A 560-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a

Question

A 560-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00-m-long, uniform, 140-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.)

(a) Find the tension T in the cable. (Give the magnitude of the tension.)
N

(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. Indicate the direction with the sign of your answer.)

Explanation / Answer

<< Find the tension, T, in the cable.>>

Take summation of moments about the hinge. Since the system is in equilibrium, then

M = 0

560(4) + 140(3) - T(cos 30)(6) = 0

where T = tension in the cable

Solving for "T"

T(cos 30)(6) = 560(4) + 140(3)

T = 511.9172387 N

<< 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.) >>

Summation of vertical forces = 0, i.e.,

V = 0

V + 511.9172387(cos 30) - 560 - 140 = 0

Solving for "V"

V = 256.6666667 N

Summation of horizontal forces = 0, i.e.,

H = 0

H - 511.9172387(sin 30) = 0

H = 255.9586193 N

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