1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Kinetic friction problem?

  1. Mar 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A system comprised blocks, a light frictionless pulley, and connecting ropes is shown in the figure. The 9.0-kg block is on a perfectly smooth horizontal table. The surfaces of the 12-kg block are rough, with muk (kinetic frict constant) = 0.30 between the block and the table. If the 5.0-kg block accelerates downward when it is released, find its acceleration.

    Here is the picture:

    2. Relevant equations
  2. jcsd
  3. Mar 19, 2012 #2
    Please show how you have approached the problem
  4. Mar 19, 2012 #3
    Thanks for the reply. I just didnt add it because it is completely wrong. But here it goes:

    (5g*9.8)-(.3*9g*9.812)/(5+9) = 1.6
  5. Mar 19, 2012 #4
    The kinetic friction constant is between the blocks or the table?
    The question states table is smooth but also provide a 'u' value
  6. Mar 19, 2012 #5
    I believe it is between the blocks...
  7. Mar 19, 2012 #6
    So your value of frictionak force is wrong.

    Remember kinetic frictional force is Nu where N is normal reaction between surfaces on which friction acts.
  8. Mar 19, 2012 #7
    would it be:
    (5g*9.8)-(.3*9g*9.812 x cos(30))/(5+9) =1.86
  9. Mar 19, 2012 #8
    Remove g from your equations.

    You are righting 9.8 . Right?

    Also why have you done cos (30 degrees).

    Its not right

    Draw FBD of 12kg (and 9kg) blocks.What is normal reaction between them?
  10. Mar 19, 2012 #9
    really not sure^^
  11. Mar 19, 2012 #10
    Where are you getting sin(30) and cos (30)?

    There is no way they will emter the equations.

    Can you post a description of the logic you are using in calculating Normal reaction?

    Just tell me what your free body diagram of 12 kg block looks like.
    (Dint worry I will help you thoroughly in case you dint't understand.At this point its imp to identify the flaw in your concept which will help you much more :-) )
  12. Mar 19, 2012 #11
    I got this mixed up with another problem thats why I added the cos... Sorry about that.
    so would just this be my fnorm?
    or just 9*9.81

    free body^^
    Last edited: Mar 19, 2012
  13. Mar 19, 2012 #12
    Unfirtunately Your FBD is wrong.(Its atleast not bery clear . No problem :-) )

    On 12 kg block two forces act in y direction

    N upwards (which is applied by 9kg block)and 12g downwards.What is acceleration of 12kg block in t direction?

    On 9kg block 3 Forces act in y direction its weight 9g downwards Normal reaction N ( due to 12kg block acting downwards.newtons third law action reaction pair)
    And normal reaction , lets call N2 due to table acting upwards
  14. Mar 19, 2012 #13
    Ok so the 12 kg blocks forces cancel each other out. The 9 kg blocks up and down forces also cancel each other out, but it also has the 12kg block pushing on it. So do I make the friction equal to: 12kg*9.81*x.3 ?

    I feel like what i Just wrote is completely wrong ^^ but it is my best attempt to comprehend it :(
  15. Mar 19, 2012 #14
    It is right :-)

    With a little more practice, all these questions will be childs play.So don't worry :-)

    I will just cite the reasons which will help you understand the full problem.

    The 5kg block pulls on the the system and starts moving with acceleration a in downward direction.

    So 5g-T=5a (where T is tension)

    Now look at 12 kg block .It is attached to table with a rope so it cant move in x direction.
    Below it lies 5kg block which can move .
    However, the friction between 12kg and the 5kg block would attempt to stop any relative motion between 12 kg and 5kg block.

    On 5kg block acts tension towards right and friction due to 12 kg block towards left.
    This friction acts to eliminate relative motion between 5kg and 12kg block (MARK THESE WORDS CAREFULLY.FRICTION PREVENTS RELATIVE MOTION BETWEEN TWO BODIES AND NOT ABSOLUTE MOTION)
    T-f=9a (accelerationof 9kg block is same as 5kg block as they are tied to same string)
    In y direction forces acting are its weight mg downwards, N downwards and Normal reaction due to table upwards.
    So N(2)=9g+N

    On 12kg block forces acting in y are Normal reaction and its weight and its acceleration in y is 0.

    So N=12g
    Also a tension force acts on it towards left and friction force acts on it towards right(reaction force of friction on 9kg block)
    Its acceleration in x axis is 0 as its tied to the table.


    So using equations which are relevant


    T-f =9a

    (also f<Nu i.e 12gu)

    Adding both equations 5g-f=14a

    So a=(5g-12ug)/14g.

    A good practice is to akways calculate other stuffs like N(2) T(2) and T and see if they are positive.
    Cause if they aren't chances are something is wrong.
    Last edited: Mar 19, 2012
  16. Mar 19, 2012 #15
    thanks so much for your help. You're a life save :)
  17. Mar 19, 2012 #16
    actually that is the friction is equal to 12 x 9.81 x .3 =35.3

    What is the acceleration of the second block then? I tried solving but my answer isn't 1 1.2 1.8 or 1.4 (those are the choices)
  18. Mar 19, 2012 #17
    Now you find the force acting on 5.0kg mass, subtract your friction force from it, and then use that F[itex]_{net}[/itex] to find acceleration.
  19. Mar 19, 2012 #18
    Hey, i edited my post on previous page that will help.
  20. Mar 19, 2012 #19
    ok so ffriction is 35.3
    the ftension from block 2 is = 5 x 9.81 = 49
    49-35.3 = 13 which is not the right answer.
  21. Mar 19, 2012 #20
    Hey.tension is not 5g.
  22. Mar 19, 2012 #21
    13 is your net force, not your acceleration. [itex]\Sigma[/itex]F = ma. And don't round so soon. 49-35.3 = 13.7, not 13.
  23. Mar 19, 2012 #22
    So a=(5g-12ug)/14g.
    the answer was 1. This formula gave me .1
    close enough I guess :)
  24. Mar 19, 2012 #23
    ... [itex]\Sigma[/itex]F = ma. 13.7 = (5+9)a. a = 13.7/14 = .98 [itex]\approx[/itex] 1. I don't see how you're getting 0.1.
  25. Mar 19, 2012 #24
    Tal144 refer to my post on previous page.

    It comes out to be 0.1m/s^2.

    So it can be 1 dm/s^2 :-)

    What are the units given?
    Last edited: Mar 19, 2012
  26. Mar 19, 2012 #25
    It said the answer was 1 m/s^2

    Thanks for your help guys!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook