Internal Forces and Acceleration: How Can We Stay with a Moving Block?

In summary, the conversation discusses the concept of action and reaction forces, specifically in regards to pushing a block with constant force. It is explained that when the block accelerates, the hand pushing it must also accelerate, and therefore there must be a net force on the hand from the rest of the body. The conversation also touches on the idea of where to draw the line between one body and another when considering action and reaction forces.
  • #1
pivoxa15
2,255
1

Homework Statement



Assume no friction.

Suppose I exert a force of 1N to a block. The block will exert 1N of force back at me. If I keep exerting the 1N than there is a constant force of 1N exerted back at me so my hand is in constant velocity? But the block I am exerting the force is accelerating forwards. So unless if I provide more than the force recquired to accelerate the block by a certain amount, the block will get in front of me. This doesn't make sense. The more force I provide, the more force the block will exert back at me. In reality I am able to push a block and stay with it. How?

I made this problem up so will not use the template.
 
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  • #2
Hmmm, think about your action reaction pair.

If you and the block are "together", that is the block is static on your hand, then that is one object. Your pair of objects cannot be the same object.

If you are pushing a block, what is the second object for the action reaction pair?
 
  • #3
When you push the block with constant force, how can your velocity be constant? All of this is wrt ground, therefore, you too must accelerate. Wrt the block, youre still.
 
  • #4
Exactly as DC said, youre one extended object. Its like the motors on a rocket, they exert the force, but the fuselage also moves. There is no relative acceleration between them.
 
  • #5
You don't account for the reaction force exerted by the object to your hand? The forces are, to the object there is a forward force exerted by my hand to the object. To my hand, there is a backward force exerted by the object to my hand. This means my hand should move backwards and the object forwards. But in reailty I can stay with the object all the way. There must be something wrong with my theory.
 
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  • #6
If that were the case, why choose the atoms of the object and the atoms of the hand to be the action/reaction pair? You could choose the atoms of your hand and the atoms of your arm... Or the atoms of one slice of the object on the atoms of another...
 
  • #7
I think I have got this problem covered. It all works out. Look at the diagram with explanation.
 

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  • #8
DieCommie said:
If that were the case, why choose the atoms of the object and the atoms of the hand to be the action/reaction pair? You could choose the atoms of your hand and the atoms of your arm... Or the atoms of one slice of the object on the atoms of another...

We are talking about classical mechanics so macroscopic objects are under analysis here.
 
  • #9
Yes I know that...I shouldn't have used to word atoms... I am asking where do you draw the line between one body and another.

You choose to draw the line between your object and your hand. Why not your hand and your arm? Or one slice of the object on another.

When you push the object, the object, your hand and forearm are all moving but your body is not. If you were in space and pushed the object, then the object your hand and forearm would still be pushing back on you. But you would feel this force in space. If you were on the space satation, then the force of the object, your hand and forearm would still be applied to you, but you wouldn't feel it because of the large mass of the station. Now what if you are on a HUGE space station, ie the earth?
 
  • #10
pivoxa15 said:
Assume no friction.

Suppose I exert a force of 1N to a block. The block will exert 1N of force back at me. If I keep exerting the 1N than there is a constant force of 1N exerted back at me so my hand is in constant velocity? But the block I am exerting the force is accelerating forwards. So unless if I provide more than the force recquired to accelerate the block by a certain amount, the block will get in front of me. This doesn't make sense. The more force I provide, the more force the block will exert back at me. In reality I am able to push a block and stay with it. How?
If the block is accelerating and your hand is keeping up with it, then your hand is accelerating also. Thus there must be a net force on your hand. The rest of your body exerts a force on your hand, and the block you are pushing pushes back on your hand. The rest of your body must be exerting a greater force on your hand than your hand is exerting on the block.
 
  • #11
Doc Al said:
If the block is accelerating and your hand is keeping up with it, then your hand is accelerating also. Thus there must be a net force on your hand. The rest of your body exerts a force on your hand, and the block you are pushing pushes back on your hand. The rest of your body must be exerting a greater force on your hand than your hand is exerting on the block.

Right and that is what I showed on my diagram in my previous post.
 
  • #12
DieCommie said:
Yes I know that...I shouldn't have used to word atoms... I am asking where do you draw the line between one body and another.

You choose to draw the line between your object and your hand. Why not your hand and your arm? Or one slice of the object on another.

