A nonuniform beam 4.60 m long and weighing 1.60 kN makes an angle of 25.0? below

Question

A nonuniform beam 4.60 m long and weighing 1.60 kN makes 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.

Part B

Find the tension T in the cable.

Part C

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

Part D

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

Explanation / Answer

here,
Angle, A = 25
length of beam, L = 4.6 m
wieght of beam, w = 1600 N

Length of cable, lc = 2.90 m
centre of gravity, c = 1.90 m
force Downwards, fd = 5050 N

Part B:
assuming tension in the cable be T ,taking moment about pivot,

w*c*Cos25 + Fd*Lc*Cos25 = T*lc

(1600*1.90*Cos25) + (5050*2.90*Cos25) =( T*2.90)

Tension, t = 5526.915 N or 5.52 kN

Part C:
let the horizontal component of the force exerted on the beam by the pivot be Fx
Then From newton Second Law, Fnet = 0
Fx = T*Sin25
Fx = 5526.915 * Sin25
Fx = 2335.775 N or 2.33 kN

Part D:
let the vertical component of the force exerted on the beam by the pivotbe Fy,
Then From newton Second Law, Fnet = 0
Fy + T*Cos25 = w + Fd
Fy + 5526.915*cos25 = 5050 + 1600
Fy = 1640.914 N or 1.64 kN

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.