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).

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.

 

[rwisz]

I would like to first clarify the question and the given information. The homework statement describes a situation where three boxes with different masses are in contact and resting on a horizontal, frictionless surface. A horizontal force of magnitude F is applied to the first box, and the three boxes slide together across the surface.

The homework equation provided, F=MA, is the fundamental equation used to describe the relationship between force, mass, and acceleration. In this situation, the force F is applied to the first box, which has a mass of m. The middle box has a mass of 2m, and the third box has a mass of 3m.

Now, to find the force that the second box exerts backward on the first box, we can use Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. In this case, the force applied to the first box (action) will result in an equal and opposite force exerted by the second box on the first box (reaction).

To determine the magnitude of this force, we can use the equation F=MA, where F is the force exerted by the second box, M is the mass of the second box, and A is the acceleration of the system. Since all three boxes are sliding together, they have the same acceleration. Therefore, we can write the equation as F=(2m)(a), where a is the acceleration of the system.

Now, we need to find the value of acceleration a. This can be done by considering the total mass of the system, which is m+2m+3m=6m. We know that the net force applied to the system is F, and using Newton's Second Law of Motion, we can write F=(6m)(a). Solving for a, we get a=F/6m.

Substituting this value of a in the equation F=(2m)(a), we get F=(2m)(F/6m)=F/3. Therefore, the force exerted by the second box on the first box is F/3. This means that the force box 2 exerts backward on box 1 as the boxes accelerate across the table is 1/3 of the applied force F.

To convert this into terms entirely of F, we can write it as 1/3F or simply F/3. Therefore, the force exerted by the