Force pushes three blocks, find mag of force exerted on last two blocks

  1. 1. The problem statement, all variables and given/known data
    Three blocks in contact with each other are
    pushed across a rough horizontal surface by a
    75 N force as shown.
    The acceleration of gravity is 9.8 m/s2 .

    If the coefficient of kinetic friction between
    each of the blocks and the surface is 0.076,
    find the magnitude of the force exerted on the
    3.5 kg block by the 4.1 kg block.
    Answer in units of N.

    2. Relevant equations
    sum of Fy=N-mg=0 => N=mg
    sum of Fx= -fk=max

    3. The attempt at a solution
    I have no clue how to figure out the force just between those two block...

    Do I add the first two blocks together to act a one force against the third block?
  2. jcsd
  3. Doc Al

    Staff: Mentor

    Hint: Start by treating the three blocks as a single system. What can you deduce?
  4. You mean like m1+m2+m3=9.6 kg

    Then the forces acting in the y direction is N1+N2+N3 and (m1+m2+m3)g
    forces acting in the x direction is fk and F

    which fk=u*(N1+N2+N3)

  5. Doc Al

    Staff: Mentor

    Right. Keep going. Apply Newton's 2nd law.
  6. so the sum of Fy= N1+N2+N3-Mg=0, where M is m1+m2+m3
    so N1+N2+N3=Mg=> where N1+N2+N3=94.08

    so the sum of Fx= -fk=Ma


    so -7.15008=9.6a
    a=-0.7448 m/s^2
    this is small... is this right?
  7. Doc Al

    Staff: Mentor


    ΣF = ma
    But ΣF ≠ -fk. Friction is not the only horizontal force acting.


    No, as explained above.
  8. Okay so where does fk play into?
    Last edited: Sep 24, 2010
  9. Doc Al

    Staff: Mentor

    Friction is one of the horizontal force acting on the blocks. What direction does it act?

    What other horizontal forces act? Look at your diagram.
  10. is it sum of Fx= fk+F=ma?
  11. Doc Al

    Staff: Mentor

    Yes, but make sure you have the correction directions and thus the correct signs for the forces. (Hint: Let 'right' be positive and 'left' be negative.)
  12. Okay, so I did -fk+F=Ma

    -7.15008+75=9.6a => a=7.0677 m/s^2 (wrong)

    what did I do wrong?
  13. Doc Al

    Staff: Mentor

    Looks fine to me. Why do you say it's wrong? Did the problem even ask you for the acceleration? (Finding the acceleration is just a step towards the required answer.)
  14. You're right I need the force between m2 and m3...

    So can I do Sum of F= (m2+m3)*a?
  15. Doc Al

    Staff: Mentor

    That won't tell you anything about the forces between m2 and m3. Hint: Analyze m3 alone or m1 + m2 together.
  16. So the only forces acting on m3 in the x direction is the force from m1 and m2...
    in the y direction it is N and m3g

    So for the force in the x direction from m1 and m2 (i'll call P12)

    I'm not sure how to calculate P12...
  17. Doc Al

    Staff: Mentor

    No. What about friction? Which block is m3? Your diagram isn't labeled.

    Once you account for all the forces, the only unknown will be the force between m2 and m3. You'll solve for it.
  18. So then sum of F = F-P12-fk=ma?

    and P12 is my force bwtween m2 and m3 right?
  19. Doc Al

    Staff: Mentor

    Right. (Assuming m3 is the leftmost block.)
  20. Sorry m3 is the right block i'm trying to figure out the right block and middle

    so if it is the right block does force even apply?

    so I did the sum of Fx = -P12-fk=m3a is that right?
  21. Doc Al

    Staff: Mentor

    OK, that's different.

    The applied force F acts on m1, not on m3.

    Which direction does m2 push on m3?
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