Why Do Objects Accelerate Despite Equal and Opposite Forces?

[zeromodz]

If Newton's law states that every action has an opposite and equal reaction, why would an object accelerate? Why does the net force not always equal zero, meaning no acceleration? For instance, a man pushes a cart with a force of 10N. According to Newton's law, the cart pushes back with 10N. So how would the system accelerate?

 

[Galileo]

The crucial thing to realize is that the pair forces in Newton's third law act on different bodies, unlike in the second law which relates the net force and acceleration of a single body.

So the cart will move simple because it is pushed. The man will be pushed back the other way if he doesn't counteract that with his feet on the floor.

 

[zeromodz]

The crucial thing to realize in that the pair forces in Newton's third law act on different bodies, unlike in the second law which relates the new force and acceleration of a single body.

So the cart will move simple because it is pushed. The man will be pushed back the other way if he doesn't counteract that with his feet on the floor.

Duhhh!

Sorry, this was a stupid question. Moderators may lock it now.

 

[BL4CKCR4Y0NS]

The man will be pushed back the other way if he doesn't counteract that with his feet on the floor.

I'm sorry... I don't understand this line... could you explain it another way, please?

 

[TurtleMeister]

Sorry, this was a stupid question. Moderators may lock it now.

No, it's not a stupid question. Lot's of people have trouble understanding Newton's laws. This forum is here to help all of us understand physics better.

I'm sorry... I don't understand this line... could you explain it another way, please?

If the man were standing on a frictionless surface, such as ice, then the man would move in the opposite direction of the cart. But if the mans feet do not slip, then the Earth itself will move in the opposite direction of the cart. But because the Earth is so massive, it's resulting movement cannot be detected. So the forces are equal in magnitude and opposite in direction. Likewise, the resulting motions (of the cart and the man/earth) are equal in momentum and opposite in direction.

 

[meichenl]

The system as a whole will accelerate when there's a force from outside the entire system.

Several other posters pointed out that in the case of the man and cart, the man pushes on the ground. Let's say the man pushes on the cart with 10N force. The cart pushes back on the man equally and opposite. The man pushes against the ground with say 9N force the other direction, and the ground pushes back equally and opposite.

There are two forces on the man - 10N and 9N - acting in opposite directions, so the man accelerates backwards just a little.

The cart has one force on it of 10N, so it accelerates forwards.

Overall the center of mass of the entire man-and-cart system accelerates forward at [strike]10[/strike]9N/(total mass). (edited after TurtleMeister's reply)

Also, because acceleration is a vector, different parts of the system can have individual accelerations that sum to zero. When a rocket in space puts a force on its fuel, accelerating it away from the rocket, the fuel puts an equal and opposite force on the rocket. The net force on the system of rocket + fuel is zero, and there's no net acceleration because the backwards acceleration of the fuel cancels the forward acceleration of the rocket, but each part considered individually is still accelerating.

 

[TurtleMeister]

There are two forces on the man - 10N and 9N - acting in opposite directions, so the man accelerates backwards just a little.

The cart has one force on it of 10N, so it accelerates forwards.

Are you assuming that the man releases the cart after pushing it? How do you get 9N? Shouldn't it be 10N for the cart and -10N for the man/earth?

Overall the center of mass of the entire man-and-cart system accelerates forward at 10N/(total mass).

Here you seem to be indicating that the man stays with the cart. Maybe he jumps onto the cart after pushing it? If that is the case then wouldn't it be 10N for the man/cart and -10N for the earth?

 

[meichenl]

Are you assuming that the man releases the cart after pushing it? How do you get 9N? Shouldn't it be 10N for the cart and -10N for the man/earth?

There was a mistake in my previous post. It said the entire system accelerates at 10N/total mass, and that should have been 9N/total mass.

In my original scenario, the man and cart would go opposite directions. Then man would be pushed slightly backwards while the cart went forwards. I was not trying to indicate that the man stays with the cart. The center of mass is still well-defined, even for objects separated by a large distance. Eventually, once the cart got more than arm's length away, the man wouldn't be able to push on it any more. Then the accelerations would have to change, or the man would need to start pushing with a pole.

