Newton's Laws with Friction: Two Blocks

[melv14]

Homework Statement

Part 1

Two blocks are in contact on a frictionless table. A horizontal force F is applied to M2, as shown. If M1 = 1.03 kg, M2 = 3.40 kg, and F = 5.10 N, find the size of the contact force between the two blocks.

http://loncapa1.fsu.edu/enc/85/c70c507c1fdfce450f24d473053587465ad535c2c20eedbf8845ed0a22afb4e439270ed6dfaf6472da2a2cac5a4ec3a06f1770c443f1ffcb6e5f0ece685f8ea7803066440f7d55b844760e01ac475fea48c6e88e4c25cdda0806fb9fcbe8bb4cfae3d0436dc5c98e9e32a9ec3bcf2bb4.gif

Part 2

If instead an equal but oppositely directed force is applied to M1 rather than M2, find the size of the contact force between the two blocks.

Homework Equations

I'm pretty sure F=MA is the equation to use.

The Attempt at a Solution

For this, I looked at the system as a whole, summed the masses and found the acceleration of the entire system.

I found this to be a=1.151m/s^2

I then isolated mass 2 and mass 1 to find the forces for part 1 and 2 respectively. I obtained the answers : 3.914N for part 1 and 1.85N for part 2.

Where am I going wrong?

 

[rl.bhat]

Net force acting on M1 is (the applied force F1 - reaction F2 from the mass M1)
Your acceleration is correct, and it is common for both.
So the acceleration of M1 = (F1 - F2)/M1 = a.
F1 is given. Find F2. Repeat the same thing in the opposite direction.

 

[melv14]

Ok, I got the first part, thanks!

The second part is still tripping me up though. What do you mean when you say do the opposite?

 

[rl.bhat]

Ok, I got the first part, thanks!

The second part is still tripping me up though. What do you mean when you say do the opposite?

Force is acted on M1 first.