What is the tension in the right cable?

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Homework Help Overview

The problem involves determining the tension in a cable system supporting a mass on a beam. The context includes a 24kg mass and an 8kg beam, with specific distances noted between the mass and the cable attachment points.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the absence of horizontal forces and suggest visual aids for clarity. They propose taking moments about a point to find the unknown tension. Questions are raised regarding the movement of the object and the implications for net force. The attachment points and orientation of the cables are also questioned.

Discussion Status

The discussion is active, with participants providing guidance on how to approach the problem. There is a focus on clarifying assumptions and understanding the forces involved, though no consensus has been reached on the specific method to solve for the tension.

Contextual Notes

There is some confusion regarding the net forces acting on the system, and it is unclear how the forces are being analyzed in relation to the beam and the mass. The assumption that the cables are vertical and attached at the ends of the beam is noted but not confirmed.

MadMcB
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Homework Statement


2 cables are used to support a 24kg mass on a 1.6m long, 8kg uniform horizontal beam. The 24kg mass is 0.6m from the right cable. What is the tension in the right cable?

Homework Equations


torque = rFsinx

The Attempt at a Solution



Fy = ma = 24(9.8) = 235.2N
Fx = ?
resultant F - ?[/B]
 
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There is no Fx forces acting if I'm visualizing it correctly.A picture would help.
Take moments about the left end, solve for the one unknown force - the rhs cable tension.
 
Is the object moving? If so or if not, what does this tell you about the net force.
 
And are the cables vertical? Where are they attached to the beam? At the ends ?
 
There is some confused thinking here:
MadMcB said:
Fy = ma = 24(9.8) = 235.2N
Fy = ma should be ##\Sigma F_y = ma_y##, i.e. the net force in the Y direction gives rise to an acceleration in the Y direction.
It is not clear whether you are considering forces on the beam, forces on the mass, or forces on the two as a system. Whichever, 24(9.8) is only one force in the Y direction, not the net force.
I assume the cables are vertical and attached to the ends of the beam. If so, follow billy_joule's advice. You can get the answer without using ##\Sigma F_y = ma_y##.
 

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