The man can push on the cart with however much force he likes, and can push on the Earth with however much force he likes. If we imagine a different scenario in which the man is going accelerate the same way as the cart, keeping his distance from the cart fixed, then he'll need to push on the ground harder than he pushes on the cart. Specifically, if the man and cart are going to accelerate at rate [itex]a[/itex], then the man needs to push on the ground with force [itex]F_{ground} = a*(m_{man} + m_{cart})[/itex] and push on the cart with force [itex]F_{cart} = a*m_{cart}[/itex].

Then the cart has one force on it, which is just [itex]a*m_{cart}[/itex], so it accelerates forwards at rate [itex]a[/itex]. The man has two forces on him. A forward force from the ground (which is the reaction to him pushing against the ground) whose magnitude is [itex]a*(m_{man} + m_{cart})[/itex], and a backward force that is the reaction force from him pushing on the cart with magnitude [itex]a*m_{cart}[/itex]. Subtracting these because they push in opposite directions, the total force on the man is [itex]a*m_{man}[/itex] in the forward direction, so the man also accelerates forward at acceleration [itex]a[/itex].

For a concrete example, suppose the cart's mass is half as much as the man's. Then the man can push on the cart with 10N and push on the ground the opposite direction with 30 N. The total force on the cart will be 10N, and the total force on the man will be 30N - 10N = 20N. There is twice as much net force on the man, but since his mass is twice as much, the man and cart have the same acceleration.

 

[TurtleMeister]

The center of mass is still well-defined, even for objects separated by a large distance.

It doesn't seem well defined for me (for the scenario you are describing). Center of mass of what? The man and cart? The man and earth? The man, cart, and earth? Edit: If the man stays with the cart then I'm assuming the center of mass should be between the man/cart and the earth.

The man can push on the cart with however much force he likes, and can push on the Earth with however much force he likes.

[strike]But they must be equal and opposite. If he pushes on the cart with 10N then he is also pushing on the Earth with -10N.[/strike] See post #13

If we imagine a different scenario in which the man is going accelerate the same way as the cart, keeping his distance from the cart fixed, then he'll need to push on the ground harder than he pushes on the cart.

That goes against what I just stated. So one of us is wrong. See post #13

Specifically, if the man and cart are going to accelerate at rate [tex]a[/tex] , then the man needs to push on the ground with force [tex] F_{ground} = a*(m_{man} + m_{cart})[/tex] and push on the cart with force [tex] F_{cart} = a*m_{cart}[/tex] .

If the man stays with the cart and accelerates with the cart then you will have two different accelerations. The accleration of the man/cart [tex]a_{mc}[/tex] and the acceleration of the Earth [tex]a_e[/tex]. [tex]F_{mc} = m_{mc} * a_{mc}[/tex] AND [tex]F_{e} = -m_e * a_e[/tex] SO [tex]F_{mc}=-F_{e}[/tex]

Then the cart has one force on it, which is just [tex] a*m_{cart}[/tex] , so it accelerates forwards at rate [tex]a[/tex]. The man has two forces on him. A forward force from the ground (which is the reaction to him pushing against the ground) whose magnitude is [tex]a*(m_{man} + m_{cart})[/tex], and a backward force that is the reaction force from him pushing on the cart with magnitude [tex]a*m_{cart}[/tex]. Subtracting these because they push in opposite directions, the total force on the man is [tex]a*m_{man}[/tex] in the forward direction, so the man also accelerates forward at acceleration [tex]a[/tex] .

[strike]The man/cart are the action and the Earth is the reaction (or vis versa, it doesn't matter). The net force on the man while he is pushing is zero. 10N from the cart, and -10N from the earth.[/strike] See post #13

For a concrete example, suppose the cart's mass is half as much as the man's. Then the man can push on the cart with 10N and push on the ground the opposite direction with 30 N. The total force on the cart will be 10N, and the total force on the man will be 30N - 10N = 20N. There is twice as much net force on the man, but since his mass is twice as much, the man and cart have the same acceleration.

How do you arrive at 30N?

Edit:
What's making this problem difficult is that we are considering three different objects instead of two. While the man is pushing he is both a part of the Earth and a part of the cart. When he finishes his push he must retract a body part to become complete with one or the other.

 

[meichenl]

In previous posts, when I referred to the center of mass, I was talking about the center of mass of the man + cart system.

There are two conceptual hurdles to this problem that seem to be troubling you.