When you push the object, the object, your hand and forearm are all moving but your body is not. If you were in space and pushed the object, then the object your hand and forearm would still be pushing back on you. But you would feel this force in space. If you were on the space satation, then the force of the object, your hand and forearm would still be applied to you, but you wouldn't feel it because of the large mass of the station. Now what if you are on a HUGE space station, ie the earth?

I could have done that with (a change on my diagram) the hand being the arm and object being the hand. The arm would still have to exert more force than the hand exerts a force back, if the arm and hand were to accelerate together.
 
  • #13
The object, your hand, and anything else that is attached and accelerating with it is only one object. What is the other object? The other object is that which is not accelerating with your object, hand and arm.
 
  • #14
DieCommie said:
The object, your hand, and anything else that is attached and accelerating with it is only one object. What is the other object? The other object is that which is not accelerating with your object, hand and arm.

I could model them as separate object. Didn't you question "You choose to draw the line between your object and your hand. Why not your hand and your arm? Or one slice of the object on another."

I am claiming that it's possible to choose differently but Newton's mechanics will work everytime although the situation may be very removed from reality. For example, the hand as a lifeless object that only moves when the arm exerts a force to it.
 
  • #15
My question was not to suggest that you treat those as separate objects. It was to make you think about why you choose the action/reaction pair that you did. You choose the wrong action/reaction pair.

If the hand is a lifeless object that only moves when the arm exerts a force... then the arm is moving too. Both the arm, and the hand are one of the action/reaction pair. What is the other? That which is not accelerating with your object, hand and arm. What would that be? It would be your shoulder, your body, the house you are in and the Earth it is on. That is the other object. So you have one object, your hand and arm, and another, your body and the earth. The forces on these two objects are equal and opposite.
 
  • #16
DieCommie said:
My question was not to suggest that you treat those as separate objects. It was to make you think about why you choose the action/reaction pair that you did. You choose the wrong action/reaction pair.

If the hand is a lifeless object that only moves when the arm exerts a force... then the arm is moving too. Both the arm, and the hand are one of the action/reaction pair. What is the other? That which is not accelerating with your object, hand and arm. What would that be? It would be your shoulder, your body, the house you are in and the Earth it is on. That is the other object. So you have one object, your hand and arm, and another, your body and the earth. The forces on these two objects are equal and opposite.

What is the correct action/reaction pair? I can choose any action/reaction pair I want couldn't I? This arbitary choice was suggested by you with "You choose to draw the line between your object and your hand. Why not your hand and your arm? Or one slice of the object on another." and I agree with it. But you are now disagreeing with your own point?

I agree that the Earth and body is another action/reaction pair. I didn't consider it in my original question but you could put it in for more completeness.
 
  • #17
Please try to understand what I said and you quoted.

"My question was not to suggest that you treat those as separate objects. It was to make you think about why you choose the action/reaction pair that you did. You choose the wrong action/reaction pair."

I was simply trying to make you think about your action/reaction pair choice. You made an arbitrary choice that was wrong, because it seems right that one object is your self and the other is that which you are pushing. I asked why you didnt choose these other pairs instead, which you never answered. It was my hope that you would ask yourself why you didnt choose these pairs and come to the realization that they are not correct pairs. You cannot choose any pair you want.

I am obviously no good at helping at this, so hopefully someone else will help you.
 
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  • #18
DieCommie said:
Please try to understand what I said and you quoted.

"My question was not to suggest that you treat those as separate objects. It was to make you think about why you choose the action/reaction pair that you did. You choose the wrong action/reaction pair."

I was simply trying to make you think about your action/reaction pair choice. You made an arbitrary choice that was wrong, because it seems right that one object is your self and the other is that which you are pushing. I asked why you didnt choose these other pairs instead, which you never answered. It was my hope that you would ask yourself why you didnt choose these pairs and come to the realization that they are not correct pairs. You cannot choose any pair you want.

I am obviously no good at helping at this, so hopefully someone else will help you.

I chose the pair I did in the OP because I didn't fully understand how they worked. However, I drew a diagram which hopefully showed that it all works. What do you think about the information in my diagram? Do you think my diagram is incorrect? Pairs like earth-hand, I always understood so I didn't consider it.

Why can't I choose any pair I want? Could you answer this question. Your help is appreciated but I may not have fully understood you.
 
