# Two boxes accelerating

1. Jan 30, 2009

### sci-doo

1. The problem statement, all variables and given/known data
Box 1 is placed on box 2, which sits on the floor. Kinetic friction constant between the boxes, and the box 2 and the floor, is equal u.

When box 1 is pushed with horizontal force F it starts to slide on box 2. At the same time, box 2 starts to move. What are the accelerations of the boxes.

2. Relevant equations
Kinetic friction Fk=ukN
F=ma

3. The attempt at a solution
The top box (box 1):
Force that accelerates (pushing minus friction)
Fa=F-Fu=F-ukN
and the acceleration
a=$$\frac{F-u_{k}N}{m_{box1}}$$

The lower box (box 2):

By the Newtons 3rd law the force that pushes the box 2 is the reaction of kinetic friction Fu between the boxes, right? Now this is where I face the problem: as the friction between box 2 and floor is the same but the weight is more (box1+box2) I get negative acceleration ?!?

2. Jan 30, 2009

### chrisk

Consider the limiting cases: The second box on a frictionless surface, and the second box fastened to the surface such that this box cannot move.

3. Jan 30, 2009

### LowlyPion

Welcome to PF.

Your kinetic friction arises because it is moving relative to the box underneath.

Now Kinetic friction is a retarding force given by the μ *m*g relationship.

The maximum force then that Box 1 can transfer to box 2 is just that.

As you note though box 2 has a retarding force available of μ *(m + m) * g.

Since μ *(2m)*g will always be > μ *m*g then I have serious doubt that the bottom box moves despite the statement of the problem. So I'd say it's not negative acceleration, just 0 movement.

4. Jan 30, 2009

### sci-doo

Makes sense.

I don't see how it could move (especially to opposite direction)... I think it'd move (to positive drctn) only if the weight was <0 ... :grumpy: