# Basic physics confusion in my structure

1. Jun 6, 2013

### x86

My physics core is a little bit weak, and this one question confuses me.

1. The problem statement, all variables and given/known data
A 3kg box sits on top of a 7 kg box. The object accelerates at 1 m/s^2. Find the force applied on the bottom block; if there are no forces besides this one.

2. Relevant equations
F= ma

3. The attempt at a solution
So, my initial attempt was F = m * a. m = 7kg + 3kg = 10 kg. F = 10 N.

However, this is incorrect. The answer is supposed to be 3 kg. They only take the bottom block into account and this is extremely confusing for me. Of course, this is from an even larger kinematics question; but this is the only step I do not understand. And it's basically exactly the same as the one I'm asking now.

Why do we use 3 kg, and not 10 kg?

2. Jun 7, 2013

### voko

What is "the object" here?

3. Jun 7, 2013

### x86

The object would be the two blocks, relative to the ground. Could one split this larger object into two smaller objects, and say the force on the bottom block is 7N and the force on the top block is 3N?

I'll tell you the bigger question, that I summarized.

So basically there's two blocks sitting on top of each other. The friction between the blocks is 15 N and the friction between the block and the floor is 10 N.

This means on the bottom block there is a force of 5 N (forward) [since the top block is the one moving, and the frictional force is causing the bottom block to move]

So I'm confused as what mass to use now to find the acceleration.

F = ma

Would it be the mass of the two blocks combined, or can I just use the mass of the bottom block? This is very confusing for me. The correct answer uses only the mass of the bottom block, but I don't understand why

Last edited: Jun 7, 2013
4. Jun 7, 2013

### voko

I suggest that you state the problem EXACTLY as it was given.

5. Jun 7, 2013

### x86

I apologize for not being clear.

Okay, so basically there is this block:

The friction between the two blocks (Us = 0.6) and between the floor and the block is Us = 0.2

I draw a freebody diagram and see for the bottom block there is a force of friction with the floor to the left and a force of friction with the block on top to the right, causing said block to move right.

We can say that Ff_block - Ff_floor = ma

Which gives me 9.8 N = ma

I'm confused as to what to use for the mass. The correct answer uses 3kg for the mass, but it should be using 7kg as thats the objects weight. So it seems as if they are splitting the block up. How can they do this? I don't understand how they can use 3 kg for the mass

6. Jun 7, 2013

### voko

What are you supposed to find? What else is given?

7. Jun 7, 2013

### Staff: Mentor

If you are analyzing forces on the bottom block then of course you'll use the mass of that block in applying Newton's laws.

8. Jun 7, 2013

### x86

Ah that's all I really wanted to know, I guess. I was confused, because I was thinking of both blocks as one object