  • #19
I have done a very poor job of explaining this and am sorry. As a person who tutors for a living, I am ashamed! But I am certainly no expert


I don't follow your diagram after looking at it carefully... the box on the bottom right... 20N applied, this is always applied. When you apply the 20 to the 1kg box, you get 20m/s^2 acceleration. At this time the box is pushing whatever is pushing it back with 20N force. This is likely to be the whole earth, as the pushing mechanism is attached to the earth. Then the box strikes the 3kg box, and they become a 4kg box. They are at the same box essentially, as they are not accelerating with respect to each other (only for a moment they did). Now they still have 20N of force applied, so they are accelerating at 5m/s^2. If the frictional force is low and they accelerate beyond your hand, so be it. Then they lose there source of force, and continue on losing whatever velocity to the friction.

Now your original post says
"If I keep exerting the 1N than there is a constant force of 1N exerted back at me so my hand is in constant velocity?" They only have a constant velocity if your hands force equals the friction force (which is easy to do with your hand)

"But the block I am exerting the force is accelerating forwards." Only if the friction is low enough so it can't overcome your 20N of force. If this happens they will gain velocity, and may overtake your hand as described above.

"The more force I provide, the more force the block will exert back at me. " Ok, so if you push it hard, and it flies away from you hand, it will have pushed on you harder too. So the force your body/earth feels would be more.

"In reality I am able to push a block and stay with it. How?" You are not pushing it hard enough to overcome friction. There is still a force against your body/earth, but this is smaller than the case above.
 
  • #20
DieCommie said:
I don't follow your diagram after looking at it carefully... the box on the bottom right... 20N applied, this is always applied.

Maybe my diagram isn't clear but I stated that the 1kg object is the hand itself, not literally a lifeless box. So the 1kg 'box'=hand is actually providing the 20N force. There is no force exerted onto the hand. So the 20N is the force pushing the hand as well as the 3kg real box.

DieCommie said:
When you apply the 20 to the 1kg box, you get 20m/s^2 acceleration. At this time the box is pushing whatever is pushing it back with 20N force.

If the 1kg box=hand was not pushing any other boxes than the acceleration would be 20m/s^2 but it is pushing another box so the acceleration itself will not be 20m/s^2. Both the hand and box must accelerate at the same acceleration which was the original condition. And if we do an experiment and use my hand to push a box, my hand can accelerate with the box.
 
  • #21
i think that one confusion that needs to be cleared right away is that "your hand is in constant velocity because of the constant 1 N that is applied."

F= ma

The acceleration is constant, NOT the velocity...
 
  • #22
pivoxa15 said:
Maybe my diagram isn't clear but I stated that the 1kg object is the hand itself, not literally a lifeless box. So the 1kg 'box'=hand is actually providing the 20N force. There is no force exerted onto the hand. So the 20N is the force pushing the hand as well as the 3kg real box.
An object, your hand included, cannot "push itself"--it requires an external force to accelerate. If it's accelerating along with the box, then a net force is acting on it:

Forces on box: The hand pushes the box with 15 N of force; the 3 kg box accelerates at 5 m/s^2.

Forces on hand: The box pushes the hand with 15 N of force to the left; Whatever is attached to the hand (your arm/body) pushes the hand with 20 N of force to the right. The net force on the hand is 5 N to the right--it accelerates at 5 m/s^2.
 
  • #23
Doc Al said:
An object, your hand included, cannot "push itself"--it requires an external force to accelerate. If it's accelerating along with the box, then a net force is acting on it:

Forces on box: The hand pushes the box with 15 N of force; the 3 kg box accelerates at 5 m/s^2.

Forces on hand: The box pushes the hand with 15 N of force to the left; Whatever is attached to the hand (your arm/body) pushes the hand with 20 N of force to the right. The net force on the hand is 5 N to the right--it accelerates at 5 m/s^2.

I specified on my digram that inside the hand, chemical energy is converted into mechanical energy so the hand is pushing itself that way. You don't think it's good? You think it is better to model the situation as an external force pushing the hand? When the external force came from within the hand itself.
 
  • #24
pivoxa15 said:
I specified on my digram that inside the hand, chemical energy is converted into mechanical energy so the hand is pushing itself that way. You don't think it's good? You think it is better to model the situation as an external force pushing the hand? When the external force came from within the hand itself.
External force cannot come from within an object. If you are concerned with the acceleration of an object, you must consider all forces on that object. A hand is no different.

Think of a car. You can step on the gas all you want (converting chemical energy to mechanical energy), but you won't go anywhere without an external force--the friction of the road pushing on the tires. An external force is required to accelerate the car, just like an external force is required to accelerate your hand.
 
