A nonuniform beam 4.49 m long and weighing 1.01 kN makes an angle of 25 below th

Question

A nonuniform beam 4.49 m long and weighing 1.01 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 2.96 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 2.01 m down the beam from the pivot. Lighting equipment exerts a downward force of 4.96 kN on the lower-left end of the beam.

Part A

Find the tension T in the cable.

T = answer in 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.

answer in 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.

answer in kN

Explanation / Answer

The tension in the cable T can be found by taking moment about the pivot

Moment = Force*distance from the pivot

Hence,

1.01k ×2.01cos25 + 4.96k×4.49cos25 = T×2.69

here angle is in degree

Solving we get T=8.187 kN

b)

Let  the vertical component of the force exerted on the beam by the pivot be Fy

hence from the diagram we can find the force equation as

Fy + Tcos25 =4.96k +1.01K

hence Fy = 1.449kN downwards

c)

Let  the vertical component of the force exerted on the beam by the pivot be Fx

hence from the diagram we can find the force equation as

Fx = Tsin25 = 3.459kN upwards

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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