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Newton's Third Law Problem

  • Thread starter Ossim777
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Homework Statement


Two Blocks, A (10kg) and B (12kg) are placed next to each other on a rough surface. An applied force acting on block A causes both blocks to accelerate at 1.2m/s^2 to the right. If the magnitude of the force of friction on Block B is 18.3N, calculate the force (magnitude and direction) exerted by block B on Block A.

Homework Equations


f=ma



The Attempt at a Solution


I've drawn out the problem and arrived at my numbered solutions. Here's what I have so far.
fnet of the entire system =26.4N.
Here's my action/reaction pairs:
Block A on Force=Block A on B
A on B=B on A
B on A=Force of B moving forward.
At first I simply multiplied 12 by 1.2 (m*a) to find B moving forward, getting 14.4 N and finding my B on A, which is what I'm searching for.
Problem is, there's no way the answer is that simple if the question has given me force of friction and the values of acceleration, block A and 3 different action/reaction pairs.
How exactly do I go about factoring these variables into the equation I've layed out?

All of my thanks for anyone who cares to help.
 

Answers and Replies

  • #2
ehild
Homework Helper
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1,883

Homework Statement


Two Blocks, A (10kg) and B (12kg) are placed next to each other on a rough surface. An applied force acting on block A causes both blocks to accelerate at 1.2m/s^2 to the right. If the magnitude of the force of friction on Block B is 18.3N, calculate the force (magnitude and direction) exerted by block B on Block A.

Homework Equations


f=ma



The Attempt at a Solution


I've drawn out the problem and arrived at my numbered solutions. Here's what I have so far.
fnet of the entire system =26.4N.
Here's my action/reaction pairs:
Block A on Force=Block A on B
A on B=B on A
B on A=Force of B moving forward.
At first I simply multiplied 12 by 1.2 (m*a) to find B moving forward, getting 14.4 N and finding my B on A, which is what I'm searching for.
Problem is, there's no way the answer is that simple if the question has given me force of friction and the values of acceleration, block A and 3 different action/reaction pairs.
How exactly do I go about factoring these variables into the equation I've layed out?

All of my thanks for anyone who cares to help.
14.4 N is the net force experienced by block B. It is the sum of forces exerted on it: the force from A FA→B and the opposite force of friction, Ff=18.3 N.

ehild
 

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