A 520-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a

Question

A 520-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 120-N rod as indicated in the figure below. The left end of 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. (Assume the cable is connected to the very end of the 6.00-m-long rod, and that there are 2.00 m separating the wall from the sign.) ICE CREAM (a) Find the (magnitude of the) tension T in the cable. (b) Find the horizontal and vertical components of the force exerted on the left end of the rod by the hinge. (Take up and to the right to be the positive directions. Indicate the direction with the sign of your answer.) horizontal component vertical component

Explanation / Answer

(a) Weight of signboard will act at its center of gravity which lies at (6-2=) 4 m distance from the hinge. Weight of rod will act at its center of mass which lies at (6/2=) 3 m distance from the hinge. Further, vertical component of tension in the cable = T Cos 30 acting at 6 m distance from the hinge.

Now taking moment about O of the forces mentioned above and equating it to zero we get,

520*4 + 120*3 - T Cos 30 * 6 = 0.

this gives T= 469.58 N

(b) Assuming net vertical force at hinge = Fv and net horizontal force at hinge = Fh.

Doing vertical force balances on the rod, Fv + T Cos 30 - (520 + 120) = 0.

This gives, Fv = 233.33 N

Similarly, doing horizontal forces balance on the rod, Fh - T Sin 30 = 0

This gives Fh = 234.79 N

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