connected objects w/ friction

[mrnastytime]

1. The problem statement, all variables and given/known data
A block of mass m1 = 2 kg and a block of mass m2 = 3 kg are tied together and are pulled from rest across the floor by a force of Fp = 20 N. The coefficient of friction of the blocks with the floor is µ = 0.23.



2. Relevant equations
What is the acceleration of the two blocks?


3. The attempt at a solution
a=F/(m1 + m2)
a=20N/(2kg + 3kg)
a=4 m/s^2 ???

I know that somehow µ = 0.23 is involved, but i dont know where to use it.

[Delphi51]

u = 0.23 is the coefficient of friction, used in the formula F = u*Fn,
where Fn is the "normal" force pressing the object against the surface it is moving on.
Begin by calculating the force of friction on each block.

[mrnastytime]

I still dont understand. What is the force of Fp=20 N? and where does that fit in

[Delphi51]

Someone is pulling the blocks with a force of 20 N.
There is a friction force opposing that, which you are about to calculate.
Once you have the two forces acting on the blocks, you can use F = ma to find the acceleration.

[mrnastytime]

I tried calculating the force of friction on each block, but my answers still werent correct. what is the correct set up? thats all i need to know to see where i made the mistake.

[Delphi51]

Use F = u*Fn to find the force of friction. Show your calc and we'll find your error.

[mrnastytime]

F1=u*Fn
-mg=u*Fn
-mg/u*=Fn
-2*9.81/0.23=-85

F2=-mg+Fp+u*Fn
mg-Fp=u*Fn
mg-Fp/u*=Fn
3*9.81-20/0.23=41

[Delphi51]

"-mg+Fp+u*Fn " is not correct.
mg, the weight, is downward while Fp is horizontal.
You can't add forces in different directions.

First, figure out the force of friction.
Ff = u*Fn = ???
Fn is the normal force, pressing the blocks against the surface. It is caused by gravity. Can you work it out?

After you get Ff, write "sum of forces = ma" for the horizontal direction.
You'll have the 20 N pulling force and the friction force opposing the motion.
Solve for acceleration.