Is this force diagram correct?

In summary, both the teacher and the car-equipment accelerate at the same rate and the muscles of the arms are not doing work in this case. The cart and the upper body of the teacher are linked by a fixed frame formed by arms-hands-handle-cart-equipment, with a force-reaction balanced pair acting between each of the links. If only analyzing the trolley with external forces of hands and ground friction, the conclusion can be reached that the trolley must be accelerating due to the magnitude of the resultant force being greater than the opposing forces. In the given situation, with the teams and rope accelerating to the right, the tension in the rope is equal to the forces applied by the two teams. However, if we consider the rope
  • #1
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Homework Statement
From this diagram I dont understand why the trolley would accelerate. I understand why the person-trolley system would accelerate from the the force diagram. But just considering the trolley system itself, it seems to have forces balanced on it?
Relevant Equations
Fnet = ma
Trolley should have zero acceleration.
1631756111214.png
 
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  • #2
Both, the teacher and the car-equipment accelerate at the same rate.
The muscles of the arms are not doing work in this case, only the muscles of the legs are.
You can consider the cart and the upper body of the teacher linked by a fixed frame formed by arms-hands-handle-cart-equipment.
There is a force-reaction balanced pair acting between each of the links or members of that frame.
 
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  • #3
Lnewqban said:
Both, the teacher and the car-equipment accelerate at the same rate.
The muscles of the arms are not doing work in this case, only the muscles of the legs are.
You can consider the cart and the upper body of the teacher linked by a fixed frame formed by arms-hands-handle-cart-equipment.
There is a force-reaction balanced pair acting between each of the links or members of that frame.
How would you come to the conclusion that trolley must be accelerating if you were to just analyse the trolley with the hands and ground friction being the external forces?
 
  • #4
pkc111 said:
How would you come to the conclusion that trolley must be accelerating if you were to just analyse the trolley with the hands and ground friction being the external forces?
The magnitude of one force is greater than the opposite one; the magnitude of the resultant force is the difference.
 
  • #5
OK so in the following situation, if the teams and rope were accelerating to the right, is the tension in the rope a reaction force (therefore equal in size but opposite in direction to any of the applied forces)?
1631764569096.png
 
  • #6
pkc111 said:
How would you come to the conclusion that trolley must be accelerating if you were to just analyse the trolley with the hands and ground friction being the external forces?
You have to assume that the teacher pushes the cart so as to keep the cart a constant distance in front. She adjusts Fteacher to keep the two accelerations the same.

I must point out something misleading in the question. It says "the force of friction, which opposes the motion". The main frictional force shown is Ffloor, and that provides the motion.
The force shown as f opposing the motion includes rolling resistance, which is different from friction. What friction there is contributing to f is static friction and is a result of kinetic friction in the wheel axle; it is not kinetic friction between the wheel and the floor.
pkc111 said:
in the following situation, if the teams and rope were accelerating to the right, is the tension in the rope a reaction force (therefore equal in size but opposite in direction to any of the applied forces)?
If we take the rope as massless then there can be no net force on it, and the tension is the same along its length. It is therefore equal to the forces applied by the two teams, so those are equal to each other.
Does that answer your question?
 
  • #7
haruspex said:
You have to assume that the teacher pushes the cart so as to keep the cart a constant distance in front. She adjusts Fteacher to keep the two accelerations the same.

I must point out something misleading in the question. It says "the force of friction, which opposes the motion". The main frictional force shown is Ffloor, and that provides the motion.
The force shown as f opposing the motion includes rolling resistance, which is different from friction. What friction there is contributing to f is static friction and is a result of kinetic friction in the wheel axle; it is not kinetic friction between the wheel and the floor.

If we take the rope as massless then there can be no net force on it, and the tension is the same along its length. It is therefore equal to the forces applied by the two teams, so those are equal to each other.
Does that answer your question?
But they are not equal forces applied by the teams. That is why the question says that the system is accelerating to the right.
 
