How Do You Calculate Tension in a Wrecking Ball Support Cable?

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To calculate the tension in a wrecking ball support cable, the weight of the wrecking ball (4800 N) and the boom (3600 N) must be considered along with the angles of the cable (32 degrees) and the boom (48 degrees). The equations of equilibrium for forces in the x and y directions are applied to solve for the tension (Ft) in the cable, resulting in a calculated tension of approximately 12455 N. Additionally, the force exerted by the hinge at point P is determined to be 10562 N in the x-direction and 875 N in the y-direction. A correction is noted regarding the torque calculation for the boom's weight, suggesting it should be based on the vertical component (3600 sin 48) rather than the horizontal. Accurate calculations are crucial for ensuring structural integrity in such scenarios.
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1. Homework Statement

A wrecking ball (weight = 4800 N) is supported by a boom, which may be assumed to be uniform and has a weight of 3600 N. A support cable runs from the top of the boom to the tractor. The angle between the support cable and the horizontal is 32, and the angle between the boom and the horizontal is 48. Find (a) the tension in the support cable and (b) the magnitude of the force exerted on the lower end of the boom by hinge at point P.


2. Homework Equations
\SigmaT= Fr = 0
Fx = -Ftcos + Rx = 0
Fy = -Wball - Wboom +Ftsin+Ry = 0

3. The Attempt at a Solution
\SigmaT= = 3600 (L/2) + 4800 L - Ft sin32 L = 0
\SigmaT= = 1800 L + 4800 L = Ftsin32L
\SigmaT= = 6600 = Ft sin 32
Ft = 12455N

Fx = -Ftcos32 + Rx = 0
Fx = -12455 cos 32= -Rx
Rx = 10562 N

Fy = -4800 - 3600sin48 + Ftsin 32 + Ry = 0
Fy = -4800 -3600sin48 + 12455sin 32 +Ry = 0
Ry = 875N
 
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One problem I see with you solution is that you used 3600 (L/2) to get the torque from the gravity on the boom. However, the force of gravity points straight down, while the L vector point at an angle of 48 to the horizontal.
 
So should it be 3600 sin 48 (L/2)?
 
yeah, sounds about right
 
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