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

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Homework Help Overview

The problem involves calculating the tension in a support cable for a wrecking ball system, including the weight of the wrecking ball and the boom. The angles of the cable and boom with respect to the horizontal are provided, and participants are tasked with determining the tension and forces acting on the system.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the torque calculations related to the weight of the boom and question the appropriateness of using the boom's weight in the torque equation. There is a suggestion to consider the vertical component of the weight in the torque calculation.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the torque calculation and suggesting adjustments to the original approach. There is no explicit consensus yet, but a potential correction to the torque calculation has been proposed.

Contextual Notes

Participants are working under the constraints of the problem statement and the provided angles, with a focus on ensuring the correct application of physics principles in their calculations.

BMK
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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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