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March 28, 2017

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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 - ,

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

  • Physics2 - ,

    it is down from horizontal drwls,

    Thank you for your time

  • Physics2 - ,

    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 - ,

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

  • Physics2 - ,

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

  • Physics2 - ,

    ok thank you drwls!

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