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Shopping Cart

  1. Jun 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A shopper pushes a 7.5 kg shopping cart up a 13 degree incline. Find the magnitude of the horizontal force needed to give the cart an acceleation of 1.41 m/s^2.



    2. Relevant equations

    F = m * a

    3. The attempt at a solution

    I am not sure how to set up the F(x) and F(y) equations. I have tried 7.5 * 1.41 = 10.6 N, but I know that is not right because I have not accounted for the the 13 degree incline. I'm stuck, Please help!
     
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  3. Jun 21, 2009 #2

    dx

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    If one applies a horizontal force of F, what is the component of this force along the incline?
     
  4. Jun 21, 2009 #3
    Consider the forces acting up and down the incline. Down the incline, you have a component of the force of gravity. Up the incline you have a component of the horizontal force of the pusher. Using F = ma, and F = (component of force) - (component of gravity), you find the force F.

    Hint: If m = 5 kg, F = 18.5 N
     
  5. Jun 21, 2009 #4

    tiny-tim

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    Hi Dave25! :smile:

    Yes, total force up the slope = ma …

    so what is the component of the F up the slope,
    and what is the component of the weight?
     
  6. Jun 21, 2009 #5
    Thanks for all of the help everyone. So are these force equations right?

    F(x) = N sin13= mg
    F(y) = mg / cos 13

    I don't think they are cause I'm not getting the right numbers. I am trying to draw a free body diagram, but I am still confused. Any more helpful hints you guys could give me would be much appreciated.
     
  7. Jun 21, 2009 #6

    ideasrule

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    The question gives you an acceleration and asks you for a horizontal force needed to provide that. This means that the right equations must have both the acceleration and the force in it, so no, your equations aren't right.

    Try expressing the force on the object (1) parallel to the incline and (2) perpendicular to the incline in terms of the other variables (mass, acceleration, etc). That's easier than working with F(x) and F(y).
     
  8. Jun 21, 2009 #7
    I have seemingly tried all possible ways and yet the harder I try the more confused I get. I have looked in the back of my book to find the answer to be 28 N, yet I do not know how to get there. If someone could tell me the general equation I would really appreciate it. Thanks.
     
  9. Jun 21, 2009 #8
    Is this correct: mg(cos13 + sin13) / cos13 + sin 13) ? I saw this equation to a similar question but it does not seem to work on this one.
     
  10. Jun 21, 2009 #9
    How about this: my normal force = 10.6 N (approx. 11N)

    Then, 7.5 kg * 9.8m/s^2 * sin13 = 16.5 (approx. 17N)

    Can I add these together to get my answer of 28 N?
     
  11. Jun 21, 2009 #10
    So, will adding these forces like I have above give me the correct answer?
     
  12. Jun 21, 2009 #11
    1) Could you sum it up in one post please? :)

    2) Alright, I'll settle this. You need to consider the forces acting up and down the slope. If you have a cart of mass m on a slope, the component down the slope is mgsin(theta), and the other component is normal to the slope. If you push a mass m horizontally against a slope, the component up the slope is F cos(theta).

    These two forces are the only ones moving the mass up/down the slope in teh problem (other components are perpendicular to the slope and so don't move the object along the slope), and so [tex] F_{resulting} = F \cos \theta - mg \sin \theta [/tex], right? Now, F = ma, and you know the mass m and the acceleration up the slope a = 1.41 m/s^2. Plug it in and solve for F.

    PS: Do you know how to break a force vector into components? eg. if you apply a force on a block at an angle [tex]\theta[/tex] to the horizontal, can you break it up into a vertical and a horizontal force? Same kind of thing we're doing here.
     

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  13. Jun 21, 2009 #12
    Sorry about all the posts. I have a test tomorrow and I am kinda freaking out.
     
  14. Jun 21, 2009 #13
    Sure, sure. Do you understand the method now, any related questions?
     
  15. Jun 21, 2009 #14
    Yeah, I understand it now. Thanks a lot.
     
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