# Object exerts an opposing force of equal magnitude

1. Jul 13, 2006

### Ukitake Jyuushirou

newton's 3rd law stated with every force u exert on an object, the object exerts an opposing force of equal magnitude. if so why is that we're able to move objects around?

2. Jul 13, 2006

### arildno

Because the forces act on DIFFERENT objects.

3. Jul 13, 2006

### andrevdh

Lets consider two cases. In the one case you push a box over the floor and in the other case you push against a wall. Obviously the answer lies in how hard the object can push back at you. If it has not moved when you reached your maximum strength, then you are in trouble, no motion of the object = no motion of you. The answer therefore lies in the system that you are considering (you + box). You need to look at the total force that the combined system are experiencing for motion (according to N2). Remember that the box will match your pushing force against it, but you are also pushing against the floor with your feet, and the floor are pushing back at you with the same force (N3 again). If this force overpowers, or matches, the matching force from the box, then we've got motion of the system. Note that you can decide how hard you want to push the system (you + box) by how hard your legs are pushing against the floor, thereby determining how quickly the system will move.

Last edited: Jul 13, 2006
4. Jul 13, 2006

### Ukitake Jyuushirou

wouldnt the force N3 generated from the feet pushing against the floor be opposed by the box?

5. Jul 13, 2006

### HallsofIvy

Staff Emeritus
No, the box can't "apply" more force than the friction force on it. If you subtract the friction force from the force N3 you are applying to the box, the result is the "net" force that causes acceleration. Once again, when you press against the box with force N3, you feel the box pressing back against your hands with force N3 but part of that is friction force, the rest ma where a is the acceleration of the box. Your hands pushing on the box and the box are the "equal and opposite" action and reaction, but part of that reaction is due to the boxes acceleration.

6. Jul 13, 2006

### Ukitake Jyuushirou

so when i push a box with my hands , the box also exerts an equal and opposing force but the reason why the box will move is becoz my feet is also exerting a force downwards on the earth that pushes me back and that force is transmitted to the box and over comes the friction and becomes the net force...am i rite?

while we're at it...

can i find the coefficient of friction if i am given the normal force and the force that is directed horizontally on the object?

assuming normal force is 236N and the force directed horizontally is 69.2 N

7. Jul 13, 2006

### arunbg

If force is to be maintained, you need to certainly use the friction to prevent you from going backward, while pushing the box forward.

If you are talking about the coefficient of static friction, then the force directed horizontally must be the force applied just before the object starts to slide.
Use $F=\mu_sN$

If you are talking about coefficient of kinetic friction, then you also need to take acceleration of the mass into consideration.

___________________

Arun

Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.
~ Albert Einstein

8. Jul 13, 2006

### arildno

No. The box moves forward because there is a net force acting upon IT.
The box couldn't care less what force acts upon your feet, or if indeed, you're wearing sneakers.

9. Jul 13, 2006

### Staff: Mentor

"N3" is andrevdh's shorthand for "Newton's 3rd Law"; it's not identifying a specific force.

If the feet push against the floor, then the floor pushes back against the feet (not on the box directly). That's what N3 tells you. (Of course, the floor pushing against your feet allows you to push harder against the box. Lot's of luck trying to push the box if you are standing on ice.)

10. Jul 14, 2006

### andrevdh

Lets consider another case. Two identical blocks resting on a frictionless surface. You push the block closest to you with your finger resulting in the blocks moving forwards together. The resultant force that the blocks are experiencing comes from your finger pushing them. The action-reaction forces in between the two blocks are equal and are not contributing to the motion of the blocks. They are called internal forces. The same can be said regarding the action-reaction pair of forces in your question. In the case of the two blocks the force coming from your finger is like the force pushing against your feet in my other example.

11. Jul 14, 2006

### andrevdh

N3 refers to Newton's third law. When you push the box it pushes back against you with a force of same magnitude (according to N3). Your feet pushing against the floor will generate another force by the floor pushing against you (N3 again). The reaction force from the box and the floor are not the same forces, they can differ in magnitude depending on how hard you push against the floor. The example in my previous post about two similar blocks clarifies the situation better.

You can see that the force from the finger and the force that the front block experiences will be different in magnitude using N2 in the situation. The force from your finger have to accelerate both blocks, while the interaction force that the block in the front experiences just have to accelerate that block alone.

Last edited: Jul 14, 2006
12. Jul 14, 2006

### Ukitake Jyuushirou

erm pardon my poor understanding of physics

so as long as the net force generated cancels the friction force, the box will move. as with all pushing motion a force is directed at the box and another force from my feet generated downwards as long as they overcome friction force, the box will move?...

13. Jul 14, 2006

### Staff: Mentor

Keep it simple: As long as the net force on the box is nonzero, the box will accelerate. If your (horizontal) pushing force on the box exceeds the static friction force of the floor on the box, then there will be a net force on the box and it will move.

14. Jul 14, 2006

### Ukitake Jyuushirou

ok i'd ignore the messier bit regarding the force generated from the feet. thanks :)

15. Jul 14, 2006

### Ukitake Jyuushirou

once the force applied to the box is greater than the static friction, i can safely ignore the static friction force right? since at this point only the kinetic friction comes into play, but if given the 2 forces and i am told to find co efficient of kinetic friction, how will i do so?

16. Jul 14, 2006

### Staff: Mentor

I'm not sure what you are asking. If the box is sitting still and you push it, static friction will match your force to keep the box from sliding--up to the point where the static friction is at its maximum value, which is $\mu_s N$. As you state, once it starts moving, kinetic friction takes over; that force equals $\mu_k N$.

How you would find $\mu_k$ depends on what you know. If you know the acceleration, you can apply Newton's 2nd law to find the friction force, and from that deduce $\mu_k$.

17. Jul 14, 2006

### Ukitake Jyuushirou

the question is actually: a 20kg sled is pulled along a horizontal surface at constant velocity. The pulling force has a magnitude of 80N and is directed at an angle of 30 degrees above the horizontal, determine the coefficient of kinetic friction

using kinematics i find the force in the positive y-component is 236N and the force of gravity is 196N. a horizontal force along the positive x-component is 69.2N , here is where i got lost

18. Jul 14, 2006

### Staff: Mentor

Let's do it step by step. First identify all the forces acting on the sled. I count four:
(1) Pulling force (what are the x & y components?)
(2) weight (what are the x & y components?)
(3) normal force (what are the x & y components?)
(4) kinetic friction (what are the x & y components?)

Note: The direction of the forces (and thus the sign of the component) counts.

19. Jul 14, 2006

### Ukitake Jyuushirou

pulling force

x = +69.2
y = +40

weight = -196

normal = +196

kinetic friction = ??

20. Jul 14, 2006

### Staff: Mentor

OK

OK (what component?)

How did you get this? (Apply equilibrium conditions.)

Apply equilibrium conditions.

The first two forces are given; the last two you solve for by applying the condition for equilibrium: The sum of the forces in any direction must be zero.