A nonuniform beam 4.50 m long and weighing 1.04 kN makes an angle of 25 below th

Question

A nonuniform beam 4.50 m long and weighing 1.04 kN makes an angle of 25 below the horizontal. It is held in position by a frictionless pivot at its upper-right end and by a cable a distance of 3.01 m farther down the beam and perpendicular to it (see Figure 11.31 in the textbook). The center of gravity of the beam is a distance of 2.07 m down the beam from the pivot. Lighting equipment exerts a downward force of 5.00 kN on the lower-left end of the beam.

Part A

Find the tension T in the cable

Part B

Find the vertical component of the force exerted on the beam by the pivot.

Assume that the positive x and y axes are directed to the right and upward respectively.

Part C

Find the horizontal component of the force exerted on the beam by the pivot.

Assume that the positive x and y axes are directed to the right and upward respectively.

Explanation / Answer

length of the beam,l=4.50 m

Weight,w=1.04 kN

angle= 25

distance = 3.01 m

downward force= 5.00 kN

a)let the tension in the cable be T

taking moment about the pivot,

1.04 k * 2.07 cos 25 + 5 k * 4.5 cos 25 = T * 3.01

T= 7.422 k N

b) let the vertical component of the force exerted on the beam by the pivot can be considered as Fy

Fy + T cos 25 = 5 k + 1.04 k

Fy = 6.04k- 6.723 k=-0.69 k

c)let the horizontal component of the force exerted on the beam by the pivot can be considered as Fx

Fx = T sin 28

Fx = 7.422 k sin 25 N =2.824 kN

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