A 600 N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a

Question

A 600 N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 80-N rod as indicated in the figure below. The left end the rod is supported by a hinge and the right end is supported by a thin cable making a 30.0° angle with the vertical.
(a) Find the tension, T, in the cable.


(b) Find the horizontal and vertical components of force exerted on the left end of the rod by the hinge. (Take up and to the right to be the positive directions.)

horizontal component

vertical component

Explanation / Answer

Given that the weight of the box is W = 600N Length of rod is L= 6.00 m Weight of rod is Wr =110 N Angle with vertical is? = 30o ----------------------------------------------------- The wieght of the box acts at its center that is at2.00m from each end ( At point C) Weight of rod acts at center of the rod that is at 3.00mfrom each end ( at poitnt B) Apply Torque about the point A Fy*0 - Wr*AB- W*AC + T sin60o*AD = 0 Wr*AB + W*AC = Tsin60o*AD Wr*(3.00m) + W*4.00m = T sin60o*6.00m T = [ Wr*(3.00m) + W*4.00m ] / sin60o*6.00m then we get T = --------- N Apply Newtons law in verticle direction Fy - Wr - W + T sin60o = 0 Fy = Wr + W - T sin60o =---------- N This is the vertical force at the hing Apply Newtons law in horizontal direction Fx = T cos60o = ----------- N

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