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Please an answer - 2 crates, a rope and friction

  1. Oct 17, 2013 #1
    please an answer -- 2 crates, a rope and friction

    here is the problem image


    http://tinypic.com/r/qxrqdu/5




    Weight A is connected via a string to Weight B. Weight B is connected to a string to Weight C. Mass of A is 21 KG. Mass of B is 5 KG. Mass of C is 10 KG. the kinetic friction between the table and the 5 KG weight is 0.22

    i got 2.75 M/S^2 for acceleration is that correct?

    i need to know whats the tension in the string between A and B. include steps plz
     
  2. jcsd
  3. Oct 17, 2013 #2

    turbo

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    Gold Member

    Can you show and explain your work? There are lots of talented people here, but they are not allowed to do your homework for you.
     
  4. Oct 17, 2013 #3
    well whats the point of this website?
     
  5. Oct 17, 2013 #4
    I figured out the acceleration part but I cant figure out the Tension!
     
  6. Oct 17, 2013 #5
    To get the tension in the string between A and B, draw a free body diagram on A, and write down the Newton's 2nd Law force balance on A.
     
  7. Oct 17, 2013 #6

    Ibix

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    Science Advisor

    As per the rules you agreed to when you created your account, you need to show your working so far before we will help.

    We will help you to do your homework. We will not do it for you.
     
  8. Oct 17, 2013 #7
    well since we don't have an angle. then 210-ft= 2.75(5)
    so 210-ft= 13.75

    -ft = 13.75-210
    ft= 196.25 ?
    is that correct!
     
  9. Oct 17, 2013 #8
    No. If you are doing a force balance on weight A, shouldn't the 5kg on the right hand side of your equation be 21kg?
     
  10. Oct 17, 2013 #9
    correct my bad got confused. using different weights in one problem does not make sense either.

    so 210-ft= 2.75(210)

    I got ft= 367.5 ?
     
  11. Oct 17, 2013 #10

    berkeman

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    Staff: Mentor

    You need to show your work on the problem before you may receive tutorial help here on the PF. Please check your PMs.
     
  12. Oct 17, 2013 #11

    berkeman

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    Staff: Mentor

    The point of this website is to help you learn how to learn. If we just spoon-feed you the answers, that will no help you to improve in your understanding of the subject, and will not help you to be able to figure out problems on your own (which is how real life works).
     
  13. Oct 17, 2013 #12
    ok I showed my work for that part. can anyone tell me if 367.5 is correct?
     
  14. Oct 17, 2013 #13
    No. It's not 2.75(210), it's 2.75(21). Also, you did the algebra wrong. You have the right idea, but you gotta be more careful with the math.

    Chet
     
  15. Oct 17, 2013 #14
    210-ft=57.75

    ft=152.25 ???
     
  16. Oct 17, 2013 #15
    can anybody check my answer?
     
  17. Oct 17, 2013 #16
    Yes, provided your solution for the acceleration is correct.
     
  18. Oct 17, 2013 #17
    how about the tension on the cord between A and B
    ft=152.25 ???
     
  19. Oct 17, 2013 #18
    That is what you calculated.
     
  20. Oct 17, 2013 #19
    yes.

    21KG(10M/S^2) -FT= ma
    210-FT=21(2.75)
    FT=152.25
     
  21. Oct 17, 2013 #20
    so? is it correct or not!
     
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