# Three Blocks Accelerating

## Homework Statement

The horizontal surface on which the objects slide is frictionless. The acceleration of gravity is 9.8 m/s2. Three blocks are sliding on a frictionless horizontal surface:

Fℓ----->[Block1-1kg][Block2-7kg][Block3-4kg]<--Fr

If Fℓ = 29 N and Fr = 3 N, what is the magnitude of the force exerted on the block with mass 7 kg by the block with mass 4 kg?

F = ma

## The Attempt at a Solution

29 N - 3 N = 26 N (total force) 1kg + 7 kg + 4 kg = 12 kg (total mass)

a= F/m = 26 N/12 kg = 2.166666667 m/s2 (acceleration of all the blocks in the right direction)

masses of Block2 + Block 3 = 11 kg

F = (11 kg)(2.166666667) = 23.83333333 N (THIS IS THE WRONG ANSWER!)

What am I doing wrong?

Doc Al
Mentor
F = (11 kg)(2.166666667) = 23.83333333 N (THIS IS THE WRONG ANSWER!)
You found the net force on the "Block2 + Block 3" system. Not sure why.

Hint: What's the net force on Block 3? How can you use that result to answer the question asked in the problem?

Net force on Block 3 = (4 kg)(2.166666667) = 8.666666668 N

So, the force of 8.67 N is being exerted on Block 3 from the left, but if I am trying to find the force being exerted on Block 2 from Block 3 (from the right), then would I have to find the net force being exerted by Block 1 and 2? I'm not quite sure.

OR:

The net force on Block 3 + Fr = 8.67 N + 3 N = 11.67 N (force exerted on block 2 by block 3)

Is that right?

Doc Al
Mentor
Net force on Block 3 = (4 kg)(2.166666667) = 8.666666668 N
Good.

So, the force of 8.67 N is being exerted on Block 3 from the left,
No, 8.67 N is the net force on block 3.

but if I am trying to find the force being exerted on Block 2 from Block 3 (from the right), then would I have to find the net force being exerted by Block 1 and 2? I'm not quite sure.
Don't forget Newton's 3rd law: If you can figure out the force that block 2 exerts on block 3, then you automatically know the force that block 3 exerts on block 2.

OR:

The net force on Block 3 + Fr = 8.67 N + 3 N = 11.67 N (force exerted on block 2 by block 3)

Is that right?
Yes. Here's how to think of it:
Net Force on Block 3 = + 8.67 N (Positive, since it points right.)
Force of Block 2 on Block 3 + Fr = + 8.67 N
Force of Block 2 on Block 3 - 3N = + 8.67 N (Note that Fr is negative, since it points left.)
Force of Block 2 on Block 3 = + 8.67 N + 3N = 11.67 N

Thus the force of Block 3 on Block 2 = -11.67 N (from Newton's 3rd law); the magnitude is just 11.67 N.

Thank you!