Tension on a string

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Homework Statement



If a string is being pulled to the left by 10N and at the same time being pulled to the right by 15N, what is the tension on the string?

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The Attempt at a Solution



I always think tension as the force towards one end of the string that balances the force exerted towards the other end of the string. So for this question, I think the tension is 5N to the left. Am I correct?

Thanks guys!
 

Answers and Replies

  • #2
kuruman
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Consider a point on the string. At this point, draw an arrow 10 units long pointing to the left and an arrow 15 units long pointing to the right. The tension in the string is the vector sum of the two arrows. What is it?
 
  • #3
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The vector sum will be 5N to the right. So the tension is 5N to the right?

AND, what if the two stretching forces are equal in magnitude but opposite in direction? 10N to the left, and 10 to the right? In that case, the vector sum is 0N, but obviously, there's still tension on the string. Then what's the tension in this case?

Thank you...
 
  • #4
kuruman
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The vector sum will be 5N to the right. So the tension is 5N to the right?
Right.
AND, what if the two stretching forces are equal in magnitude but opposite in direction? 10N to the left, and 10 to the right? In that case, the vector sum is 0N, but obviously, there's still tension on the string. Then what's the tension in this case?
Thank you...
The net force on the string is zero which means that the string does not accelerate. Nevertheless, the tension is 10N. It is a common misconception that the tension in this case could be zero or 20N. Don't fall in that trap.

Consider a 10N weight hanging from the ceiling by a string. The string needs to provide 10N up to cancel the 10N down provided by gravity. The string is at rest and does not accelerate which means that any point on the string has zero net force acting on it. If you put a pencil mark anywhere on the string, at the mark there will be a force 10N up and another force 10N down. (I assume that the string is massless.) At the last (highest) point of the string you have run out of string but you have a ceiling to which the string is attached. To keep the last string point at rest, the ceiling must exert 10N up on it. That is how the 10N force of the weight is transmitted to the ceiling which, after all, has to support the weight.
 
  • #5
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Right.

The net force on the string is zero which means that the string does not accelerate. Nevertheless, the tension is 10N. It is a common misconception that the tension in this case could be zero or 20N. Don't fall in that trap.

Consider a 10N weight hanging from the ceiling by a string. The string needs to provide 10N up to cancel the 10N down provided by gravity. The string is at rest and does not accelerate which means that any point on the string has zero net force acting on it. If you put a pencil mark anywhere on the string, at the mark there will be a force 10N up and another force 10N down. (I assume that the string is massless.) At the last (highest) point of the string you have run out of string but you have a ceiling to which the string is attached. To keep the last string point at rest, the ceiling must exert 10N up on it. That is how the 10N force of the weight is transmitted to the ceiling which, after all, has to support the weight.

Ahh I see. That helps! Thank you so much Dr.! :)
 
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Excuse me , I have a question about this, could this problem be discussed by assuming the string is massless? 5N to the right is the sum of vectors, but is it also the tension? My opinion is that, if we don't take the mass of the string (or the linear density or anything similar) into consideration, this problem is meaningless
 
  • #7
kuruman
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My response to the OP was based on the assumption that this is an idealized string, the textbook definition of which is a massless string that transmits tension from one end to the other without changing its magnitude. According to you, idealized strings do not exist much like the ever present disembodied hand do not exist. I agree. Nevertheless, idealizations are necessary for enabling novices to understand the appropriate physical principles to some zeroth order approximation before proceeding to a higher order of understanding.

I took the question, as posted by the OP, as a test of the application of the ideal string definition and I answered accordingly. As you can see from OP's final response, that was the kind of help that OP was looking for.
 
  • #8
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My response to the OP was based on the assumption that this is an idealized string, the textbook definition of which is a massless string that transmits tension from one end to the other without changing its magnitude. According to you, idealized strings do not exist much like the ever present disembodied hand do not exist. I agree. Nevertheless, idealizations are necessary for enabling novices to understand the appropriate physical principles to some zeroth order approximation before proceeding to a higher order of understanding.

I took the question, as posted by the OP, as a test of the application of the ideal string definition and I answered accordingly. As you can see from OP's final response, that was the kind of help that OP was looking for.

Thanks Kuruman, I got your idea~ I'm also new to physics, maybe days after it'll be me asking new comer's problems~
 
  • #9
kuruman
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Just so that there be no confusion about this.

You can either have an idealized string in which case it transmits tension from one end to the other without changing its magnitude OR you can have a string that has different forces at its ends in which case it cannot be idealized and must have some linear density.

