As shown, a rope is attached to a l=5.00 m high shed that is to be relocated. A

Question

As shown, a rope is attached to a l=5.00 m high shed that is to be relocated. A man pulls on the end of the rope at point A; the rope is attached to the shed at point B. (Intro 1 figure) As the man pulls on the rope, it creates an angle theta with the horizontal. The end of the rope is located at x=2.90 m from the shed and y=1.00 m off the ground. The man pulls on the rope with a force of magnitude F=125 N.

What is the contribution to the moment about point O made by the x component of the force F at point A? What is the contribution to the moment about point O made by the y component of the force F at point A? What is the total moment M due to the force F about point O? Assume that moments acting counterclockwise about point O are positive whereas moments acting clockwise are negative.

Explanation / Answer

Given Data: L=5.00 m x=2.90 m y=1.00 m Force of magnitude F=125 N. Angle Phi = Tan -1 ( ( x / L-y) = Tan -1( 2.9 / 5-1) = 35.94 deg Moment about point O made by the x component of the force F at point A M fx = F * Sin Phi * L = 125* sin(35.94)*5 = 366.85 N-m Moment about point O made by the y component of the force F at point A M fy = F * Cos Phi * 0 = 0 therefore the total moment M

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