# Homework Help: Newton's law

1. Feb 18, 2007

### pivoxa15

1. The problem statement, all variables and given/known data

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.

2. Feb 18, 2007

### DieCommie

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. Feb 18, 2007

### chaoseverlasting

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. Feb 18, 2007

### chaoseverlasting

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. Feb 19, 2007

### pivoxa15

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.

Last edited: Feb 19, 2007
6. Feb 19, 2007

### DieCommie

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. Feb 19, 2007

### pivoxa15

I think I have got this problem covered. It all works out. Look at the diagram with explanation.

#### Attached Files:

• ###### newton's third law.GIF
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8. Feb 19, 2007

### pivoxa15

We are talking about classical mechanics so macroscopic objects are under analysis here.

9. Feb 19, 2007

### DieCommie

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 wouldnt feel it because of the large mass of the station. Now what if you are on a HUGE space station, ie the earth?

10. Feb 19, 2007

### Staff: Mentor

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. Feb 19, 2007

### pivoxa15

Right and that is what I showed on my diagram in my previous post.

12. Feb 19, 2007

### pivoxa15

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. Feb 19, 2007

### DieCommie

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. Feb 19, 2007

### pivoxa15

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. Feb 19, 2007

### DieCommie

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. Feb 19, 2007

### pivoxa15

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. Feb 19, 2007

### DieCommie

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.

Last edited: Feb 19, 2007
18. Feb 19, 2007

### pivoxa15

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. Feb 20, 2007

### DieCommie

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 :grumpy:

I dont 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.

"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 cant 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. Feb 21, 2007

### pivoxa15

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.

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. Feb 21, 2007

### sidrox

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. Feb 21, 2007

### Staff: Mentor

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. Feb 22, 2007

### pivoxa15

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. Feb 22, 2007

### Staff: Mentor

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 wont 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. Feb 23, 2007

### pivoxa15

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.