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Math Help - Forces Question

  1. #1
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    Forces Question

    Here is the question. I missed this day in class, so now I'm totally lost. Once I figure out how to get it set up, plugging things in and such won't be a problem I don't think.

    Two boxes are stacked on top of each other, each box has a mass of 15 kg. You are pushing on the bottom box with a force of 50 N. What is the acceleration of each box if the static and kinematic coefficient of friction between the box and floor is (i'm not sure how to make the right symbols work, but the letters after the u, which would be a mu, are subscripts) us = 0.45 and uk = 0.25 respectively?

    Help setting it up would be most appreciated.
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  2. #2
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    Quote Originally Posted by charitysmama View Post
    Here is the question. I missed this day in class, so now I'm totally lost. Once I figure out how to get it set up, plugging things in and such won't be a problem I don't think.

    Two boxes are stacked on top of each other, each box has a mass of 15 kg. You are pushing on the bottom box with a force of 50 N. What is the acceleration of each box if the static and kinematic coefficient of friction between the box and floor is (i'm not sure how to make the right symbols work, but the letters after the u, which would be a mu, are subscripts) us = 0.45 and uk = 0.25 respectively?

    Help setting it up would be most appreciated.
    Trick question, as the maximum static friction for the bottom box is 0.45*30*g~=132N is greater than the applied force neither box moves.

    Which is just as well as if it did the problem cannot be solved as the box-box coefficients of friction are not given.

    RonL
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  3. #3
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    So the answer would be that both boxes have an acceleration of 0? Say it wasn't greater than the applied force. How would I go about solving for that?
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  4. #4
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    Quote Originally Posted by charitysmama View Post
    So the answer would be that both boxes have an acceleration of 0? Say it wasn't greater than the applied force. How would I go about solving for that?
    1. You can't as there is insufficient information about box-box friction.

    2. But suppose you did have the friction information.

    Case1: No box-box slip = treat the stack as a single block of mass 30kg.

    Case2: Box-box slipping:

    You know the frictional force at the box-box contact so:

    horizontal force on lower box = applied force - box floor friction - box box friction.

    horizontal force on upper box = box box friction

    The box floor friction uses a mass of 30kg the box box friction uses a mass
    of 15kg.

    RonL
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  5. #5
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    ok... so to show my work for my homework on that problem.... would it be

    Fs = 0.45*30*9.8 = 132 N ?

    Do I do anything with the kinematic coefficient or no since the static force is more than the applied force?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by charitysmama View Post
    ok... so to show my work for my homework on that problem.... would it be

    Fs = 0.45*30*9.8 = 132 N ?

    Do I do anything with the kinematic coefficient or no since the static force is more than the applied force?
    Thats right as 132N>50N the boxes do not move and so there is no kinetic
    friction to worry about

    RonL
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