# Two Bars Friction Physics Problem

1. Jan 25, 2009

### cordyceps

1. The problem statement, all variables and given/known data
Two bars of masses m1 and m2 connected by a non-deformed light spring rest on a horizontal plane. The coefficient of friction between the bars and the surface is equal to k. What minimum constant force has to be applied in the horizontal direction to the bar of mass m1 in order to shift the other bar?

2. Relevant equations
Friction = mgk

3. The attempt at a solution
Since the spring transfers the force applied to m1, I assumed I could treat the two masses as a single mass m1+m2. Thus the minimum force required is (m1+m2)gk.
This is the wrong answer- but I don't know what I did wrong. Any suggestions? Thanks.

2. Jan 26, 2009

### ritwik06

Dont consider m1 and m2 as a single mass.
First find out how much extension you need in the spring in order that the second bar just starts moving.
Then apply the work energy theorem. The force would be minimum when m1 does not acquire any kinetic energy in the process.

3. Jan 26, 2009

### cordyceps

Isn't the extension needed in the spring in order to just move the 2nd bar equal to the friction of the second bar, m2gk?

4. Jan 27, 2009

### ritwik06

Yes, thats right.
The extension x required= (km2g)/c
c is assumed to be the spring constant

Notice that the bar2 is just about to move, it hasn't started moving.

And the work done by the constant force=the work done on the spring+the work done against friction in moving the 1st block
Tell me if you get you answer now.

The spring constant c will automatically vanish in the expression. Give it a try :D

regards,
Ritwik

5. Jan 27, 2009

### cordyceps

So Fx= (1/2)c(x^2) + m1gkx
F= (1/2)cx + m1gk
F= (1/2)m2gk + m1gk
Wow. Thanks a bunch!