# Newton's Third Law problem

1. Jan 9, 2008

### Precursor

[SOLVED] Newton's Third Law problem

1. The problem statement, all variables and given/known data
Three boxes, A, B, and C, are positioned next to each other on a horizontal, frictionless surface. An applied force acting on box A causes all the boxes to accelerate at $$1.5m/s^{2}$$. Calculate the force that box B exerts on box A.

2. Relevant equations
$$F_{A on B}= -F_{B on A}$$
F = ma

3. The attempt at a solution
What seems to be the issue is how my teacher solved this problem. What she did is isolate box B and find the $$F_{net}$$ acting on it. This is the calculation she did:

$$F_{net}= F_{A on B} - F_{C on B}$$
$$F_{net}= m_{B}a - (-m_{C}a)$$
$$F_{net}= (10 Kg)(1.5m/s^{2}) + (5.0 Kg)(1.5m/s^{2})$$
$$F_{net}= 23 N$$

Therefore, she says that $$F_{B on A} = 23 N$$.

However, what confused me is the two negative signs she put infront of $$m_{C}a$$. I think only one would suffice to take into consideration that $$F_{C on B}$$ is a negative value, being that its direction is left.

I solved this problem taking a different approach. I found $$F_{net}$$ of box B and C, and added them together. Here are my calculations:

$$F_{net}= m_{B}a + m_{C}a$$
$$F_{net}= (10 Kg)(1.5m/s^{2}) + (5.0 Kg)(1.5m/s^{2})$$
$$F_{net}= 23 N$$

Could someone please confirm whether my method of solving the problem is correct or not?​

2. Jan 9, 2008

### Shooting Star

You can find the total force. You know the mass of A. So, you know the force B applies back on A.

3. Jan 9, 2008

### Precursor

I found the total force that box B and C exerted on box A.

4. Jan 9, 2008

### Shooting Star

Why does C have to come into the picture?

F_A = F_net - F_B_on_A =>

M_A*a = M_tot*a - F_B_on_A.

I think this is conceptually the simplest. Ultimately, after algebra, all the methods are the same.

5. Jan 9, 2008

### kamerling

your teacher has $$F_{B on A}$$ pointing in the opposite direction as $$F_{A on B}$$
This is rather confusing. First you need a minus sign in the first formula, because the two forces acting on B now point in opposite directions, and then you need another minus sign in $$F_{C on B}=-m_{C}a$$. It's Easier to have positive=rightwards for all forces and accelerations.
What's worse is that between the first and the second line of her calculations, The meaning of $$F_{net}$$ changes, at first it's the net force on B, but then it's the net force on B and C as you had. It seems she changed her mind in mid-calculation.

Your calculation is ok.​

6. Jan 9, 2008

### Shooting Star

Regardless of the value you have got, why are you calling the sum of the forces on B and C as Fnet(left). Fnet is the force which is making all the boxes go.​

7. Jan 9, 2008

### HallsofIvy

Staff Emeritus
Boxes A, B, and C have total mass 23 kg. If they are acceletated at 1.5 m/s2, the force must F= ma= (23)(1.5)= 34.5 Newtons- and it is exerted upon A directly. A alone has mass 8 kg. In order to accelerate it alone at 1.5m /s2, it must be pushed by a net force of Fnet= (8)(1.5)= 12 Newtons. What force must B be exerting on A?

8. Jan 9, 2008

### Shooting Star

Hi Halls,

That's exactly what I've given in post #4 (without the littersome numericals, of course).

9. Jan 9, 2008

### Precursor

So my mistake was solving for $$F_{net}$$? What if instead I had $$F_{A on B} = 23N$$, and according to Newton's Third Law, $$F_{B on A} = 23N$$?​

10. Jan 9, 2008

### Shooting Star

Would you do it leaving C out of it? Or you just can't give up your method? It happens to all of us.

11. Jan 9, 2008

### Precursor

I suppose your method of solving the problem is easier. It makes me realize just how many ways one can go about solving such problems. Thanks for the help.