Posted by Jim on .
2 blocks are in contact on a horizontal table. A force is applied on the first block at a 30* angle. The blocks move to the right side.
If m1 = 2kg and
m2 = 1kg and
F = 6 N
kinetic friction at the middle of the 2 blocks and the table = 0.250
Find the "contact force" between the 2 blocks.

Physics2 
drwls,
You need to specify if the 30 degree angle is up or down from horizontal. That will affect the friction force.

Physics2 
Jim,
it is down from horizontal drwls,
Thank you for your time 
Physics2 
drwls,
Forward applied force = F cos 30 = 5.196 N
Friction force on m1 = (m1*g+Fsin30)*0.25 = 5.65 N
Friction force on m2 = (m2*g*0.25)
= 2.45 N
Acceleration of m1 and m2:
a = (5.65+2.45)/(m1 + m2) = 2.7 m/s^2
The force between the blocks, F', minus friction on m2, is what accelerates m2.
F'  2.45 = m2*a
Solve for F' 
Physics2 
Jim,
the angle is at the top left corner of the first block

Physics2 
drwls,
It should not matter where it was applied, as long as it was on the first block, in the downward direction you said.

Physics2 
Tom,
ok thank you drwls!