Stumped on a box on cart action/reaction pair question

[KrazySocoKid]

Homework Statement

Consider a box sitting on cart. When the cart is pulled with a force, P, the box A moves with the cart without slipping. In all questions, assume there is a negligible amount of drag. There is, however, kinetic friction between the cart and ground.
What are the two sets of action/reaction pairs between the cart and Box A?

Homework Equations

I don't think there is any equations that really relate to this. This is just testing my knowledge of action/reaction pairs.

The Attempt at a Solution

So I thought I definitely had one, weight and normal force. The weight of the box acting on the cart and normal force of the cart pushing up on the box. But now I'm questioning if that's even an action/reaction pair because weight is between the Earth and the box right? And then for the 2nd one, I'm thinking static friction, because the box isn't slipping. But then what is the friction's partner? Is it the air pushing on it from the movement? I know this problem is very simple, it's just stumping me to no end. I don't know if I'm making it too complicated in my head or what. Can someone help clear it up for me?
Thanks![/B]

 

[KrazySocoKid]

Okay, so I think I have one nailed down. One pair is the force of the cart acting upward on the box (normal force) and the equal and opposite force of the box acting downward on the cart. Now I just need help determining the second action-reaction pair, and I believe it has something to do with static friction!

 

[KrazySocoKid]

Okay actually I think I got it. Now it would be nice if someone just doubled checked my answer. Is the second action reaction pair: Friction force of box acting on cart pointing to the right, and friction force of cart acting on box pointing to the left?

 

[SweetieK11]

Hi,
I am actually working on this problem too. I can't tell for sure if you're right, but I think your first one adds up because it's acting on 2 different objects and it's in the opposite direction. For the second one, I was just wondering where the pulling force is supposed to come in? Could there be a pulling force on the cart and friction on the box to keep it from moving?

 

[KrazySocoKid]

I'm not sure if you are actually in my class haha, but if you look at the handout the hint says that the force, P is only applied to the cart, so it doesn't pull Box A. So we can disregard the pull, because it only acts on the cart, not the box.

 

[SweetieK11]

Or is there pulling force on the cart and static friction on cart to the ground?

 

[KrazySocoKid]

The handout specifies action/reaction pairs between the cart and box, so I don't think we can bring the ground into it. I think the two pairs are: Force of cart pushing up on box (normal force), force of box pushing down on cart (NOT weight, that's important) and then friction of cart acting on box pointing to the left, friction of box acting on cart pointing to the right.

 

[SweetieK11]

Okay, you're probably right about us not being able to bring the ground into this. Okay so now I am wondering why the force of the box pushing down on the cart can't be weight? What force would that be if it can't be weight?

 

[KrazySocoKid]

It's like the exact same thing as normal force, but just the opposite and equal reaction. Weight has to do with Earth acting on an object. Normal force and weight aren't action/reaction pairs because the Earth has nothing to do with normal force, but everything to do with weight. So the force I'm talking about, the box acting on cart, is just the opposite of normal force.

 

[KrazySocoKid]

It would be cool if I could get a confirmation on this too, but I'm 90% sure that's right.

 

[SweetieK11]

Okay thank you for explaining that. I'm sorry that this is your question and I'm not being completely helpful. Hopefully someone can confirm this though.

 

[KrazySocoKid]

No problem. So I guess we are in the same class? I'll see you tomorrow then, internet stranger haha

 

[SweetieK11]

Haha you wouldn't even know it's me, but I did have the class today at 9 a.m.

 

[Chestermiller]

Your answers in posts 2 and 3 are both correct, except, in post 3, you have the directions reversed. The cart exerts a friction force to the right on the box (it is trying to accelerate the box to the right), and the box exerts a friction force to the left on the cart.

Chet