A 219 pound sign is hanging from the end of a hinged boom, supported by a cable inclined at 52 ∘ with the horizontal. Find the tension in the cable and the compression in the boom. Round to lbs

Answer 1

Using the method of joints on where the cable is attached to and where the weight is hung from, we can apply force equilibrium equations in the x- and y-directions to determine the cable tension and the boom compression.

Given:

W = 219 lbs.

θ = 52°

Find:

T = ? lbs.

B = ? lbs.

Solution:

The tension force in the cable, T, will point UP and to the RIGHT as "observed" from joint A, raised up at an angle θ. Also, the force vector arrow should start at joint A and extend outwards from joint A

The compression force in the boom, B, will point to the LEFT on the RIGHT SIDE OF JOINT A (essentially pointing "inwards" towards joint A on the joint A's right side)

The sign's weight, W, will be straight down from joint A.

With all of this set-up information, the following can be achieved:

∑F_y = 0

0 = T sin θ - W

T sin θ = W

T = W csc θ = 219 lbs. csc( 52 ) ∴ T ≈ 277.91 lbs.

∑F_x = 0

0 = T cos θ - B

B = T cos θ

B = W cos θ / sin θ

B = W cot θ = 219 lbs. cot( 52 ) ∴ B ≈ 171.10 lbs.