# Newton's Third Law problem

[SOLVED] Newton's Third Law problem

## Homework Statement

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.

http://img70.imageshack.us/img70/8836/thirdlaw2cf8.png [Broken]​
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## Homework Equations

$$F_{A on B}= -F_{B on A}$$
F = ma

## 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?​

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

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

Shooting Star
Homework Helper
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.

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 $$m_{C}a$$.​

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.

Shooting Star
Homework Helper
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?​

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.​

HallsofIvy
Homework Helper
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?

Shooting Star
Homework Helper
Hi Halls,

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

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, $$F_{B on A} = 23N$$?​

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

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.