Newton's Third Law of pushing a box

[spidey64]

Ok, I'm getting a lot of mixed messages about the reactionary forces involved with the 3rd law. if i push a box at a constant speed with 100N of force, is the friction that resists the motion the opposite but equal (100N) reaction force? isn't the friction force LESS than the push force because the box is moving? is the reaction force something different (i.e. the box itself pushing back)?

 

[G01]

Yes, the reaction force is NOT the friction force. As you said, the reaction force of you pushing on the block is the force of the block pushing back on you.

Remember that the action force and the reaction force act on different objects. The action force acts on the block, while the reaction force acts on the person, the source of the action force. Consider this, and tell me why the force of friction can't be the reaction force for the pushing force.

 

[Cyrus]

No. I would draw a free body diagram, and review your class notes one more time and think about it, and post any new thoughts you might have.

Hint: Think about things one at a time. You are confusing yourself because you are trying to think of many senarios all at once. Think methodically.

 

[spidey64]

i guess my main hangup is that i don't see how any object moves at all if an equal force opposes the push force, plus friction and inertia

 

[spidey64]

scratch that, i understand when the object pushing is greater in mass, but otherwise, I'm still confused

 

[spidey64]

is the motion of the object simply a byproduct when the force of the push becomes greater than the force of friction, and regardless of the motion, the box will always push back at the same force as the push action force?

 

[Cyrus]

i guess my main hangup is that i don't see how any object moves at all if an equal force opposes the push force, plus friction and inertia

I know for a fact your physics book does not make this statement. Crack it open.

Slow down speed racer. One thing at a time. Look for the answers to your questions in your book one at a time. Not all at once.

 

[spidey64]

i have been reading, I've read every chapter up to this question and the only reason I've come here is because I've exhausted myself pouring over the pages and not finding answers pertaining to this example and i need a human source that is more flexible and able than text.

 

[Cyrus]

Tackle your questions one at a time. Type what your book says about action/reaction forces. Look at the picture next to the description very carefully.

 

[Cyrus]

Here, this might help. I scaned my old physics book:

At least we will be on the same page. Har-har-har.

http://img219.imageshack.us/my.php?image=physicszf6.png

 

[spidey64]

i think this excerpt relates the best to the concept "the tires of a car push against the road while the road pushes back on the tires...the reaction force is what accounts for motion in this example. This force depends on friction; a person or car on ice, for example, may be unable to exert the action force to produce the needed reaction force. Neither force exists without the other." there is no picture.

 

[spidey64]

so obviously the action-reaction is where two objects (me and the box) touch, with our forces pushing at each other. But my problem is with the sustained motion and the force of friction. The friction is needed in order for the action-reaction to take place. Maybe the friction merely reduces the force involved in the action-reaction, but when the force is no longer canceled out, it moves...

 

[Cyrus]

You should review Newtons FIRST law.

 

[spidey64]

yeah, isn't what i just said basically about inertia? where the box has moving equilibrium once the pushing force equals the friction force. It's in equilibrium at that point because no net force is acting on the box because friction and push cancel each other.

 

[spidey64]

ok, so the box is in dynamic equilibrium...so what's the difference between the force of the box pushing back and the force of friction?

 

[russ_watters]

You just asnwered that question. It's in dynamic equilibrium.

 

[Cyrus]

ok, so the box is in dynamic equilibrium...so what's the difference between the force of the box pushing back and the force of friction?

What do you mean?

 

[spidey64]

both the force of friction and the reaction force equal and oppose (essentially cancel out) the push force, how can there be two opposing forces (one acting on the box and one acting on the pusher)?

 

[Cyrus]

This statement is not true. What is the push force?

I know you saw a picture of that football being punted. I expect you to answer this on your own.

 

[spidey64]

the push force is the force of the interaction between the box and the pusher going both ways. the friction force is the force of the interaction between the box and the ground.

 

[Cyrus]

Yes, that's correct.

 

[spidey64]

so is the frictional force unrelated to the push force in terms of action-reaction then?

 

[Cyrus]

The handout I gave you explicitly tells you the answer to this.

 

[spidey64]

well since it doesn't mention friction, i'll just assume "no"

 

[Cyrus]

Its the big bold sentence. Now apply what that sentence says to your specific question, and see if it makes any sense.

I have to go, ill post more tomorrow.

 

[spidey64]

action-reactions are in pairs of course, the pusher and the box is one, so then the friction is not part of that but it's own pair with box, is that the idea?

 

[kdv]

ok, so the box is in dynamic equilibrium...so what's the difference between the force of the box pushing back and the force of friction?

I don't know if this may be part of the confusion but note that the forces forming an action-reaction pair never act on the same object! Therefore, they never cancel out! (Unless you consider the two objects as a combined object). For a car, there is a friction force on the car exerted by the pavement and a reaction force on the pavement produced by the car. They act on different objects.

 

[awvvu]

The box and the ground are action-reaction pairs for the friction force.

Static friction is necessary for us to walk, or for tires to work, but the friction between a moving box and the ground is kinetic friction. You seem to be getting these confused.

 

[spidey64]

alright, i think I'm getting it now, i got to go to bed, thanks for your help everyone!

 

[Laus102]

I am having the same problem, I am taking Into. to Physics and Chemistry and all the other paradigms for the 3rd law make sense to me (e.g. the rocket propulsion, car pushing against the ground etc...), but the example of someone pushing on a box with a force of 100N and the box pushing back on the person with an equal and opposite force always frustrates me. It seems that if I exerted a force of 100N on the box that the force of the box pushing back on me would make the net force in the interaction 0 N. But then how can the box move then, is the box exerting the 100N back on my body and my force of 100 N is still causing the box to move? Does friction play a part in the matter?