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Determine the force to enable the object to move down

  1. Nov 28, 2015 #1
    1. The problem statement, all variables and given/known data
    A force P acts as shown on 200kg block placed on a inclined plane . The static coefficient between the block and the plane is 0.5 . Determine the force P to ensure the block is in the verge of moving down.

    2. Relevant equations


    3. The attempt at a solution
    Wsin10 +P cos 20 = Fs
    P cos 20 = 0.5( P sin20 - W cos 10 ) -Wsin 10
    P cos 20 = 0.5( P sin20 - (200x9.81) cos 10 ) -(200x9.81)sin 10
    0.17P = 0.5( 0.34P - 1932) - 340.7
    0.94P = 0.17P - 966- 340.7
    0.17P - 0.94P = 1306.7N
    P =-1697N , but the ans given is P = 560N , which part of my working is wrong ?
     

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  3. Nov 28, 2015 #2
    20° angle is not reference to the inclined plane but to ground level.
     
  4. Nov 28, 2015 #3

    SteamKing

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    For one thing, the force P is applied 20° above the horizontal. You must calculate the components of P relative to the block on the incline to determine the correct normal force of the block.
     
  5. Nov 29, 2015 #4
    P cos 20 = 0.5( P sin20cos10 - (200x9.81) cos 10 ) -(200x9.81)sin 10
    0.17P = 0.5( 0.34P - 1932) - 340.7
    0.94P = 0.17P - 966- 340.7
    0.17P - 0.94P = 1306.7N
    P =-1697N , but the ans given is P = 560N , which part of my working is wrong ? my ANS STILL THE SAME
     
  6. Nov 29, 2015 #5
    It should be PCos(20 +10) not PCos(20)....and you to adjust that Sin 20 too.
     
  7. Nov 29, 2015 #6
    THANKS , I NOTED MY MISTAKE , I GT THE ANS FINALLY
     
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