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  • #8
pkc111 said:
But they are not equal forces applied by the teams. That is why the question says that the system is accelerating to the right.
Don't confuse the forces the teams exert on each other with the force each exerts on the ground.
Draw the free body diagram for one team, showing the horizontal forces on it, and write the Fnet=ma equation.
 
  • #9
My question is about the forces on the accelerating rope and the tension in the rope, its not about the forces on the ground.
So I am not understanding your statement: "If we take the rope as massless then there can be no net force on it, and the tension is the same along its length. It is therefore equal to the forces applied by the two teams, so those are equal to each other".
 
  • #10
pkc111 said:
My question is about the forces on the accelerating rope and the tension in the rope, its not about the forces on the ground.
I understand that, but to explain how it is that the forces the teams exert on each other can be the same even though they accelerate you need to consider the net force on each team (since that is what causes the acceleration).
pkc111 said:
"If we take the rope as massless then there can be no net force on it, and the tension is the same along its length. It is therefore equal to the forces applied by the two teams, so those are equal to each other".
The net force on a massless object is necessarily zero; otherwise it would have infinite acceleration.
Of course, in reality all objects have mass. If we allow the rope some very small mass and it is accelerating then the force on one end will be a little more than the force on the other, but the difference will be small compared to the the two forces (because the rope's mass is small compared to the masses of the teams).
 
  • #11
Again, the forces on the ends of the rope are not at all equal because the rope is accelerating. I am guessing that the size of the tension force is always equal to the size of the force supplied to the rope by the weaker team. Is this considered a reaction force by the rope to the applied force by the weaker team?
 
  • #12
pkc111 said:
Again, the forces on the ends of the rope are not at all equal because the rope is accelerating. I am guessing that the size of the tension force is always equal to the size of the force supplied to the rope by the weaker team. Is this considered a reaction force to the applied force by the weaker team?
For the rope, as for any object, ##F_{net}=ma##. If the rope is massless m=0, so ##F_{net}=0##. Tensions at the ends are equal. The teams exert equal forces on each other.

Why do they accelerate?
Since you refuse to do what I advise is the way to figure it out I'll do it for you:
For a team, as for any object, ##F_{net}=ma##. ##F_{net}## for a team is the difference between the tension in the rope and the force of friction the ground exerts on the team.
Suppose the teams have masses ##m_L, m_R##, that the frictional forces from the ground are ##F_L, F_R## measured away from the rope, the tension in the rope is T, and the teams accelerate to the right at rate ##a##.
##T-F_L=m_La##, ##F_R-T=m_Ra##.
Solve to find ##a##.
 
  • #13
Haruspex, The question is very clear, and repeated a number of times now, is the force of the weaker team on the rope considred a reaction force to the weaker team applying a force to the rope? If you cannot answer that question then that's fine, but please have the decency to let somebody who does know provide an answer, without supplying the extra attitude.
This question was originally posted in the advanced section for a reason.
 
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  • #14
pkc111 said:
is the force of the weaker team on the rope considred a reaction force to the weaker team applying a force to the rope?
You've lost me. How is the force the weaker team applies to the rope different to the force the weaker team applies to the rope?
If you meant "is the force of the weaker team on the rope considred a reaction force to the force the rope applies to the weaker team?" I answered that in the affirmative in post #6:
haruspex said:
[the tension in the rope] is therefore equal to the forces applied by the two teams
That is, the force the tension exerts on one team is equal and opposite to the force that team exerts on the rope; and if the rope is massless those forces are the same for the other team.

But in post #7 you asserted that because there is acceleration the teams must be exerting different forces on the rope (even if it is massless). That is a major misunderstanding and needs discussion.
 
  • #15
Ok let's go back to basics:
Typo from me in last question..is the force of the weaker team on the rope considered a reaction force to the force of the rope on the weaker team? ie are they an action/reaction pair?
 
