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.
6 answers
You need to specify if the 30 degree angle is up or down from horizontal. That will affect the friction force.
it is down from horizontal drwls,
Thank you for your time
Thank you for your time
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'
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'
the angle is at the top left corner of the first block
It should not matter where it was applied, as long as it was on the first block, in the downward direction you said.
ok thank you drwls!
0 answers