Question

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.

Answer 1

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

Answer 2

it is down from horizontal drwls,

Thank you for your time

Answer 3

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'

Answer 4

the angle is at the top left corner of the first block

Answer 5

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

Answer 6

ok thank you drwls!