A 400-N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a

Question

A 400-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 (magnitude of the) tension T in the cable. Correct: Your answer is correct. N (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 Incorrect:

Your answer is incorrect. Your incorrect answer may have resulted from roundoff error. Make sure you keep extra significant figures in intermediate steps of your calculation.

N vertical component Incorrect:

Your answer is incorrect. Your incorrect answer may have resulted from roundoff error. Make sure you keep extra significant figures in intermediate steps of your calculation. N

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

400(4) + 140(3) - T(cos 30)(6) = 0

where T = tension in the cable

Solving for "T"

T(cos 30)(6) = 400(4) + 140(3)

T = 388.74 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 + 388.74(cos 30) - 400 - 140 = 0

Solving for "V"

V = 203.34 N

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

H = 0

H - 388.74(sin 30) = 0

H = 194.37 N

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