Force problem PLEASE guide me in the right direction

[Byrne]

Cart A is locate din front of cart B and is being pulled by a force of 20 N

. The two carts have the same mass and are connected by a rope. There isn't any friction involved in this problem. Determine the tension in the rope.

What exactly do I need to find here?

I mean, if I assume a mass of 5 kg for each cart, then the force of gravity and normal force for each would be 49 N... the entire system would have a mass of 10 kg then. I just don't know where to go from here. Help!​

 

[Imo]

Well let's work through this...
What causes the tension? Cart A pulling on Cart B (think about you pulling on a string attatched to a cart. Your pulling causes the tension)

So what is the force of Cart A pulling on Cart B? There are 4 forces that you are given to work with. 1) Friction, which is zero 2) Normal Force 3) Gravity 4) The force on Cart B

We know that gravity and the normal cancel, so their insignificant. What about the force on Cart B? How does it relate to Cart A? (I suggest remebering what Newton's third law is...the one about actions and reactions)

Hope that helps

 

[HallsofIvy]

Since you are not told the mass, call it m (and hope it cancels out in the end!- If it does then using "5 kg" or whatever you like doesn't hurt- but you should first show that it cancels).

(And, by the way, this is horizontal motion with no friction- the force of gravity has nothing whatsoever to do with this problem!)

First think of the two carts as a single object of mass 2m. You know the force so you can calculate the acceleration. Since the two carts are attached by a rope that must have the same acceleration and what you just calculated is it.

But the second cart, alone, has mass m. What force pulling on it alone would give it that acceleration? THAT is the tension in the rope. (Did the m's cancel?)

 

[patrioticindia]

You can think like this...

I am assuming that u r applying the force on Cart B (doesnt matter much :D)

You are pulling cart B, and the tension in the string is opposing ur force...however, tension being an internal force of the string, simultaneously helps Cart A to accelerate.

Now if we assume the string to b inextensible, then both the carts have same acc.

So ((F(applied)-Tension)/(Mass of Cart B)) = (Tension/Mass of Cart A)

Solve this to get the tension