The first is that we can be very flexible about defining our system when using Newton's second law. You can make the system whatever you like, in fact. It's still true that the net external force on the system is equal to the system's mass times the acceleration of the system's center of mass. So the net force on the cart is equal to its mass times its acceleration, and the net force on the man is equal to the mass of the man times his acceleration. In addition to both of those, the net force on the man and cart together is equal to the mass of the man and cart together times the acceleration of the center of mass of the man/cart system. This even applies for the man/earth/cart system, or the earth/man system, or the earth/cart system, or the Earth alone. You can even divide the man up into little parts and consider them as individual systems and do your calculations with the forces of the various parts of the man on each other. You can slice it up however you want, and in my previous posts I took advantage of this by sometimes talking about the man and cart individually, and sometimes talking about them as one system.

An interesting point about this is that it works precisely because of Newton's Third Law. When you go from considering the man and cart separately to considering them as one system, the forces between them go from being external forces (which do count in Newton's Second Law) to internal forces (which don't). If Newton's Third Law didn't hold, then those internal forces would be non-zero, and the scheme would be logically inconsistent. We could calculate the acceleration of the center of mass of the man/cart system by finding their individual accelerations and then adding them vectorially (weighted by the masses), or we could do it by finding the total external force on the man/cart system. If Newton's third law didn't hold, those methods might give different answers.

The second conceptual difficulty confronting you is recognizing when the third law does and does not apply. The third law describes the relationship between the force of A on B and the force of B on A. For example, the force of the cart on the man and the force of the man on the cart are an action/reaction pair under the third law. However, the force of the cart on the man is not necessarily equal and opposite to the force of the ground on the man. This is not a third law pair because there are three different agents involved - the man, the cart, and the ground. That's why, if the man pushes on the cart with 10N, the cart pushes back with 10N, but then the man can choose to push on the ground with however much force he chooses. If he chooses 0N, he'll be pushed backwards and the center of mass of the man/cart system will be stationary. If he chooses to push back on the ground with 10N, he'll stay put himself and the cart will accelerate away. If he pushes with 30N (assuming he is twice as massive as the cart) he will accelerate the same way as the cart.

In answer to your final question, I got the 30N figure by intuition and then showed it worked in equations in the previous post.

 

[TurtleMeister]

Sorry meichenl, your answer of "intuition" does not help me. I guess we'll just have to disagree. Maybe a homework helper or science adviser can straighten this out.

 

[meichenl]

Physics is not a matter of opinion.

I am qualified to answer this question since I have a physics degree from a top school and have taught physics as a private tutor, teaching assistant, and classroom instructor for five years. I spent considerable effort writing that reply, and your flippant summary, implying that all I said was that I used intuition, is inaccurate and disrespectful. If you reread the post, you'll see that I spelled out all the thought processes behind the answer. The intuition was only used to come up with the number; I then actually did the calculation to see that the answer worked. Further, I provided the general formula for this number in a previous post.

No one can actively put ideas in your head. If you continue to assume others are explaining poorly and shifting the blame for your lack of comprehension to external sources, you're handicapping your own learning ability.

 

[TurtleMeister]

Corrections:

But they must be equal and opposite. If he pushes on the cart with 10N then he is also pushing on the Earth with -10N.

This would only be true if the man had zero mass.

If we imagine a different scenario in which the man is going accelerate the same way as the cart, keeping his distance from the cart fixed, then he'll need to push on the ground harder than he pushes on the cart.

That goes against what I just stated. So one of us is wrong.

I was wrong. Since the man has mass, the force between the man and the Earth must be greater than the force between the man and the cart.

The man/cart are the action and the Earth is the reaction (or vis versa, it doesn't matter). The net force on the man while he is pushing is zero. 10N from the cart, and -10N from the earth.

The net force on the man will be the force between the man and the Earth minus the force between the man and the cart.

If the man stays with the cart and accelerates with the cart then you will have two different accelerations. The accleration of the man/cart [tex]a_{mc}[/tex] and the acceleration of the Earth [tex]a_e[/tex]. [tex]F_{mc} = m_{mc} * a_{mc}[/tex] AND [tex]F_{e} = -m_e * a_e[/tex] SO [tex]F_{mc}=-F_{e}[/tex]

But I did get that one right. :)
Not that your comment was wrong. I was just looking at it in a different way.

meichenl, I must apologize for my quick response. I should have given this more thought. The introduction of a third body confused me.

 

[meichenl]

hey Turtle,

I was snappier than the situation really called for, as well. Glad we're on the same page now.