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**[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.

[tex]F_{A on B}= -F_{B on A}[/tex]

F = ma

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

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

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?

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

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