How is a system in equilibrium when net force and net work are both zero?

  • #1
NoFaceJack
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Homework Statement
A mass m hangs from two strings attached to the ceiling such that they make the same angle with the vertical (as shown in Figure 2.69: See Attached Image). The strings are shortened very slowly so that the mass is raised a distance Δh above its original position. Determine the work done by the tension in each string as the mass is raised.
Relevant Equations
Work done by gravity (Gravitational Potential Energy)
In the solving portion of the textbook, the reasoning of the author in solving this problem is that the net work done on the system is zero because the net force of the system is zero.

So my question is how was the system in equilibrium (net force=0)?

My thinking is that since it is stated in the problem that the object was raised very slowly, so the time it takes to move the object is very large, that the acceleration can be estimated as zero. And thus, we can estimate that the net force, and the net work are both zero.

Is my logic correct? or is there any other reasoning behind this?

Thank you very much in advance.
 

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  • #2
The work done by the tension on what? Tension in and off itself is not a force, it is a state in the string. Once you consider a part of the diagram and draw the free-body diagram for that, this may cause tensile forces to appear due to the tension in the string.

If the assumption is "on the mass", then the answer is obviously non-zero. The gravitational force will do negative work on the mass and the total work on the mass needs to be zero based on the work-energy theorem. The only other forces that can do work on the mass are the tensile forces from the strings on the mass.
 
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  • #3
The net work on the mass is the change in kinetic energy, which is zero. This is exactly zero, not approximately zero. The mass must have accelerated to to begin with, but then it would have decelerated to rest. The total acceleration, therefore, is precisely zero; not approximately zero.

There's no need to assume that the net force is approximately zero throughout, or that the mass is raised slowly. The work done is the same as long as the mass starts and ends at rest.

It's true that in some problems there must be a small kinetic energy at the end that is assumed to be negligible, but not in this case.
 
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  • #4
Thank you. I understand now.
 
  • #5
PeroK said:
The net work on the mass is the change in kinetic energy, which is zero. This is exactly zero, not approximately zero. The mass must have accelerated to to begin with, but then it would have decelerated to rest. The total acceleration, therefore, is precisely zero; not approximately zero.

There's no need to assume that the net force is approximately zero throughout, or that the mass is raised slowly. The work done is the same as long as the mass starts and ends at rest.

It's true that in some problems there must be a small kinetic energy at the end that is assumed to be negligible, but not in this case.
But it does not ask for the net work done on the mass; it asks for the work done by the tension.
Since it is, presumably, inelastic, the tension cannot do any work of its own accord, but it could convey work done on the string by a pull at one end to work it does on the mass at the other. Whether that should be considered work done by the tension seems a matter of taste.
The author's reasoning is unclear since the system is undefined.
 
  • #6
The way I see it, there is work done or energy entering or within the system, as the gravitational potential energy of the mass has increased.

The horizontal components of the tensions of the strings are symmetrical and opposed, and because of that, cancel each other and are not able to do any work on the suspended mass.

The vertical components of the tensions of the strings are symmetrical as well but not opposed, and because of that, there is a resultant vertical force able to do work on the suspended mass by moving it upwards a distance h against gravity.

How quickly that work was done only tells us that that movement took a long time, not that the resultant vertical force or the distance was zero.
NoFaceJack said:
In the solving portion of the textbook, the reasoning of the author in solving this problem is that the net work done on the system is zero because the net force of the system is zero.

So my question is how was the system in equilibrium (net force=0)?

My thinking is that since it is stated in the problem that the object was raised very slowly, so the time it takes to move the object is very large, that the acceleration can be estimated as zero. And thus, we can estimate that the net force, and the net work are both zero.

Is my logic correct? or is there any other reasoning behind this?

Thank you very much in advance.
Welcome, NoFacejack!

The way I see it, there is work done or energy entering (external agent pulling the strings) or within the system (strings contracting due to cooling), as the gravitational potential energy of the mass has increased.

The horizontal components of the tensions of the strings are symmetrical and opposed, and because of that, cancel each other and are not able to do any work on the suspended mass.

The vertical components of the tensions of the strings are symmetrical as well but not opposed, and because of that, there is a resultant vertical force able to do work on the suspended mass by moving it upwards a distance h against gravity.

How quickly that work was done only tells us that that movement took a long time, not that the resultant vertical force or the distance was zero.
 
  • #7
Lnewqban said:
The horizontal components of the tensions of the strings are symmetrical and opposed, and because of that, cancel each other and are not able to do any work on the suspended mass.
Correct result, wrong argument. The correct argument is that the horizontal components are orthogonal to the displacement. Otherwise they would each do work of the same magnitude but opposite sign.
 
  • #8
Lnewqban said:
The way I see it, there is work done or energy entering or within the system, as the gravitational potential energy of the mass has increased
This statement makes no sense. Its not even wrong.
 
  • #9
hutchphd said:
This statement makes no sense. Its not even wrong.
It seems that I have not understood the problem, then.
Would you mind explaining your reasoning, please?
 
  • #10
I would need to diagram the sentence (is English your native tongue? perhaps that is part of it)
"There is work done or energy entering or within the system" is a statement that conveys no information therefore cannot be evaluated.
 
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