In the figure, a 50.8 kg uniform square sign, of edge length L = 2.14 m, is hung

Question

In the figure, a 50.8 kg uniform square sign, of edge length L = 2.14 m, is hung from a horizontal rod of length dh-3.27 m and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance dy 3.82 m above the point where the rod is hinged to the wall. (a) What is the tension in the cable? (b) What is the horizontal component of the force on the rod from the wall? Take the positive direction to be to the right. (c) What is the vertical component of this force? Take the positive direction to be upward. Cable dy Hinge Rod H. PerezL UnitsTN (a) NumberTT112.704 (b) NumberTT-67.62 (c) Number UnitsT N Units

Explanation / Answer

(a) Consider the point where the rod attaches to the wall as the point A.

The point where the cable is attached to the other end of the rod point B and the point where the cable is attached to the wall 3.82 m above the level of the rod as the point C.

Now, take the rod as a free body diagram, and cut off the cable and simulate the unknown tention in the cable by T and break down this unknown tension into a vertical Component Ty and a horizontal component Tx.
The weight of the sign is acting at its center 1.07 meter from point A
Take the moment of forces about point A (Clockwise is positive, coounter clockwise is negative) and set =0
(50.8x9.8) x 1.07 - Ty (2.14) =0 (The horizontal component of the tension Tx has no moment about point A)
=> Ty x 2.14 = 532.69

=> Ty = 248.92 N

Now, define the angle that the cablle makes with the rod as theta so that theta = Arc tan (3.82 / 2.14)

= 60.7 degrees
Therefore, the total tension in the cable T = Ty/Sin theta = 248.92/Sin 60.7 = 285.32 N

(b) The horizontal force in the rod = T Cos theta = 285.32 Cos 60.7 = 139.63 N

(c) The vertial force in the rod = -50.8 x 9.8 + 248.92 = - 248.92 N (The force is in downward direction).

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