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Math Help - vertical motion mechanics problem

  1. #1
    Senior Member furor celtica's Avatar
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    vertical motion mechanics problem

    a lift starting from rest moves downward with constant acceleration. It covers a distance s in time t, where s = (1/6)(g)(t^2). A box of mass m is on the floor of the lift. find, in terms of m and g, an expression for the normal contact force on the box from the lift floor.
    i got 5/6(mg) and 11/12(mg) using a= d/t^2 and v^2 = u^2 + 2as but the answer is supposed to be 2/3(mg)
    what is wrong with these formulae? i dont think i made any calculation mistakes
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  2. #2
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    Using the kinematic equation
    s = vo*t + 1/2*g*t^2
    you can see that the acceleration of the lift is 1/3*g.
    The apparent acceleration of the mass inside the falling lift is
    g' = (g -1/3*g) = 2g/3.
    So the reaction of the floor on the mass is 2mg/3
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  3. #3
    Senior Member furor celtica's Avatar
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    how were my formulae inappropriate?
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  4. #4
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    Quote Originally Posted by furor celtica View Post
    how were my formulae inappropriate?
    How did you get 5/6 mg and 11/12mg? Show your calculations.
    From where did you get a = d/t^2?
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  5. #5
    Senior Member furor celtica's Avatar
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    good point, where the hell did i get that?
    i guess i just developed from the general form of acceleration values, you know distance/time^2 but i guess thats wrong huh
    how exactly?
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  6. #6
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    \vec{P} = m\vec{g} and \vec{N} are the 2 forces applying on the box

    m\vec{a} = m\vec{g}+\vec{N}

    Using an y-axis oriented downwards with y=0 at time t=0

    ma_y = mg-N

    The acceleration of the box is the same as the lift one \frac{d^2s}{dt^2} = \frac{1}{3}g

    \frac{1}{3}mg = mg-N

    N = \frac{2}{3}mg
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