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  • #16
pkc111 said:
Ok let's go back to basics:
From your statement "How is the force the weaker team applies to the rope different to the force the weaker team applies to the rope?" You seem to be saying that they would necessarily be equal? Is this the action/reaction pair? Is the tension force in the rope at that point one of these forces?
You don't seem to have noticed I wrote the same phrase twice, since that is how I read your question in post #13. I did not write
"How is the force the weaker team applies to the rope different to the force the rope applies to the weaker team?" Those would be an action/reaction pair.
Oh, and I do not think of tension as a force. It is better to think of tension as pairs of action and reaction along the rope.
 
  • #17
OK so the weaker team and rope tension at that point are an action/reaction pair?
 
  • #18
pkc111 said:
OK so the weaker team and rope tension at that point are an action/reaction pair?

Yes, but see my update to post #16.
 
  • #19
OK so is there a similar action/reaction pair (applied force/rope tension) at the stronger team end?
 
  • #20
pkc111 said:
OK so is there a similar action/reaction pair (applied force/rope tension) at the stronger team end?
Of course.
 
  • #21
So I don't undertand how it can be said that the tension is the same throughout the rope (post#6) if they are very different at either end?
I also don't understand how the net force on the rope must be zero just because the mass of the rope is zero (post#6).
 
  • #22
pkc111 said:
So I don't undertand how it can be said that the tension is the same throughout the rope (post#6) if they are very different at either end?
I also don't understand how the net force on the rope must be zero just because the mass of the rope is zero (post#6).
The inviolable formula is ##F_{net}=ma##. It follows immediately that if m is zero then F is zero.
As I wrote in posts #6 and #12, the forces the teams apply to the rope are also equal and opposite. The difference leading to the acceleration is between the forces the teams exert on the ground and consequent difference in the reaction forces from the ground.
Try solving the equations in post #12.
 
  • #23
The inviolable formula (as you put it) is actually a=Fnet/m.
m=0 does not prove Fnet = 0.
To say that any rope accelerates (but has the same force applied to each end) seems very wrong.
 
  • #25
If so, is it saying that the tension in the rope at one end is 240N but 236N at the other?
 
  • #26
pkc111 said:
If so, is it saying that the tension in the rope at one end is 240N but 236N at the other?
As I have written repeatedly (posts #6, #10, #12, #22), the forces at the end of the rope are the same if the rope is taken as massless. As soon as you assign a mass to the rope, acceleration means the forces will be a little different.
 
  • #27
pkc111 said:
The inviolable formula (as you put it) is actually a=Fnet/m.
m=0 does not prove Fnet = 0.
To say that any rope accelerates (but has the same force applied to each end) seems very wrong.

It does imply that, because if m=0 and the net force is not zero then a=F/m implies a is infinite.
It only seems wrong because in practice everything has mass. That's one of the problems with idealisations.
 
  • #28
I have never asked about a massless rope.
A description of the simple situation of the real rope shown above would have been all that was necessary.
Thank you.
 

1. What is a force diagram?

A force diagram is a visual representation of the forces acting on an object. It uses arrows to show the direction and magnitude of each force, and can help to analyze how the forces are balanced or unbalanced.

2. How do I know if my force diagram is correct?

To determine if a force diagram is correct, you should check that all forces acting on the object are included and that their directions and magnitudes are accurate. You should also make sure that the forces are balanced according to Newton's Third Law of Motion.

3. What are some common mistakes in force diagrams?

Some common mistakes in force diagrams include forgetting to include all forces, drawing forces in the wrong direction, and drawing forces with incorrect magnitudes. It is also important to remember that forces should be drawn as vectors, with both direction and magnitude represented.

4. Can a force diagram be used for any object?

Yes, a force diagram can be used for any object, as long as there are forces acting on it. Force diagrams are commonly used in physics and engineering to analyze the motion and stability of objects.

5. How can I improve my force diagram skills?

The best way to improve your force diagram skills is through practice and understanding of the principles of forces and motion. You can also seek guidance from a teacher or tutor, and use online resources and simulations to further develop your understanding.

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