Solving Two-Block Friction Problem - Force Calculation

[TexasCow]

Homework Statement

http://img219.imageshack.us/img219/4953/physicsyo5.png
That's the best that I can do. As you can see, the two blocks are touching and both share the same coefficient of friction. There's an applied force on the first, lighter block. I've done all of my homework but just can't figure out:
-Force exerted by the 80kg block on the 210kg block
-Force exerted by the 210kg block on the 80kg block
-I have already determined the net acceleration, which I believe is 1.410m/s^2

Homework Equations

F=ma
Ff=(Fn)(mu)

The Attempt at a Solution

F=ma
Fnet=(m)(anet)
Fnet=(210kg)(1.410)=296.1N

F=Fa-Fnet
F=750-296.1=453.9N I believe this is the answer for the "Force of the 80kg block on the 210kg block".

However, I don't know how to calculate the other. Any help is appreciated!

 

[Joshrk22]

Okay well F - Ff = m*a.

F = 750 N
Ff = 341
750 - 341 = 290 * a
a = 1.41 m/s

Then the force of each blocking pushing on each other is equal because of Newton's 3rd law. So the Force on the first block minus the force of friction is equal to the force of one block on another.

F - Ff = 409 N

Or you could do Fnet = m * a and plug in 290 kg times 1.41 in which you will get 409 N.

Therefore,

A) 409 N
B) -409 N

 

[TexasCow]

Thanks!

Quick question though. Since the masses are different and thus having different friciton forces, wouldn't the forces exerted by one block on the other be different?

 

[Joshrk22]

Nope. If you push against the wall with a force of 50 Newtons, the wall pushes back on you with a force of 50 Newtons. This is how a scale works.