OP's initial question is a bit ambiguous about which of the two cases it refers to. I assumed the first case, but genxium is correct in pointing out the second possibility. A complete answer to OP's initial question "If a string is being pulled to the left by 10N and at the same time being pulled to the right by 15N, what is the tension on the string?" could be

If the string is idealized, the forces at the two ends cannot be different as stated; they must be the same and the tension on the string is that same value. If the string is not idealized, then the forces at the two ends can be different, but then the "tension on the string" is a poorly defined quantity because it depends on the position along the string.
 
  • #10
PeterO
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Homework Statement



If a string is being pulled to the left by 10N and at the same time being pulled to the right by 15N, what is the tension on the string?

Homework Equations



N/A

The Attempt at a Solution



I always think tension as the force towards one end of the string that balances the force exerted towards the other end of the string. So for this question, I think the tension is 5N to the left. Am I correct?

Thanks guys!

I don't believe it is possible to apply a 10 N force to one end of a string while applying a 15 N force to the other end. Now if you are trying to do this by hanging the string over a massless, frictionless pulley with a 1 kg mass on one side and a 1.5 kg mass on the other [taking g=10 for simplicity; if you need g=9.8 then the masses are a little bigger so the weight forces are 10N and 15N]
In that case you would see the tension in the string is 12 N.

1kg mass: 10N down, 12 N up net force 2N up: F=ma --> a = 2
1.5kg mass: 15 N down, 12 N up net force 3 N down: F=ma ---> a =2
They have to accelerate at the same rate, since they are tied together.
[if you want g=9.8, the acceleration is a little less]

[And also I understand the tension to be in the string, not on the string.]
 
  • #11
kuruman
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I don't believe it is possible to apply a 10 N force to one end of a string while applying a 15 N force to the other end. Now if you are trying to do this by hanging the string over a massless, frictionless pulley with a 1 kg mass on one side and a 1.5 kg mass on the other (taking g=10 for simplicity; if you need g=9.8 then the masses are a little bigger so the weight forces are 10N and 15N)
In that case you would see the tension in the string is 12 N.

1kg mass: 10N down, 12 N up net force 2N up: F=ma --> a = 2
1.5kg mass: 15 N down, 12 N up net force 3 N down: F=ma ---> a =2
They have to accelerate at the same rate, since they are tied together.
[if you want g=9.8, the acceleration is a little less]
All this is correct but does not constitute proof that what I said is incorrect. See below.
[And also I understand the tension to be in the string, not on the string.]
You understand correctly. In fact I initially had "in" instead of "on", but I changed it to match OP's initial posting.

This is how you can have 15N at one end of the string and 10N at the other.

Take a string of mass m and tie a mass 2m at one end. Put the mass on a horizontal frictionless plane. Pull on the free end of the string with a force of 15N. What is the force exerted by the mass on the string at the other end?

Answer: The force exerted by mass 2m on the string is equal and opposite to the force exerted by the string on the mass (Newton's 3rd law). Furthermore, the string is the only thing that exerts a horizontal force on the mass, therefore it is the net horizontal force. By Newton's 2nd law then it is equal to mass times horizontal acceleration. So we need to show that 2m*a is 10N.

Now a = 15N/(m+2m) = 5N/m. Therefore 2m*a = 2m*5N/m = 10N. Q.E.D.
 
  • #12
PeterO
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All this is correct but does not constitute proof that what I said is incorrect. See below.

You understand correctly. In fact I initially had "in" instead of "on", but I changed it to match OP's initial posting.

This is how you can have 15N at one end of the string and 10N at the other.

Take a string of mass m and tie a mass 2m at one end. Put the mass on a horizontal frictionless plane. Pull on the free end of the string with a force of 15N. What is the force exerted by the mass on the string at the other end?

Answer: The force exerted by mass 2m on the string is equal and opposite to the force exerted by the string on the mass (Newton's 3rd law). Furthermore, the string is the only thing that exerts a horizontal force on the mass, therefore it is the net horizontal force. By Newton's 2nd law then it is equal to mass times horizontal acceleration. So we need to show that 2m*a is 10N.

Now a = 15N/(m+2m) = 5N/m. Therefore 2m*a = 2m*5N/m = 10N. Q.E.D.

Firstly my response was to the original post, not to any of the discussions/analyses offered.

Secondly
If you do have a string of mass m, then you have to consider where in the string you want to measure the tension. It will be 15N at the end you are pulling, but will be only 10N down at the end attached to the 2m mass, and [provided the string has a constant mass/unit length] a gradient from 15 to 10 along the string. For example, the tension half way along the string will be 12.5N.

It indeed is the Newton's Third law problem.

Take a plastic slinky which is 1 metre long when relaxed and 4m long when subjected to a force of 10N. Attach one end to the wall and hold the other end. It is not possible for any human alive to stand 2m from the wall, holding the end of the slinky, then apply a force of 20N to the slinky while keeping his/her feet fixed on the floor.
 

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