  • #25
Doc Al said:
External force cannot come from within an object. If you are concerned with the acceleration of an object, you must consider all forces on that object. A hand is no different.

Think of a car. You can step on the gas all you want (converting chemical energy to mechanical energy), but you won't go anywhere without an external force--the friction of the road pushing on the tires. An external force is required to accelerate the car, just like an external force is required to accelerate your hand.

I see that an external force is recquired to move a car (which is from the road in the form of friction). But what is the external force that is recquired to move a hand? Force from the arms? What will move the arms, force from the shoulders? What will move the shoulders? Force from the body? What will move the body? ...

In this situation it seems like an infinite descent where does it end? However, with the car example, things are simple, force of friction via the road is exerted to the car to make it move forwards. This friction force comes as the result of rubbing the tyres with the road. The action-reaction pair is straight forward in this case.
 
  • #26
pivoxa15 said:
I see that an external force is recquired to move a car (which is from the road in the form of friction). But what is the external force that is recquired to move a hand? Force from the arms? What will move the arms, force from the shoulders? What will move the shoulders? Force from the body? What will move the body? ...

In this situation it seems like an infinite descent where does it end?
Yet, if you want your hand to accelerate there must be external forces on it. The net force must be such to give it the desired acceleration. Yes, the wrist pushes on the hand!

It might seem an infinite regress, but it's not. At some point, you are pushing against the earth--that's as far as we need to go. If the acceleration you wish to create is small enough, and you only need it for a brief while, your body may itself have enough mass to push-off against.
However, with the car example, things are simple, force of friction via the road is exerted to the car to make it move forwards. This friction force comes as the result of rubbing the tyres with the road. The action-reaction pair is straight forward in this case.
"Action-reaction" pairs are always straight forward. Your hand pushes the block, the block pushes your hand; your wrist/arm/shoulder pushes your hand, your hand pushes your wrist/arm/shoulder.

To analyze the motion of the block, all you need are the forces on the block. But if you want to analyze the motion of the hand, you'd better include all the forces on the hand.
 
  • #27
Doc Al said:
Yet, if you want your hand to accelerate there must be external forces on it. The net force must be such to give it the desired acceleration. Yes, the wrist pushes on the hand!

It might seem an infinite regress, but it's not. At some point, you are pushing against the earth--that's as far as we need to go. If the acceleration you wish to create is small enough, and you only need it for a brief while, your body may itself have enough mass to push-off against.

I could still accelerate my bodily parts in free fall in space, without touching anything else. In that case would you say, a part of my body i.e. arms is pushed forwards and the body experiences a force backwards as a reaction.
 
  • #28
pivoxa15 said:
I could still accelerate my bodily parts in free fall in space, without touching anything else.
Absolutely! For example, you can accelerate--briefly--your arm with respect to your body. That means you can exert forces between one body part and another. But you cannot accelerate your center of mass--that requires an external force from something outside your body.
In that case would you say, a part of my body i.e. arms is pushed forwards and the body experiences a force backwards as a reaction.
Yes! Your body pushes your arms forward; your arms push your body backward.

But in all cases, whatever is being accelerated requires an external force. (Your arm can't accelerate itself; something external to your arm--your torso, perhaps--must push on it.)
 

Related to Internal Forces and Acceleration: How Can We Stay with a Moving Block?

1. What is Newton's law of block force?

Newton's law of block force states that the force exerted on a block is directly proportional to its mass and acceleration. This means that the larger the mass of the block and the faster it accelerates, the greater the force applied to it.

2. How does Newton's law of block force apply to everyday life?

Newton's law of block force applies to everyday life in many ways, such as when pushing or pulling objects, driving a car, or playing sports. It helps us understand the relationship between mass, acceleration, and force in various situations.

3. What are the three laws of motion that make up Newton's law of block force?

The three laws of motion that make up Newton's law of block force are: 1) an object at rest will remain at rest, and an object in motion will remain in motion at a constant velocity unless acted upon by an external force, 2) force equals mass times acceleration, and 3) for every action, there is an equal and opposite reaction.

4. How did Newton come up with his law of block force?

Newton's law of block force was discovered by Sir Isaac Newton in the late 17th century. He developed this law based on his observations of objects in motion, as well as his understanding of calculus and other mathematical principles.

5. Can Newton's law of block force be applied to non-block objects?

Yes, Newton's law of block force can be applied to any object, not just blocks. As long as an object has mass and is experiencing acceleration, this law can be used to calculate the force acting on it.

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