What is the tension in the right cable?

AI Thread Summary
To determine the tension in the right cable supporting a 24kg mass on a horizontal beam, one must analyze the forces and torques acting on the system. The total weight acting downward is calculated as 235.2N. It is essential to consider the moments about the left end of the beam to isolate the tension in the right cable. The discussion highlights confusion regarding the net forces and whether the cables are vertical and attached at the ends of the beam. Proper visualization and application of torque principles will lead to the correct solution for the tension in the right cable.
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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