• Support PF! Buy your school textbooks, materials and every day products Here!

Newton's Third Law problem

  • Thread starter Precursor
  • Start date
  • #1
222
0
[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 [tex]1.5m/s^{2}
[/tex]. Calculate the force that box B exerts on box A.

http://img70.imageshack.us/img70/8836/thirdlaw2cf8.png [Broken]​
[/URL]

Homework Equations


[tex]F_{A on B}= -F_{B on A}[/tex]
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 [tex]F_{net}[/tex] acting on it. This is the calculation she did:

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

Therefore, she says that [tex]F_{B on A} = 23 N
[/tex].

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

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

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

Could someone please confirm whether my method of solving the problem is correct or not?​
 
Last edited by a moderator:

Answers and Replies

  • #2
Shooting Star
Homework Helper
1,977
4
You can find the total force. You know the mass of A. So, you know the force B applies back on A.
 
  • #3
222
0
I found the total force that box B and C exerted on box A.
 
  • #4
Shooting Star
Homework Helper
1,977
4
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
454
0
What seems to be the issue is how my teacher solved this problem. What she did is isolate box B and find the [tex]F_{net}[/tex] acting on it. This is the calculation she did:

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

Therefore, she says that [tex]F_{B on A} = 23 N
[/tex].

However, what confused me is the two negative signs she put infront of [tex]m_{C}a[/tex].​


your teacher has [tex]F_{B on A}[/tex] pointing in the opposite direction as [tex]F_{A on B}[/tex]
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 [tex] F_{C on B}=-m_{C}a[/tex]. 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 [tex]F_{net}[/tex] 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
Shooting Star
Homework Helper
1,977
4
I solved this problem taking a different approach. I found [tex]F_{net}[/tex] of box B and C, and added them together. Here are my calculations:

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

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.​
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,833
955
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
Shooting Star
Homework Helper
1,977
4
Hi Halls,

That's exactly what I've given in post #4 (without the littersome numericals, of course).
 
  • #9
222
0
So my mistake was solving for [tex]F_{net}[/tex]? What if instead I had [tex]F_{A on B} = 23N
[/tex], and according to Newton's Third Law, [tex]F_{B on A} = 23N
[/tex]?​
 
  • #10
Shooting Star
Homework Helper
1,977
4
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
222
0
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.
 
Top