Tension, Net Force, and Acceleration relationship?

[Riman643]

Homework Statement

What is the relationship between Tension, Net Force, and Acceleration?

Relevant Equations

F = ma

Okay so I am having trouble understanding the relationship between Tension, Net Force, and Acceleration. For example, if two objects in a system are connected by a tension force and accelerating at the same speed, but with different forces acting on each object, would:

1. The net force be equal to the sum of the net forces of the two objects?

2. The acceleration of the net force be same for the acceleration of the two objects?

3. The tension force on the first object be equal to the tension force on the second object?

 

[kuruman]

1. Is your system the two objects taken together? If so, there is only one net force, the sum of all the forces acting on both masses.
2. It's not clear what you mean by "the acceleration of the net force". The net force does not accelerate, the system on which it acts does. Again, what is your system?
3. Once more, what is your system? If the two objects are connected by a massless string with nothing in between then,
1. If your system is $m_1$ then the tension $T_1$ acting on it can be represented by an arrow that has its tail on $m_1$ and points away from $m_1$ along the string.
2. If your system is $m_2$ then the tension $T_2$ acting on it can be represented by an arrow that has its tail on $m_2$ and points away from $m_2$ along the string.
Note: Newton's 3rd law guarantees that the magnitudes are equal, $T_1=T_2$ and their directions opposite.
3. If your system is both $m_1$ and $m_2$, the tension becomes irrelevant to the dynamics because it is an internal force, i.e. the 3rd law action-reaction pair is within the system.

 

[Riman643]

Yes the string would be massless. Each object would have the same forces acting on them differently (the x & y component of gravity would be different for each, friction different for each, etc.). So the net force of the system would be sum of the net force on the first object and the net force of the second object right? And the acceleration of the system would be equal to the acceleration of the objects?

 

[kuruman]

And the acceleration of the system would be equal to the acceleration of the objects?

More precisely the acceleration of the center of mass of the system will be equal to the net force acting on the system divided by the total mass of the system. The term "the acceleration of the objects" is ambiguous. How do you understand it?

 

[Riman643]

The term "the acceleration of the objects" is ambiguous. How do you understand it?

The objects are accelerating at the same speed in this scenario. The net force of the system has an acceleration, would it be equal to the acceleration of the objects?

 

[kuruman]

The objects are accelerating at the same speed in this scenario. The net force of the system has an acceleration, would it be equal to the acceleration of the objects?

The net force on, not of, the system is never equal to the acceleration of the objects. You cannot say that a force is equal to an acceleration. They are two different entities with different dimensions.

If the objects accelerate as one, then the net force (or sum of all the forces) on them is the total mass of the objects times their common acceleration.

 

[Riman643]

The net force on, not of, the system is never equal to the acceleration of the objects. You cannot say that a force is equal to an acceleration. They are two different entities with different dimensions.

If the objects accelerate as one, then the net force (or sum of all the forces) on them is the total mass of the objects times their common acceleration.

The net force on the system must have an acceleration though, right (F = ma)? Would the acceleration of the net force on the system be equal to the acceleration of the objects?

 

[Chestermiller]

Are you familiar with the concept of free body diagrams? Do you feel that you have advanced to the point where you no longer need to use free body diagrams?

 

[Riman643]

Are you familiar with the concept of free body diagrams? Do you feel that you have advanced to the point where you no longer need to use free body diagrams?

Yes I am familiar with free body diagrams. I still need to use them. I am able to make a free body diagram of this situation but it still does not help with my struggles.

 

[Chestermiller]

Yes I am familiar with free body diagrams. I still need to use them. I am able to make a free body diagram of this situation but it still does not help with my struggles.

Try using 2 separate free body diagrams, one for each mass. What are the separate force balance equations that these free body diagrams yield?

 

[kuruman]

The net force on the system must have an acceleration though, right (F = ma)? Would the acceleration of the net force on the system be equal to the acceleration of the objects?

The net force does not "have" an acceleration, perhaps you are struggling as you say because you are looking at this incorrectly. In a nutshell, this is how to think of free body diagrams and what they are used for
1. Choose a system, it can be one mass or several masses.
2. Enclose your system with a dotted line to separate it from the rest of the world.
3. Draw arrows representing the forces that act on one or more components of the system. The entites that exert these forces must be outside the dotted line. For example, if your system is two masses connected with a string, the weight $mg$ of each mass is such a force because the Earth exerts it and is not part of the system. The tension $T$ in the rope is not a force acting on the system because it is part of the system and the system cannot exert a force on itself.
4. Add all the arrows that you have drawn inside the dotted line as vectors. The resultant is a single arrow representing the net force, $F_{net}$. That's the left side of Newton's second law.
5. Multiply each individual mass making up your system with its acceleration and add all such products as vectors to get a resultant. Draw this resultant outside the dotted line because it is not a force acting on the system.
6. Newton's second law asserts that the resultant vector inside the dotted line (the net force vector $\vec F_{net})$ is the same as the resultant mass times acceleration ($\sum_{i} m_i \vec a_i$).

I hope this helps.

 

[Riman643]

Try using 2 separate free body diagrams, one for each mass. What are the separate force balance equations that these free body diagrams yield?

Here are the equations I have for my free body diagrams:

1st Object: Fnet = g_x - f - T
2nd Object: Fnet = g_x + T - f

The Net Force will be the sum of these two forces right?

 

[kuruman]

These equations don't mean much to us without a description of the physical situation and the free body diagrams that accompany it. Can you post these?

Regardless of the pictures, it looks like you have written separate expressions for the net force acting on each object. Assuming that the equations are written correctly, if you were to draw a third free body diagram of the two masses together as your system, then yes the net force on the combined two-mass system will be the sum of the two separate net forces. This is known as "superposition."

