Static equilibrium and rope tension

In summary, the conversation discusses a problem involving a rope stretched horizontally between two supports and an object hanging at the center of the rope. The weight of the object and the amount of sag in the rope are given, and the question asks for the tension in the rope. The conversation includes equations for sum of forces and torques, but it is noted that a moment equation is unnecessary in this case.
  • #1
Lord Anoobis
131
22

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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  • #2
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.
 
  • #3
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.
 

1. What is static equilibrium?

Static equilibrium is a state of balance in which all forces acting on an object are equal and opposite, resulting in no net force and no acceleration.

2. How is rope tension related to static equilibrium?

Rope tension is one of the forces that can act on an object in static equilibrium. In order for an object to be in static equilibrium, the sum of all forces acting on it, including rope tension, must equal zero.

3. What factors affect the rope tension in a system?

The rope tension in a system is affected by the weight of the object being supported, the angle at which the rope is being pulled, and the strength and elasticity of the rope itself.

4. How can I calculate the rope tension in a system?

The rope tension can be calculated using the equation T=mg, where T is the tension in the rope, m is the mass of the object being supported, and g is the acceleration due to gravity.

5. What happens to the rope tension if the system is not in static equilibrium?

If the system is not in static equilibrium, the forces acting on the object will not be balanced and there will be a net force. This can cause the rope tension to change and potentially cause the object to move or accelerate.

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