In the figure (Figure 1) a 6.00-m-long, uniform beam is hanging from a point 1.0

Question

In the figure (Figure 1) a 6.00-m-long, uniform beam is hanging from a point 1.00m to the right of its center. The beam weighs 130N and makes an angle of 30.0?  with the vertical. At the right-hand end of the beam a concrete block weighing 100N  is hung; an unknown weight whangs at the other end.

Part A

If the system is in equilibrium, what is w? You can ignore the thickness of the beam.

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Part B

If the beam makes, instead, an angle of 50.0? with the vertical, what is w?

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Figure 1 of 1

w =   N   In the figure (Figure 1) a 6.00-m-long, uniform beam is hanging from a point 1.00m to the right of its center. The beam weighs 130N and makes an angle of 30.0? with the vertical. At the right-hand end of the beam a concrete block weighing 100N is hung; an unknown weight whangs at the other end. Part A If the system is in equilibrium, what is w? You can ignore the thickness of the beam. w = N SubmitMy AnswersGive Up Part B If the beam makes, instead, an angle of 50.0? with the vertical, what is w? w = N SubmitMy AnswersGive Up Provide FeedbackContinue Figure 1 of 1

Explanation / Answer

given,

length of the beam = 6 m

weight of beam = 130 N

weight on the right hand side = 100 N

angle = 30 degree

hanging point is 1 m from its center

let W = weight of beam

L = length of the beam

x = hanging point from the center

Wr = weight on right side

Wl = weight on left side

theta = angle

The moment of the weight 100 N on the right side is

moment_right = Wr(L/2 - x) sin(theta)

moment_right = 100 (6/2 - 1) * sin(30)

The weight of the uniform beam can be considered as acting through its mid-point.

So

The moments of the weights on the left side are

moment_left = W x sin(theta) + Wl (L/2 + x) sin(theta)

moment_left = 130 * 1 * sin(30) + Wl (6/2 + 1) sin(30)

In equilibrium, these must balance one another, hence :

Wr(L/2 - x) sin(theta) = W x sin(theta) + Wl (L/2 + x) sin(theta)

100 (6/2 - 1) * sin(30) = 130 * sin(30) + Wl (6/2 + 1) sin(30)

100 (6/2 - 1) = 130 + Wl (6/2 + 1)

100 * 2 = 130 + Wl * 4

Wl = 17.5 N

w = 17.5 N

since the weight is not dependent on the angle as we can see in the final equation,

so,

weight when beam will make 50 degree angle will be = 17.5 N

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