 

[Riman643]

These equations don't mean much to us without a description of the physical situation and the free body diagrams that accompany it. Can you post these?

Regardless of the pictures, it looks like you have written separate expressions for the net force acting on each object. Assuming that the equations are written correctly, if you were to draw a third free body diagram of the two masses together as your system, then yes the net force on the combined two-mass system will be the sum of the two separate net forces. This is known as "superposition."

Sweet thanks! I am very certain I have done the equations correctly. So then I could dived the net force of the combined two-mass system by both masses and get the acceleration of the objects correct?

 

[Chestermiller]

Here are the equations I have for my free body diagrams:

1st Object: Fnet = g_x - f - T
2nd Object: Fnet = g_x + T - f

The Net Force will be the sum of these two forces right?

Shouldn't the left-hand sides have ma's? Shouldn't the the right hand sides have mg's? What do you get when you add the two equations, given that the accelerations of the two masses is the same?

 

[Chestermiller]

Please state the exact problem you are trying to solve, word for word.

 

[kuruman]

Sweet thanks! I am very certain I have done the equations correctly. So then I could dived the net force of the combined two-mass system by both masses and get the acceleration of the objects correct?

Before I say "correct" or "incorrect", I echo @Chestermiller's response

Please state the exact problem you are trying to solve, word for word.

 

[Chestermiller]

In a modified Atwood machine, assume the pulley is mass less and the string is mass less. The lower inline is 20 degrees above the horizontal and the upper incline is 8 degrees above the horizontal. the lower object is on the 20 degree plane and has a mass of 12 kg, the upper object is on the 8 degree plane and has a mass of 2 kg. Find the acceleration of the objects and the tension if the objects are sliding down the planes. The kinetic friction on plane to the object is 0.1.

OK. Let’s see your diagram of the apparatus.

 

[kuruman]

Thanks for the diagrams. Here is a key question: Are the strings parallel to their respective inclines or at an angle relative to them as suggested in your figure?

 

[Riman643]

They are parallel to the inclines.

 

[kuruman]

They are parallel to the inclines.

Good, that simplifies things.
Since you are drawing two separate FBDs, you can have separate coordinate axes in each; they don't need to be the same. Draw them so that in each the x-axis is along the incline and the y-axis is perpendicular to the incline. Then only the weight will have x and y components which you will have to figure out. The tension and friction have only an x-component and the normal force only a y-component. The acceleration is along the x-direction only.

The most important thing to realize is that from each free body diagram you will get two equations, one for the x-direction and one for the y-direction which means a total of four equations. So at this point I will say that your equations in post #12 are incorrect. You only show two and you have a force "f" which is the same in both. If (as I suspect) it represents the force of kinetic friction, please note that, because the angles of the two inclines are different, the normal forces are different, therefore the forces of kinetic friction will be different even though the coefficient of kinetic friction is the same.

Consider all this and see if you can make some progress. We are here to help, but please post your work.

 

[Chestermiller]

 

[Chestermiller]

This is the rest of the problem worked out.

I really don't like the way that this was done. It is totally wild-ass (undisciplined).

Here is how I would have set up the problem: Let M be the 12 kg mass and m be the 2 kg mass. Then the force balances parallel to the inclines are:

$$Ma=Mg\sin{20}-Mg(\cos{20})f-T$$
$$ma=T-mg\sin{8}-mg(\cos{8})f$$
where a is the acceleration of M down its incline and also the acceleration of m up its incline.
Note that there is no overall "net force" in these equations. Do these in any way match the equations you wrote (I can't tell because you didn't leave it algebraic)? What do you get if you add these two equations together?

 

[Riman643]

Yes these are my equations I had, I just neglected to put all the algebra in. Adding the forces together I get 29.96 N. I guess net force is bad terminology then for the forces acting on both objects.

 

[Chestermiller]

Yes these are my equations I had, I just neglected to put all the algebra in. Adding the forces together I get 29.96 N. I guess net force is bad terminology then for the forces acting on both objects.

Yes, because the net forces on each of the objects are pointing in different directions, and force is a vector. So if someone taught it to you in this way, they used bad judgment.

 

[Riman643]

Yes, because the net forces on each of the objects are pointing in different directions, and force is a vector. So if someone taught it to you in this way, they used bad judgment.

Thanks. I think I understand now where my confusion was coming from.

 

[kuruman]

$$ma=T-mg\sin{8}-mg(\cos{8})f$$

Easy to miss, but the angle of the upper incline relative to the horizontal is $\theta_2=20^o+8^o$ which affects the components of the weight.

On edit: Wait a minute, what happened to OP's drawing where both inclines were on the same side with the pulley where the slope changes? I am confused.

 

[haruspex]

it looks like you have written separate expressions for the net force acting on each object.

Please use the reply and quote features so that readers know to which text your comment is directed.

 

[haruspex]

Problem Statement: What is the relationship between Tension, Net Force, and Acceleration?

The tension force on the first object be equal to the tension force on the second object?

Not always.

In pulley problems, the usual assumption is that the string does not slip on the pulley. If the question statement refers to a frictionless pulley it means the axle, not the contact with the string.

If the pulley is massless and has no axial friction then the tension must be the same on each side of the pulley; otherwise there would be a net nonzero torque on the pulley, leading to infinite angular acceleration.

If the system is static and there is no axial friction then, again, the two tensions must be equal, even if the pulley has mass.

In all other cases you need to allow two different tensions and should draw a separate FBD for the pulley.

Finally, there is the rope/string. If that has mass and the system is accelerating then the two ends of the same section of rope will have different tensions. Draw an FBD for each section.