# Homework Help: Board pulled out from under a box

1. Oct 21, 2008

### aligass2004

1. The problem statement, all variables and given/known data
http://i241.photobucket.com/albums/ff4/alg5045/MLD_2l_15_001.jpg

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box in us. The coefficient of kinetic friction between the board and the box is, as usual, less than us. Throughout the problem, use g for the magnitude of the acceleration due to gravity. In the hints, use Ff for the magnitude of the friction force between the board and the box.

Find Fmin, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the box (which will then fall off of the opposite end of the board.) Express you answer in terms of some or all of the variable us, m1, m2, g, and L. Do not include Ff in your answer.

u = mu

2. Relevant equations

F=ma
fs = N(us)

3. The attempt at a solution

So I know the free body diagram for m1 has the weight down and a normal force up. The free body diagram for m2 has the weight1 and weight2 down and a normal force up. It also has F going to the right and fs going to the left.

I'm not sure if these free body diagrams are completely right or not.

Last edited: Oct 21, 2008
2. Oct 21, 2008

### einstein_a_go_go

For the free body diagram for m1, do you have a friction force on it?

Otherwise sounds right...if I am reading it right

3. Oct 21, 2008

### aligass2004

I don't have a friction force on m1 because I'm not sure where it would go. Would it go to the right? Opposite the friction force for m2? And if that's right, I know the vertical equation is N = m1g and I'm not sure what the horizontal equation would be. I know for m2, it's moving to the right so the horizontal equation would be F - fs = ma, but I'm not sure about the horizontal equation for m1.

4. Oct 21, 2008

### nasu

The existence of a force does not depend on your knowledge about its direction.
The problem states explicitly that there is friction between the box and the board.
There is a friction force acting on the board and an equal and opposite one acting on the box.
Otherwise, no matter how you pull the board, the box will remain at rest and eventually will fall at the other end.

5. Oct 21, 2008

### aligass2004

Ok, but that still doesn't help me figure out the horizontal equation for m1.

6. Oct 22, 2008

### nasu

There is only one horizontal force acting on m1: the friction.
The direction of the friction force on m1 is the same as the direction you pull the board.