A nonuniform beam 4.41m long and weighing 1.05kN makes an angle of 25 ? below th

Question

A nonuniform beam 4.41m long and weighing 1.05kN 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 2.95m 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 2.06m down the beam from the pivot. Lighting equipment exerts a downward force of 5.10kN on the lower-left end of the beam.

A) Find the tension T in the cable.

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

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

part A:

the tenstion in the cable be T and using the concept of momets about the axis

F1s1cos theta + F2s2 cos theta = T *S

1050 * 2.06 cos 25 + 5100 * 4.41 cos 25 = T * 2.95

1960.343 + 20383.768 = T * 2.95

T = 7.574 kN--------------------<<<<<<<<<<<<<Answer

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b) let the vertical component of the force exerted on the beam

by the pivotbe Fy

then

Fy + T cos 25 = 5100 + 1050

Fy = 6150 - (7574 cos 25)

Fy = 6.15k -(7.54 cos 25)   or

Fy = -714 k N-----------------------<<<<<<<<<<<<<<Answer

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c)let the horizontal component of the force exerted on the beam by

the pivot be Fx, then Fx = T sin 28

Fx = 7.574 k sin 25 N

or Fx =3.2 kN ----------------------<<<<<<<<<<<<<Answer

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