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Forces on ramp

  1. Jan 9, 2016 #1
    1.
    The block of 50kg with m =30kg is in the verge of moving down the slope where theta = 20degree, determine the new value of m when the mass of block is increased more to 20kg .






    2. Relevant equations


    3. The attempt at a solution
    I have done in this way , is it okay ?

    (50-30)(9.81) = Fs

    196.2 = Us (50 x 9.81 cos20)

    Us= 0.43

    (70-m)(9.81) = 0.43(70 x 9.81 cos20 )

    m = 41.7kg
     

    Attached Files:

  2. jcsd
  3. Jan 9, 2016 #2
    Your way of solving it isn't correct! You cannot add two vertical vectors (weight of M and weight of m) and get a vector with horizontal AND vertical components! In the first equation, you are omitting the angle of the incline, you can't do that.

    From Newton's Second Law (assuming acceleration is zero and hanging block is moving down)

    ∑Fx = 0
    ⇒ mg - Mgsinθ - Fs - FT + FT = 0
    ⇒ Fs = g(m - Msinθ)

    Since the system is on the verge of moving, the static friction has reached its maximum and thus:

    → Fs(max) = Fs = μsFN

    Therefore;

    μsFN=g(m - Msinθ)
     
  4. Jan 9, 2016 #3
    using mg-Mg(sin theta) = Fs
    i have 30(9.81) -5099.81)(sin20) =Fs
    Fs= 126.5 N
    126.5=Us(50x9.81xsin20)
    Us= 0.75

    (70-m)(9.81)= 483.9 , m = 20.7kg , but thge ans given is 41.7 kg ,
    is the ans given wrong ?
     
  5. Jan 9, 2016 #4
    My mistake, I forgot to correct your second formula. Like I said, ##(M-m)g## is incorrect, so is ##(70-m)(9.8)##.

    This is the equation you should use, as derived previously from Fs=g(m-Msinθ):

    →μsFN=g(m - Msinθ)
    ⇒μsMgcosθ=g(m - Msinθ)

    This time M=70kg and m is unknown, you already have μs.

    Also, be careful with your trigonometry. The normal force is ##(50)(9.8)cos(20°)## not ##(50)(9.8)sin(20°)##

    Do you understand why?
     
    Last edited: Jan 9, 2016
  6. Jan 9, 2016 #5
    so , is my method shown above correct?
     
  7. Jan 9, 2016 #6
    No it is incorrect since you forgot to include the angle of the incline into your equations. Can you show me how you got your previous answer? Maybe it will help if I point out some mistakes there.
     
  8. Jan 9, 2016 #7
    this is what i gt
    30(9.81) -5099.81)(sin20) =Fs
    Fs= 126.5 N
    126.5=Us(50x9.81xsin20)
    Us= 0.75

    (70-m)(9.81)= 483.9 , m = 20.7kg
     
  9. Jan 9, 2016 #8
    which part of my qorking that i forgot to include the inclined angle?
     
  10. Jan 9, 2016 #9
    can you take at look on my another thread , why the tension from 100kg to the roller(up) and the tension from point B to roller doesnt cancel each other?

    in the above question , the tension cancel each other?
    https://www.physicsforums.com/threads/total-forces-acting-downward-on-the-plane.851538/
     
  11. Jan 9, 2016 #10
    First of all, like I said, normal force is
    $$Mgcosθ$$
    And NOT
    $$Mgsinθ$$

    So your answer of ##126.5N## is incorrect as well as your value of the coefficient of static friction of ##0.75##.

    Second of all, you should follow my derivation because this equation you are using

    ##(M-m)g = μMgcosθ##

    Is incorrect as well, it should be

    ##(Msinθ-m)g = μMgcosθ##

    Could you please show me the work you did by your own and how you got the correct answer? Please show how you used Newton's Second Law too so I could help you understand better!
     
  12. Jan 9, 2016 #11
    30(9.81)-50(9.81)sin20 = Fs
    Fs=126.5N,
    126.5= Us(50x9.81xcos20)
    Us= 0.27

    (70-m)(9.81) = 0.27 ( 70x9.81xcos20)

    m=51.7kg , is it correct?
     
  13. Jan 9, 2016 #12
    Not quite, where are you getting (70-m) from?
     
  14. Jan 9, 2016 #13
    when the mass of block is increased more 20kg , caqn you post your working again ?i'm confused now
     
  15. Jan 9, 2016 #14
    nvm , focus on the Fs first , you said that my method of finding Fs is incorrect... how to get it correctly ?
     
  16. Jan 9, 2016 #15
    Ok, my bad. I uploaded a picture of my work, hopefully this will help.
     

    Attached Files:

  17. Jan 9, 2016 #16
    that proves my Fs is correct
     
  18. Jan 10, 2016 #17
    ##196.2N## is not the value of Fs. It does work to find the mass of ##m##, but you get incorrect values for the coefficient of static friction and frictional force.

    Don't forget to use the correct sign for your forces... you should use ##-126.5N## instead of ##126.5N##.
     
  19. Jan 10, 2016 #18

    haruspex

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    You have that step right now, but it doesn't match what you did in the 70kg case:
    Restructure that expression so that it follows the same form as in the 50kg version. (there are two differences.)
     
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