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Question

A 1200 kg, 17 m long crane boom is holding up a 3000 kg at its far end. The boom pivots at point A, and is attached to a support cable at point B. The support cable makes an angle of 10∘ with the boom, and the boom makes an angle of 30∘ with the horizontal.

a) Determine the magnitude of tension in the support cable, in N.

b) Determine the magnitude of force at the pivot point, in N.

c) Determine the angle of the force at the pivot point above the horizontal, in ∘

Diagram: https:// usb06ca.theexpertta.com/images/xqj1ezuz.h1q.jpg

Answer 1

This is quite a question.

To solve this problem we will use the sum of forces and the sum of torques.

a) For the tension choosing a point to measure torques from can eliminate some of the forces that seem difficult to calculate right now. Also notice there are two cables here and the tension is not the same in them.

For the torque we can use the equation τ = F r sinθ. Remember θ is the angle between the r and F vectors when both of their tails are at the origin.

For the tension T the torque is

τ_T = T (17m) sin(170°)

The angle from r to F comes from T being in quadrant 3 and 10° clockwise from -r. (try to graph this)

For the weight attached W the torque is

τ_W = (3000kg)(9.8m/s²)(17m) sin(-120°) = - (3000kg)(9.8m/s²)(17m) sin(120°)

The last torque comes from the weight of the beam B.

τ_B = (1200kg)(9.8m/s²)(8.5m) sin(-120°) = - (1200kg)(9.8m/s²)(8.5m) sin(120°)

Notice the r vector has half the magnitude as gravity acts at the center of mass of the beam.

There is no angular acceleration so

τ_T + τ_W + τ_B = 0

T = ((3000kg)(9.8m/s²)(17m) sin(120°)+(1200kg)(9.8m/s²)(8.5m) sin(120°)) / ((17m) sin(170°))

T ≈ 146,918 N

b) A sum of torques will give us the perpendicular component of torque so we can't use that. The beam is in equilibrium so the some of forces will still work. Remember F=ma is a vector equation so it has an x and y component.

x-component of forces on beam

-T cos(20°) + F_ax = 0

F_ax = 138,058 N

y-component of forces on beam

-T sin(20) - B - W + F_ay = 0

F_{ay ≈} 91,409 N

F ≈ √(138,058² + 91,409²) N ≈ 165,576N

c) From the positive x-axis

θ ≈ arctan(138,058/ 91,409) = 56.49°

Make sure to double check my algebra.