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Statics - Friction problem

  1. Mar 8, 2014 #1

    Jud

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    1. The problem statement, all variables and given/known data

    Transtutors001_0f7b69b2-1e48-4ca4-8d27-89c487b0202a.PNG

    Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 35kg and the coefficient of static friction between the crate and the plane is μs = 0.40

    2. The attempt at a solution

    Fx = P cos 30 - 343.35 sin 30 + F = 0
    Fy = -P sin 30 - 343.35 cos 30 + N = 0

    F = (μs)(lNl) = 0.40 (N)

    so,

    Fx = P cos 30 - 343.35 sin 30 + 0.40(N)

    therefore,

    Fy = -P sin 30 - 343.35 cos 30 = -N

    so, F = 0.40 (P sin 30 + 343.35 cos 30)


    Sub into Fx,

    P cos 30 - 343.35 sin 30 + 0.40 (P sin 30 + 343.35 cos 30) = 0


    Now solving for P I get 51.6 N, which is incorrect.
    The answer is given. P = 49.5 N

    Can anyone give me an indication of where I have gone wrong please.

    Thank You.
     
  2. jcsd
  3. Mar 8, 2014 #2
    You went wrong when you plugged numbers in your equations very early. Work out the result symbolically, then plug in the numbers.
     
  4. Mar 8, 2014 #3

    Jud

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    P cos θ - mg sin θ + μs ( P sin θ + mg cos θ) = 0

    Still yields the same answer?
     
  5. Mar 8, 2014 #4
    This is not the result, this is just a step toward it. The result would have the form P = ... where the right hand side would not contain P.
     
  6. Mar 8, 2014 #5

    Jud

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    P ( cos θ - mg sin θ + μs (sin θ + mg cos θ)) = 0

    I'm having major trouble transposing.

    P(-mg sin θ + μs (sin θ + mg cos θ)) = -cos θ

    P(μs(sin θ + mg cos θ)) = -cos θ + mg sin θ

    P(sin θ + mg cos θ) = -cos θ + mg sin θ / μs

    P (sin θ) = (-cos θ + mg sin θ / μs) - mg cos θ

    P = (-cos θ + mg sin θ / μs sin θ) - mg cos θ)
     
  7. Mar 8, 2014 #6
    Open the brackets.

    P cos θ - mg sin θ + μs P sin θ + μs mg cos θ = 0

    Then collect like terms.

    P (cos θ + μs sin θ) - mg (sin θ - μs cos θ) = 0

    And this is just one step away from P = ...
     
  8. Mar 8, 2014 #7

    Jud

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    Sorry but I need bludgeoning in the head because I get 51.6N again, and I cannot see only one step available.

    -P = (cos 30 + 0.4 sin 30) - 343.35 (sin 30 - 0.4 cos 30)
    -P = -51.669N
    P = 51.669N
     
  9. Mar 8, 2014 #8

    SammyS

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    Solving

    P cos 30° - 343.35 sin 30°+ 0.40 (P sin 30° + 343.35 cos 30°) = 0

    for P gives 49.4689 .

    Check your algebra.
     
  10. Mar 8, 2014 #9
    P (cos θ + μs sin θ) - mg (sin θ - μs cos θ) = 0 does not lead to that. P (cos θ + μs sin θ) means P multiplied by (cos θ + μs sin θ), not P plus (cos θ + μs sin θ).
     
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