Tension of a rope over a pulley

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Homework Statement


the cord shown have supports a force of 500N and warps over the frictionless pulley , determine the tension in the cord at A and the horizontal and vertical component of reactions at A .

Homework Equations




The Attempt at a Solution


for the moment about A , the author gave 500(0.2) -T(0.2) =0
But if i resolve the T into vertical and horizontal components , my ans would be difrent .. why cant I resolve the forces into 2 components ?
my working :
500(0.2)- Tsin30(0.2) -Tcos30(0.2) = ) , T=386N
 

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Answers and Replies

  • #2
andrewkirk
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This is a strangely phrased question.

Unless I am missing something, the tension in the cord is everywhere 500N, so I don't know why they feel the need to ask about the 'tension in the cord at A'. But what can 'tension in the cord at A' mean anyway, given that A is the axle of the pulley, which is 20cm away from the cord?

It seems to me that only the second part of the question makes sense - about what the reaction force is at the axle A. To do that bit, consider the pulley and axle as a unit and think of the three forces exerted on it, by the diagonal pull on the cord from the right, the vertical downwards pull on the cord at the left, and the diagonal upwards pull of the bracket on the axle, which must balance the other two.
 
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  • #3
billy_joule
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for the moment about A , the author gave 500(0.2) -T(0.2) =0
But if i resolve the T into vertical and horizontal components , my ans would be difrent .. why cant I resolve the forces into 2 components ?
my working :
500(0.2)- Tsin30(0.2) -Tcos30(0.2) = ) , T=386N

You can resolve in x and y but there's no need. Your error is that you've used the wrong lever arm length for both.
Look closely, the lever arm length for Tx & Ty is not the pulley radius. they are both different and both less than 0.2 m.
 
  • #4
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You can resolve in x and y but there's no need. Your error is that you've used the wrong lever arm length for both.
Look closely, the lever arm length for Tx & Ty is not the pulley radius. they are both different and both less than 0.2 m.
I have resolved the tension into 2 components , however i managed to get the r for the force of T cos30 , it is more than 0.2 m , right ?
for T sin30 , since it is parallel to A , so the no moment occured ?
 
  • #5
billy_joule
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I have resolved the tension into 2 components , however i managed to get the r for the force of T cos30 , it is more than 0.2 m , right ?
for T sin30 , since it is parallel to A , so the no moment occured ?
'A' is a point and a point cannot be parallel to anything.
I think you mean the line of action of Tx passes through A in which case there would be no moment. But the line of action of Tx does not pass through A, it would if θ were zero.

The point where T acts on the pulley is not horizontally across from A, it appears you've assumed it is which is root of your problems.
You can resolve into Tx and Ty and solve if you get the geometry right*, but, like I said, there's no need. andrewkirk's method is easier and faster.
(*As a hint - to find the lever arm length for Tx you need to use similar triangles and the fact that the rope is tangential to the pulley at the final point of contact.)
 
  • #6
haruspex
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what can 'tension in the cord at A' mean anyway, given that A is the axle of the pulley, which is 20cm away from the cord?
judging from the posted solution, the intent was to ask for the tension at C. Yes, it's trivial, but is here used to illustrate an application of balance of moments.
 
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do u mean if the angle has direct contact with the roller , then i can only resolve it into 2 components ?
 

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  • #8
billy_joule
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do u mean if the angle has direct contact with the roller , then i can only resolve it into 2 components ?

I'm not sure what you mean. The angle doesn't 'contact' the roller, the angle can be drawn anywhere along that straight section of string, drawing it away from the point of contact like the given diagram does is a bit clearer than your diagram where it's touching the roller.

If you show us you can solve for the x and y reaction force at A via andrewkirks method I will show you how to solve the moments about A via the unnecessarily complicated method where T is resolved into Tx and Ty and the moment due to each component is found :smile:
 
  • #9
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I'm not sure what you mean. The angle doesn't 'contact' the roller, the angle can be drawn anywhere along that straight section of string, drawing it away from the point of contact like the given diagram does is a bit clearer than your diagram where it's touching the roller.

If you show us you can solve for the x and y reaction force at A via andrewkirks method I will show you how to solve the moments about A via the unnecessarily complicated method where T is resolved into Tx and Ty and the moment due to each component is found :smile:
500(0.2)- Tsin30(0.2) -Tcos30(0.2) = ) , T=386N
shouldn't the equation look like this ?
 
