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Pressing a Block against a Wall

  1. Sep 30, 2014 #1
    1. The problem statement, all variables and given/known data
    I have this homework question and was able to set up the free body diagrams for the two blocks but got stuck forming the equations in part a. Please help if you can and give explanations - I want to understand the concept for my test coming up.
    A block with mass M is pushed with a force F0 at an angle theta against a frictionless wall. It is connected to another hanging block with mass m by an ideal string and pulleys.
    a. Apply Newton's Law in each direction for the two blocks to find 4 equations using the symbols given above.
    b. Find an expression for the acceleration of m (the hanging block).
    c. What does your answer in part b predict for the acceleration "a" if the hand stops pushing? Does this make sense?
    d. What is the acceleration "a" of m if theta = 23.5 degrees, m = 3.2 kg, M = 2.1 , and F0 = 4.9 N?

    2. Relevant equations
    Fnet = ma
    FG = mg

    3. The attempt at a solution
    a. *a1 = acceleration of block M a2 = acceleration of block m
    For M
    x: Fnet = 0 y: Mg - T - F0(cos theta) = Ma1
    For m
    x: Fnet = 0 y: T - mg = ma2
    b. a2 = (Mg - F0 cos theta - Ma1 - mg) / m
    * I think something went wrong here *
    **** Is the acceleration of block M and m different or the same? I cannot figure that one out.
    Is it perhaps that their accelerations are different with the hand present and the same with the hand not present?
    ----------------------------------------------------------------------------------------------
    --> I could not solve c or d because I don't know if "a" is the acceleration of block M or m or if their accelerations are equal.
     

    Attached Files:

  2. jcsd
  3. Sep 30, 2014 #2
    Is the string changing in length, or is its length constant? If its length is constant, what can you say about a1 and a2?

    Chet
     
  4. Sep 30, 2014 #3
    The length of the string is constant. Does this mean that a1 and a2 are equal?
     
  5. Sep 30, 2014 #4
    At every moment in time, the amount that M moves up has to match the amount that m moves down if the length of the string is constant.
     
  6. Sep 30, 2014 #5
    I was thinking that a1 and a2 are not equal because the while the same tension acts on both blocks, block M has the additional pressing force.
     
  7. Sep 30, 2014 #6
    What does the kinematics of the motion tell you?

    Chet
     
  8. Sep 30, 2014 #7
    I suppose that the two blocks have to move the same amount within the same amount of time since they are attached to the same string - therefore their accelerations are equal regardless of the forces.
     
  9. Sep 30, 2014 #8
    You suppose correctly.

    Chet
     
  10. Sep 30, 2014 #9
    It is not "regarless of forces". The tension in the string adjusts so that the net forces and the the two accelerations will be the same.
     
  11. Oct 1, 2014 #10
    I am still having difficulty forming the correct equations for part a. Here is my revised attempt.

    a.
    Mg - T - F0(cos theta) = Ma
    and
    T - mg = ma

    b.
    a = (Mg - mg - F0cos theta) / (m + M)

    Can someone perhaps point out what mistake I made? I think it has to do with the direction of the forces (positive versus negative), but I have reworked it and do not know how to fix these.
     
  12. Oct 1, 2014 #11
    Perhaps what I don't understand is which direction the acceleration of the two blocks is. If I knew that I would know which forces to set positive and negative.
     
  13. Oct 1, 2014 #12
    a. Apply Newton's Law in each direction for the two blocks to find 4 equations using the symbols given above.
    b. Find an expression for the acceleration of m (the hanging block).
    c. What does your answer in part b predict for the acceleration "a" if the hand stops pushing? Does this make sense?
    d. What is the acceleration "a" of m if theta = 23.5 degrees, m = 3.2 kg, M = 2.1 , and F0 = 4.9 N?

    Actually, I think I have solved the question. I have posted my newest solutions to parts b and d.

    b. a = (mg + F0cos theta - Mg) / (M + m)
    d. a = 2.903 m/s^2
     
  14. Oct 1, 2014 #13
    There is nothing wrong with what you originally did. You just defined "a" with the opposite direction than the way they did. Very nice job.

    Chet
     
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