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Tension in ropes

  • Thread starter sp3sp2sp
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  • #1
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Homework Statement


This is not HW problem specific, but more of a lack of understanding for an equation in answer.
A pair of students are lifting a heavy trunk on move-in day. (Figure 1) Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity v⃗ . Each rope makes an angle θ with respect to the vertical. The gravitational force acting on the trunk has magnitude FG.

(the figure is just 2 people pulling rope on either side of the trunk)
What is the force of tension in each rope? The answer is T = (FG) / (2cos(θ))

Homework Equations




The Attempt at a Solution


I undertand that the y-components of the tensions will be T_y = T*cosθ (because θ is here measured from the vertical)
I also get that F_mg is the downward force which points along negative-y axis.
I also get that all forces must balance because it says constant v, so no acceleration. So mg + T_y = 0 and T_1x + T_x2 = 0
But Im still not understanding how the tension in each rope, T = (FG) / (2cos(θ))? Dont we have to factor in the x-components of the tension also? If so where are the x-components in the equation?

thanks for any help
 

Answers and Replies

  • #2
haruspex
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mg + T_y = 0
There are two sections of rope, each exerting Ty upwards on the ring.
Dont we have to factor in the x-components of the tension also?
That has been done by dividing by cos(θ).
 
  • #3
CWatters
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The Attempt at a Solution


I undertand that the y-components of the tensions will be T_y = T*cosθ (because θ is here measured from the vertical)
I also get that F_mg is the downward force which points along negative-y axis.
I also get that all forces must balance because it says constant v, so no acceleration. So mg + T_y = 0 ....
That should be mg + 2T_y = 0

Then substitute T_y = TCos(θ).

That's all you need to do.

and T_1x + T_x2 = 0
True but not needed to solve for T.

Dont we have to factor in the x-components of the tension also? If so where are the x-components in the equation?

thanks for any help
The x-components are orthogonal (aka 90 degrees) to mg. So the x-components of the tension doesnt carry any of the force mg.
 
  • #4
kuruman
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Dont we have to factor in the x-components of the tension also? If so where are the x-components in the equation?
The x-components must be equal and opposite because the crate is rising straight up as stated by the problem. They do not affect the motion and therefore are not mentioned.

P.S. I do not see a figure. Does everybody else see it?
 
  • #5
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Homework Statement


This is not HW problem specific, but more of a lack of understanding for an equation in answer.
A pair of students are lifting a heavy trunk on move-in day. (Figure 1) Using two ropes tied to a small ring at the center of the top of the trunk, they pull the trunk straight up at a constant velocity v⃗ . Each rope makes an angle θ with respect to the vertical. The gravitational force acting on the trunk has magnitude FG.

(the figure is just 2 people pulling rope on either side of the trunk)
What is the force of tension in each rope? The answer is T = (FG) / (2cos(θ))

Homework Equations




The Attempt at a Solution


I undertand that the y-components of the tensions will be T_y = T*cosθ (because θ is here measured from the vertical)
I also get that F_mg is the downward force which points along negative-y axis.
I also get that all forces must balance because it says constant v, so no acceleration. So mg + T_y = 0 and T_1x + T_x2 = 0
But Im still not understanding how the tension in each rope, T = (FG) / (2cos(θ))? Dont we have to factor in the x-components of the tension also? If so where are the x-components in the equation?

thanks for any help
You know the tensions in the two ropes. So what are the components of these two tensions in the x direction? Are they equal in magnitude and opposite in direction?
 
  • #6
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OK I now get that they are pulling straight up, so its all y-components, but why are dividing by cos θ? The question asks about force in one rope so we divide by 2, but why are we dividing by cos θ? thanks for any more help
 
  • #7
CWatters
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Please upload the diagram.

At the moment all the replies above, including mine, assume the ropes are pulling up at an angle (eg the ropes are NOT pulling vertically). On that basis the vertical component is TCos(theta) per rope or 2TCos(theta) in total. Where T is the tension in one rope. When you rearrange that you get 2Cos(theta) on the bottom. Simple geometry.

However if your diagram is different then our replies might all be wrong.
 
  • #8
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OK I now get that they are pulling straight up, so its all y-components, but why are dividing by cos θ? The question asks about force in one rope so we divide by 2, but why are we dividing by cos θ? thanks for any more help
Here are the force balances in the vertical and horizontal directions, respectively:
$$T\cos{\theta}+T\cos{\theta}=mg$$
$$T\sin{\theta}-T\sin{\theta}=0$$
Please tell me if these balances make sense to you. Have you learned how to resolve a vector into horizontal and vertical components?
 

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