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Finding acceleration given mass and tension

  1. Oct 21, 2015 #1
    If there is a 70N tension in a rope attached to a 4.0 kg bucket, what will be the vertical acceleration of the bucket?

    I'm not really sure how to do this. I can find tension but not acceleration. I think I'm missing something simple. I'm also confused about the direction because it doesn't state whether it is going up or down.

    For the formula I got a=Ft-g and m+mg-Ft=a. Idk either are right
     
  2. jcsd
  3. Oct 21, 2015 #2

    mfb

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    Staff: Mentor

    Where do those formulas come from? Their units do not match.

    Did you draw a diagram with all relevant forces in it? That should always be the first step if you are unsure how to proceed.
     
  4. Oct 21, 2015 #3
    I drew an FBD with Fg downwards and Ft upwards
     
  5. Oct 21, 2015 #4

    jbriggs444

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    If those are the only two forces on the bucket, what is the net force on the bucket?
     
  6. Oct 21, 2015 #5
    What did your equation look like before you got this equation from it? It looks like you started out with the right equation but then made mistakes rearranging it.

    You know, for example, that Ft = 70 N and g = 9.8 m/s2, but subtracting them doesn't make sense because they have different units.
     
  7. Oct 21, 2015 #6
    Fnet=Ft-Fg then I turned fnet into ma and Fg into mg so it looked like
    Ma=Ft-mg then rearranged it to a=Ft-g by dividing m on both sides
     
  8. Oct 21, 2015 #7

    jbriggs444

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    That should be a lower case m on the left ##ma = F_t - mg##
    Now, when you divide the right hand side by m, what do you get again?
     
  9. Oct 21, 2015 #8
    You mean,

    ##ma=F_t-mg##,

    right?

    Then dividing the left side by ##m## does give you ##a##, but you got the wrong result when you divided the right side by ##m##.
     
  10. Oct 21, 2015 #9
    Who says that the rope is pulling upward on the bucket? How do we know that the rope isn't pulling downward on the bucket?

    Chet
     
  11. Oct 21, 2015 #10
    Doesn't the m on the right side cancel?
    And maybe this question doesn't require a direction because it doesn't tell me if it's going upwards or downwards
     
  12. Oct 21, 2015 #11
    Plug the numbers in and find out! Tell us the magnitude of each force, subtract them, and see if you get a net force that's equal to the mass times the acceleration.
     
  13. Oct 21, 2015 #12

    jbriggs444

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    There are two terms on the right hand side.
     
  14. Oct 21, 2015 #13
    Ohh so it's a=Ft/m - g?
     
  15. Oct 21, 2015 #14
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