A 13,850 -N crane pivots around a friction-free axle at its base and is supporte

Question

A 13,850-N crane pivots around a friction-free axle at its base and is supported by a cable making a 25

A 13,850-N crane pivots around a friction-free axle at its base and is supported by a cable making a 25 degree angle with the crane (see figure). The crane is 16 m long and is not uniform, its center of gravity being 7.0 m from the axle as measured along the crane. The cable is attached 3.0 m from the upper end of the crane. When the crane is raised to 55 degree above the horizontal holding an 11,600-N pallet of bricks by a 2.2-m very light cord, find the following. the horizontal and vertical components of the force that the axle exerts on the crane

Explanation / Answer

(a)
For the system to be in balance the sum of the moments should equal zero.

Let the tension force on the cable = F

F horizontal = F * sin 60

F vertical = F * cos 60

Moments around the axle:

[13850 * 7 cos 55] + [11600 * 16 cos 55] + [F cos 60 * 13 cos 55] - [F sin 60 * 13 sin 55] = 0

F = [13850 * 7 cos 55] + [11600 * 16 cos 55]/13 * ( [sin 60 * sin 55] - [cos 60 * cos 55] )

= 55608.24 + 8188.91 *(0.709 - 0.287)

= 55608.24 + 3455.72

= 59063.96 N

This is the tension force in cable.

(b)
Again for the system to be balanced the sum of the horizontal forces must equal zero.

Let the horizontal component of force on the axle be H

H - F(h) = 0 (taking forces left to right to be positive)

H = F(h)

F(h) = F sin 60
     = 59063.96 * sin 60
    = 51151 N

H = 51151 N

For the system to be in balance the sum of vertical forces must equal zero

Let force on axel to be V.

V - 13850 - 11600 - Fv = 0

F(v) = F cos 60
     = 59063.96 * cos 60
    = 29531.98 N

V - 25450 - 29531.98 = 0

V = 54981.98 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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