A nonuniform beam 4.60 m long and weighing 1.54 kNmakes an angle of 25.0? below

Question

A nonuniform beam  4.60 m long and weighing 1.54 kNmakes an angle of 25.0? below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 2.90 m farther down the beam and perpendicular to it in (Figure 1) . The center of gravity of the beam is a distance of 1.90 m down the beam from the pivot. Lighting equipment exerts a downward force of 5.05 kN on the end of the beam.

A)Sketch a free-body diagram of the beam.

B) Find the tension T in the cable

C) Find the x component of the force exerted on the beam by the pivot.

D) Find the y component of the force exerted on the beam by the pivot.

Thank you very much in advance!!!

Explanation / Answer

b) let the tension in the cable be T
taking moment abt pivot,
1.54 k * 1.90cos 25 + 5.05 k * 4.60 cos 25 = T * 2.90

T= 8.174271199 k N---answer


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

Fx = T sin 25
Fx = 8.174271199 k sin 25 N ---answer

d) Let the vertical component of the force exerted on the beam by the pivotbe Fy
Fy + T cos 25 = 5.05 k + 1.54 k
Fy = 6.59 k - 8.174271199 k cos 25 N = -0.818405641 ----answer

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