  • #10
billy_joule
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500(0.2)- Tsin30(0.2) -Tcos30(0.2) = ) , T=386N
shouldn't the equation look like this ?
No. That is the same incorrect equation you posted in the OP which I've attempted to address in my previous posts. If there's something specific you don't understand please point it out.

Moments won't help you find the x and y reactions at A. Do you understand andrewkirks method? Do you know what equations to use for static equilibrium problems?
 
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No. That is the same incorrect equation you posted in the OP which I've attempted to address in my previous posts. If there's something specific you don't understand please point it out.

Moments won't help you find the x and y reactions at A. Do you understand andrewkirks method? Do you know what equations to use for static equilibrium problems?
to be honest , i dont understand andrewkirks method . can you please explain in your way ?
 
  • #12
haruspex
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to be honest , i dont understand andrewkirks method . can you please explain in your way ?
It's just the standard balance of linear forces in a static set up, ##\Sigma F = 0##.
What are the forces acting on the pulley?
What is the sum of them in vertical direction?
For this purpose, consider the section of rope in contact with the pulley as part of the pulley. It makes life a lot easier.
 
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It's just the standard balance of linear forces in a static set up, ##\Sigma F = 0##.
What are the forces acting on the pulley?
What is the sum of them in vertical direction?
For this purpose, consider the section of rope in contact with the pulley as part of the pulley. It makes life a lot easier.
in vertical direction ,shouldn't the force = 500(0.2)-Tcos30(0.2) = 0 , T = 577N ? ( i have resolved the force into vertical direction)

the author found the T = 500N first (500x0.2 - T(0.2) = 0 ), then
the reaction at Ax =
-Ax +500sin30 = 0
the reaction of Ay =
Ay-500N-500cos30 = 0 , Ay = 933N

the author found the T first , then he only proceed to find the value of Ax and Ay
 
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  • #14
BvU
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in vertical direction ,shouldn't the force = 500(0.2)-Tcos30(0.2) = 0 , T = 577N ? ( i have resolved the force into vertical direction)
No, they work in the same direction. And there is a third vertical force that keeps the pulley in place ...

Haru said:
For this purpose, consider the section of rope in contact with the pulley as part of the pulley. It makes life a lot easier.
so the point where the rope acts is the point where it touches the pulley. In other words: where the tangent to the pulley makes an angle of ... with the vertical. You fill in the dots. That point helps you find the lever arms for the torques due to the separate components of T. You won't be surprised (won't you ?) that the sum of these two is 0.2 T ...
 
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andrewkirk
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to be honest , i dont understand andrewkirks method . can you please explain in your way ?
It's what haruspex said. Just expanding a little on that: label the points where the rope first makes contact with the pulley on the left and where it leaves the pulley on the right as X and Y. Then do the problem as if, instead of the rope going freely around the pulley, it was knotted tightly to a cleat at X and again at Y. Then what you have is the pulley assembly being subject to linear forces applied at points X, Y and A. You know the forces at X and Y. You need to work out the force at A that balances it out to keep it immobile.
 
  • #16
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No, they work in the same direction. And there is a third vertical force that keeps the pulley in place ...

so the point where the rope acts is the point where it touches the pulley. In other words: where the tangent to the pulley makes an angle of ... with the vertical. You fill in the dots. That point helps you find the lever arms for the torques due to the separate components of T. You won't be surprised (won't you ?) that the sum of these two is 0.2 T ...
in vertical direction , shouldn't the force = 100(0.2)-Tcos30(0.2) = 0 , T = 577N ?
if I directly use 100(0.2)-T(0.2) = 0 , then i didnt the T is not exactly in vertical direction...
 
  • #17
haruspex
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in vertical direction , shouldn't the force = 100(0.2)-Tcos30(0.2) = 0 , T = 577N ?
We need to separate two discussions:

1. The torque balance
In your original post you correctly wrote
500(0.2) -T(0.2) =0
But you found that if you separated T into horizontal and vertical components you got the wrong answer.
As has been pointed out to you, it was because the lever arm distance (0.2 in the above equation) is different for the components than it is for the force T taken as a whole.
If you draw horizontal and vertical lines from the point where the angled rope meets the pulley, you will find that these lines pass closer than 0.2 to A. The lever arm is the shortest distance between the axis and the line of action of the force.
A little geometry shows that these lever arms are L cos(theta) and L sin(theta), where L=0.2 and theta = 30 degrees. Correspondingly the components of T are T cos(theta) and T sin(theta). So the total torque on the pulley from T is (T cos(theta) L cos(theta)) + (T sin(theta) L sin(theta)) = TL (cos2(theta)+sin2(theta)) = TL.

