How Do Newton's Laws Apply to Sliding Blocks with Different Masses and Friction?

[jehan4141]

Two blocks are sliding to the right across a horizontal surface, as the drawing shows. In Case A the mass of each block is 3.0 kg. In Case B the mass of block 1 (the block behind) is 6.0 kg, and the mass of block 2 is 3.0 kg. No frictional force acts on block 1 in either Case A or Case B. However, a kinetic frictional force of 5.8 N does act on block 2 in both cases and opposes the motion. For both Case A and Case B determine (a) the magnitude of the forces with which the blocks push against each other and (b) the magnitude of the acceleration of the blocks.

I'm in intro physics...I've been looking at this problem for a while. I understand that I am supposed to set up a free body diagram for both boxes. Let's just look at part A first.

This is my reasoning. The two boxes are contiguous and thus accelerate at the same magnitude. On box 1 there is a normal force and mg working on it. Then I THINK there is a force pointing to the left. Let's call this force F.

I THINK force F is caused by the equal and opposite reaction of pressing it pressing on block B. Is my logic correct?

If this is true, then for block 1: -F=ma

Then on Block B, we have friction working on it towards the left and the force F put on it. This is in the +x direction. We also have normal force and weight.

And for block B: F - Ffriction = ma?

 

[Doc Al]

Looks good, just be careful with signs. (For example, what's the direction of the acceleration?)

 

[jehan4141]

Looks good, just be careful with signs. (For example, what's the direction of the acceleration?)

I'm not sure about the accelerations?

So for Box 1 we have normal force and weight that cancel out. And we also have the force of friction and the force F, BOTH in the -x direction...This acceleration should be in the negative direction for box 1?
F₁ = m(-a)
a = -F/m = -F/3

And for Box 2 we have normal force and weight that cancel. We also have friction in the left direction and F in the +x direction. So acceleration is positive? This is just a gut feeling that acceleration is in the +x direction. How do I know if the net force on box 2 is in the + or - direction? But if my assumption is correct that a is in the +x direction, we have:

F-Ffriction = ma
F-Ffriction = 3a =3(-F/3)
F-Ffriction = -F
-Ffriction = -2F
-5.8 = -2F
F = 2.9 N?

F = ma
2.9 = 3a
a=0.967 m/s²?

 

[Doc Al]

I'm not sure about the accelerations?

So for Box 1 we have normal force and weight that cancel out. And we also have the force of friction and the force F, BOTH in the -x direction...This acceleration should be in the negative direction for box 1?
F₁ = m(-a)
a = -F/m = -F/3

Looks OK. There's no friction acting on block 1. The only horizontal force is that from block 2. (Both blocks accelerate together, to the left.)

And for Box 2 we have normal force and weight that cancel. We also have friction in the left direction and F in the +x direction. So acceleration is positive? This is just a gut feeling that acceleration is in the +x direction. How do I know if the net force on box 2 is in the + or - direction?

The only external force on the two block system is the friction acting to the left, so both blocks accelerate to the left. But you need not assume that, the equations will tell you.

But if my assumption is correct that a is in the +x direction, we have:

F-Ffriction = ma
F-Ffriction = 3a =3(-F/3)
F-Ffriction = -F
-Ffriction = -2F
-5.8 = -2F
F = 2.9 N?

Good. This is the force exerted by block 1. It acts to the right. Note that the friction force, which acts to the left, is twice as much.

F = ma
2.9 = 3a
a=0.967 m/s²?

For block 1 you had: a = -F/m, which will give you the correct sign for the acceleration.

For block 2, you need the net force = -5.8 + 2.9 = -2.9. Thus a = F/m, which will give you the correct sign.

 

[jehan4141]

Thanks Doc! My quiz is Friday and that helped me so very much!