A nonuniform beam 4.51 m long and weighing 1.10 kN makes an angle of 25 below th

Question

A nonuniform beam 4.51 m long and weighing 1.10 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 3.02 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 2.10 m down the beam from the pivot. Lighting equipment exerts a downward force of 4.91 kN on the lower-left end of the beam.

(NEED PARTS A-C) PLEASE

PART A:

Find the tension

T in the cable.

T = _____________kN

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.

Fv = __________kN

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.

Fh =_______kN

Explanation / Answer

a)let the tension in the cable be T
taking moment abt pivot,
1.10 kN * 2.10 cos 25 + 4.91 kN * 4.51 cos 25 = T * 3.02
T= 8.02 k N---answer

b) let the vertical component of the force exerted on the beam by the pivotbe Fy
Fy + T cos 25 = 4.91 kN + 1.10 kN
Fy = 6.01 kN - 8.02 kN cos 25 KN

Fy=(-1.93 kN)

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

Fx = T sin 28
Fx = 8.02 kN sin 28

=2.17 KN

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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