the weight of the block on the table is 380 N and that of the hanging block is 215 N. Ignore all frictional effects, and assuming the pulley to be massless.
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