Question

A 480-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.)(a) Find the (magnitude of the) tension T in the cable.___________ 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______________Nvertical component __________N

Answer 1

Answer:

a. T = 438.8 N

b.  Fx = 219.4 N

Fy = 219.98 N

Explanation:

Since the system is in equilibrium the sum of forces:

ΣM = 0

480N*(4m) + 120N*(3m) - T*(cos 30)*(6m) = 0

Solve to "T"

T(cos 30)*(6) = 480*(4) + 120*(3)

T = 438.8 N

Summation of forces in axis y'

ΣFy = 0

Fy + 438.8*(cos 30) - 480N - 120N = 0

Solving for "F"

Fy = 219.98 N

Summation of forces in axis x'

ΣFx = 0

Fx - 438.8*(sin 30) = 0

Fx = 219.4 N

Answer 2

A)  The magnitude of the tension T in the cable is; T = 438.8 N

B) The horizontal and vertical components of the force exerted on the left end of the rod are;

Horizontal component; F_x = 219.4 N

Vertical Component; F_y = 220 N

Let us take moments about the right end of the rectangular sign. Thus;

ΣM;

480(4) + 120(3) - (T cos 30)(6) = 0

1920 + 360 - 5.196T = 0

2280 - 5.196T = 0

5.196T = 2280

T = 2280/5.196

T = 438.8 N

B) We want to find the vertical and horizontal components of the force exerted on the left end of the rod.

Sum of vertical forces;

ΣFy = 0

F_ y + 438.8(cos 30) - 480 - 120 = 0

Where F_ y  is the vertical component of the force exerted on the left end of the rod.

Thus;

F_ y = 600 - 380

F_ y = 220 N

Similarly;

Sum of horizontal forces;

ΣFx = 0

F_ x - 438.8(sin 30) = 0

Where F_ x  is the horizontal component of the force exerted on the left end of the rod.

F_ x = 219.4 N

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