# Exploring the Equilibrium of a System: Net Force & Net Work

• NoFaceJack
In summary: I see it, there is work done or energy entering or within the system, as the gravitational potential energy of the mass has increased.
NoFaceJack
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.

#### Attachments

• TSOKOS 2 37.png
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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.

NoFaceJack
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.

NoFaceJack
Thank you. I understand now.

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.

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.

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.

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.

hutchphd said:
This statement makes no sense. Its not even wrong.
It seems that I have not understood the problem, then.

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.

## 1. What is meant by equilibrium in a system?

Equilibrium in a system refers to a state in which all forces acting on the system are balanced, resulting in no net force. This means that the system will remain in a constant state of motion or rest.

## 2. How do you determine the net force in a system?

To determine the net force in a system, you must first identify all the individual forces acting on the system. Then, you can use vector addition to find the sum of all the forces. The net force will be the resulting vector from this addition.

## 3. What is the relationship between net force and net work?

Net force and net work are closely related as they both involve the overall effect of forces on a system. Net force is the sum of all forces acting on a system, while net work is the total amount of energy transferred to or from a system. In other words, net force determines the motion of a system, while net work determines the change in energy of a system.

## 4. How does the equilibrium of a system affect its motion?

When a system is in equilibrium, there is no net force acting on it. This means that the system will either remain at rest or continue moving at a constant velocity. If the system is not in equilibrium, there will be a net force acting on it, causing it to accelerate in the direction of the force.

## 5. What are some real-life examples of systems in equilibrium?

Some real-life examples of systems in equilibrium include a book sitting on a table, a car traveling at a constant speed on a straight road, and a pendulum at its lowest point. In all of these cases, the forces acting on the system are balanced, resulting in no net force and a state of equilibrium.

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