So whichever way you do it, T = 500N
(Where did the 100 come from in your last post?)

2. Proceeding to the next part of the question, this is where Andrew's method comes in. It involves linear forces, not torques. The radius of the pulley is no longer relevant (if it ever was).
On that understanding, what is the force balance equation for the pulley in the vertical direction?
 
  • #18
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We need to separate two discussions:

1. The torque balance
In your original post you correctly wrote

But you found that if you separated T into horizontal and vertical components you got the wrong answer.
As has been pointed out to you, it was because the lever arm distance (0.2 in the above equation) is different for the components than it is for the force T taken as a whole.
If you draw horizontal and vertical lines from the point where the angled rope meets the pulley, you will find that these lines pass closer than 0.2 to A. The lever arm is the shortest distance between the axis and the line of action of the force.
A little geometry shows that these lever arms are L cos(theta) and L sin(theta), where L=0.2 and theta = 30 degrees. Correspondingly the components of T are T cos(theta) and T sin(theta). So the total torque on the pulley from T is (T cos(theta) L cos(theta)) + (T sin(theta) L sin(theta)) = TL (cos2(theta)+sin2(theta)) = TL.

So whichever way you do it, T = 500N
(Where did the 100 come from in your last post?)

2. Proceeding to the next part of the question, this is where Andrew's method comes in. It involves linear forces, not torques. The radius of the pulley is no longer relevant (if it ever was).
On that understanding, what is the force balance equation for the pulley in the vertical direction?
total torque on the pulley from T is (T cos(theta) L cos(theta)) + (T sin(theta) L sin(theta)) = TL (cos2(theta)+sin2(theta)) = TL.

ok , understand nw
 
  • #19
haruspex
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when i resolve the T into horizontal component , it passes thru A ,
No.
Look carefully at the bottom right diagram in the image you attached in the OP.
An arrow goes up to the right from A to the point of contact of the T force with the pulley.
A horizontal line drawn through that point passes above A, not through it.
 
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  • #20
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No.
Look carefully at the bottom right diagram in the image you attached in the OP.
An arrow goes up to the right from A to the point of contact of the T force with the pulley.
A horizontal line drawn through that point passes above A, not through it.
sorry , here's what i mean , when i resolved the T into horizontal component , the Tcos30 passes thru A
 

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  • #21
haruspex
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sorry , here's what i mean , when i resolved the T into horizontal component , the Tcos30 passes thru A
Well, the way you have drawn it it's the T sin(30) that passes through A, so I'll assume that's what you meant.
Now, you have not drawn T as acting at the point where T contacts the pulley. That's ok, but then you have to be more careful with the lever arm for T cos(30). If you look carefully you will see it is now more than 0.2. Specifically, it is 0.2/cos(30).
 
  • #22
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Well, the way you have drawn it it's the T sin(30) that passes through A, so I'll assume that's what you meant.
Now, you have not drawn T as acting at the point where T contacts the pulley. That's ok, but then you have to be more careful with the lever arm for T cos(30). If you look carefully you will see it is now more than 0.2. Specifically, it is 0.2/cos(30).
ya , i mean T sin30 passes thru A, and f for T cos 30 is more than 0.2 , ( i also noticed that 0.2 cos 30 is less than 0.2) , so the correct r for T cos 30 is 0.2/ Tcos 30 ?

can you please draw me a diagram that show (T cos(theta) L cos(theta)) + (T sin(theta) L sin(theta)) = TL (cos2(theta)+sin2(theta)) = TL ?
 
  • #23
haruspex
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so the correct r for T cos 30 is 0.2/ Tcos 30 ?
0.2/cos(30), yes.
can you please draw me a diagram that show (T cos(theta) L cos(theta)) + (T sin(theta) L sin(theta)) = TL (cos2(theta)+sin2(theta)) = TL ?
Just put the point of application of T where the rope meets the pulley.
 

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