Static equilibrium and rope tension

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SUMMARY

The discussion centers on calculating the tension in a rope with negligible mass, stretched horizontally between two supports, when a weight of 3160N is hung at its center, causing a sag of 35.0cm. The correct tension in the rope is determined to be 7920N. The participant initially struggled with applying torque equations but was advised that a moment equation is unnecessary for this scenario, as the rope only supports tensile forces. This clarification emphasizes the importance of understanding static equilibrium principles in physics.

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  • Understanding of static equilibrium principles
  • Knowledge of forces and torques in physics
  • Familiarity with tension calculations in ropes
  • Basic trigonometry for angle calculations
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  • Learn about tension forces in different configurations
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Lord Anoobis
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Homework Statement


A rope of negligible mass is stretched horizontally between two supports. When an object of weight 3160N is hung at the centre of the rope, the rope is observed to sag by 35.0cm. What is the tension in the rope?

Problem 5.png

Homework Equations

The Attempt at a Solution


[/B]
Sum of forces in x-direction: ∑Fx = Fx2 - Fx1 = 0

Sum of forces in y-direction: ∑Fy = Fy1 + Fy2 - W = 0

Sum of torques: I chose the axis to be through the point attached to the left support, resulting in

L.Fy2 - 1/2L.W - 0.35Fx2 = 0

And at this point my efforts literally come to nought, because with Fy1 = Fy2, the terms in L in the torques equation cancel each other. I can see that by finding the angle between the rope and the horizontal it's possible to get the tension, which is T = 7920N. However this approach will not suffice for more advanced problems. So my question is, where am I going wrong here?
 
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Lord Anoobis said:

Homework Statement


A rope of negligible mass is stretched horizontally between two supports. When an object of weight 3160N is hung at the centre of the rope, the rope is observed to sag by 35.0cm. What is the tension in the rope?

View attachment 74720

Homework Equations

The Attempt at a Solution


[/B]
Sum of forces in x-direction: ∑Fx = Fx2 - Fx1 = 0

Sum of forces in y-direction: ∑Fy = Fy1 + Fy2 - W = 0

Sum of torques: I chose the axis to be through the point attached to the left support, resulting in

L.Fy2 - 1/2L.W - 0.35Fx2 = 0

And at this point my efforts literally come to nought, because with Fy1 = Fy2, the terms in L in the torques equation cancel each other. I can see that by finding the angle between the rope and the horizontal it's possible to get the tension, which is T = 7920N. However this approach will not suffice for more advanced problems. So my question is, where am I going wrong here?

It's not clear what 'more advanced problems' you are worried about.

A rope can only support a tensile force; therefore writing a moment equation in this case is unnecessary.
 
Ah, I see. This is why learning on one's own can be such a pain. Good thing Physics Forms is here to clear things up. Thank you.
 

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