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Posted by on Thursday, November 18, 2010 at 8:25pm.

Block B in the figure below weighs 713 N. The coefficient of static friction between block and table is 0.20. Assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary.

  • Physics - , Thursday, November 18, 2010 at 8:36pm

    Where is the knot? where is block A

  • Physics - , Thursday, November 18, 2010 at 8:47pm

    the knot is inbetween block b and the wall. block a hangs from the knot. the knot is connected to the wall by a rope attached to the wall at 30 degrees from the horizontal

  • Physics - , Thursday, November 18, 2010 at 8:56pm

    OK, break the problem into two parts, the rope tension into vertical and horizontal parts.

    Since the wall and block both support the vertical equally, the vertical down force on the blockB is 1/2 weightA.

    So forcefrictionB= mu*forcenormal
    =mu(massB*g+1/2 massA*g)

    now, that force of friction is opposing the horizontal part of tension.

    horizontal: Tension*cosTheta, but
    tension/.5Ma*g=sinTheta
    or tension= .5Ma*g*sinTheta

    horiztonal=friction force
    .5Ma*g*sinTheta*cosTheta=mu*(MassB*g+.5Ma*g)

    solve for MassA

  • Physics - , Thursday, November 18, 2010 at 9:01pm

    is Ma the mass of block a?

  • Physics - , Thursday, November 18, 2010 at 9:09pm

    i still couldn't get the answer either

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