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Dynamics Coursework; Velocity/Acceleration/Kinetic Friction

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data
    The 125kg concrete block A is released from rest in the position shown and pulls the 200kg log up the theta=30degrees ramp.

    Case 1: if the coefficient of kinetic friction between the log and the ramp is Uk=0.5
    (a) Find the magnitudes of velocity and acceleration of the block A as it hits the ground at B

    Case 2: If it is changed to a log with 400kg, consider kinetic friction between the log and ramp Uk=0.5. Determine the value of the critical slope (theta), that if theta is larger than this value, the log will slip down when the concrete block A is released.

    Case 3: If theta=30degrees and it is changed to a log of 400kg, determine the value of the friction coefficient (the critical friction coefficient) that would result in a zero acceleration of the block.


    3. The attempt at a solution
    Attachments:
    'pic' shows a diagram of the problem
    '001' shows my cable and FBD for how i have set up my equations
    '002' shows my working so far for the first question

    Attempts:
    Case 1: I have achieved 2 equations as shown in 002
    0.5(173.205)-2T+200sin30=(200/9.8)Ac
    or 186.6025-2T=(200/9.8)Ac
    and 125-T=(125/9.8)Aa

    I currently have 2 equations with 3 unknowns, and dont quite know how to approach that.
    For Case 1, I simply have to work out those 3 variables and then find velocity of Block A using Aa and fall distance, but really need a little assistance working out those set of equations.

    Am currently working on other problems until I can get Case 1 done, then I will post attempts at Case 2 and 3, but if anyone can allude as to how I should approach them when I get there, that would be fantastic too.

    Thankyou!
     
  2. jcsd
  3. Apr 25, 2012 #2
    Replying because it doesnt seem that the attachments worked.
     

    Attached Files:

    • pic.jpg
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    • 001.jpg
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    • 002.jpg
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  4. Apr 26, 2012 #3
    I also found the third equation to use as 0=2Aa+2Ac.

    Using this I found T, and substituted it in to find Aa and Ac.
    However, by this method Aa and Ac and equal with opposite signs (as I should have predicted from the above formula). This is incorrect, so i'm stumped again.
     
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