How Is Force Distributed Between Three Boxes on a Frictionless Surface?

[dlthompson81]

Homework Statement

Three boxes are in contact and resting on a horizontal, frictionless surface. One box has mass m, the middle box has mass 2m, and the third box has mass 3m. A horizontal force of magnitude F is applied to the first box, and the three boxes slide together across the surface.

Find the force box 2 exerts backward on box 1 as the boxes accelerate across the table, in terms entirely of F.

Homework Equations

F=MA

The Attempt at a Solution

I already know the answer to this is 5/6. The teacher went over it in class, but I don't fully understand it. Can anyone help me out with an explanation of how exactly it works? It would be much appreciated. I have a midterm Monday, and I'm sure this will be on it. Thanks for the help.

 

[PhanthomJay]

The answer is 5/6 (F).

Can you find the acceleration of the system of blocks in terms of F and m? Then draw a free body diagram of block 1. This is essential. In a free body diagram, you show all forces acting on the block. These forces are contact forces like applied forces, friction, normal forces, as well as action at a distance forces like gravity (weight) forces. Once you identify these forces, use Newton's laws. In this FBD of block 1, apply Newton's 2nd law in the x direction to solve for the force of block 2 on block 1.

 

[dlthompson81]

Ok I think I may have this figured out.

The acceleration is A=F/6M

The force exerted back on block 1 is 5/6.

So that is the mass of 2 and 3 which is 5M * A

5M(F/6M) = 5/6F

So, for say the force exerted back on block 2 from block 3 it would be:

3M(F/6M)= 3/6F Is that right or am I still lost?

 

[PhanthomJay]

Ok I think I may have this figured out.

The acceleration is A=F/6M

yes, good.

The force exerted back on block 1 is (5/6)(F)[/color].

That answer was given you by the teacher..you are trying to find it, not know it.

So that is the mass of 2 and 3 which is 5M * A

5M(F/6M) = 5/6F

So, for say the force exerted back on block 2 from block 3 it would be:

3M(F/6M)= 3/6F Is that right or am I still lost?

You went from trying to find the force of block 2 on 1, to finding the force of block 3 on 2? You must pretend no answer was given you. Look at block 1. There is an applied force F acting on it to the right, and a force of block 2 on 1 acting to the left. What is the net force acting on block 1? Then F_net =mA, where you calculated A correctly. Solve for the force of 2 on 1. Now see if you get the correct answer, as was given to you by the teacher.