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Newtons law - cart question

  1. Oct 25, 2010 #1
    1. The problem statement, all variables and given/known data
    a grocery cart is being oused with a force of 450 N at an angel of 30degree to the horizontal. if the mass of the cart and groceries is 42kg.

    a) calculate the force of friction if the coeffcient of friction is 0.60.

    b) determine the acclecration of the cart.

    2. Relevant equations

    a) it is the 30 degree angel from the handle to the cart.
    but i dont understand why you need to do 450sin(30) and add it by Fn=412.02N

    3. The attempt at a solution
    Ff= muFn
    450sin (30)= 225N

    Fn= (42)(9.81)
    = 412.02N

    Fn= 225N + 412.02N
    Fn= 637.02N
    ff= 637.02N * 0.60
    = 3.8 x 10^2 N

    i got the answer but why do u need to find the vertical component and add it to Fn . what does the vertical component represent? and why does this give you FN?
    please help me :-)
  2. jcsd
  3. Oct 25, 2010 #2
    Since you're on an incline, Fn no longer equals mg, but rather mgcosx, use that in your Ff equation.
  4. Oct 25, 2010 #3


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    Vertically there is the weight acting downwards, the normal reaction upwards and the vertical component.

    Since the resultant in the vertical direction is zero, then Fn+Fy-component=mg
  5. Oct 25, 2010 #4
    wait so is it like the force going downward + normal force = Ff?
    oh so we are finding the force acting downward because the handle is floating . not on the surface.
    do you understand what im saying? so it this number like replacing Fg?
    i think im understand it but still little confused. thanks for help everyone.
  6. Oct 25, 2010 #5


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    Not at all.

    In your free body diagram, vertically, there would be the normal force acting upwards and the weight acting downwards.

    Since you did not state whether or not the 30 degrees was above or below the horizontal, I am assuming it is above which means that the vertical component points upwards as well.

    The sum of the forces point up = sum of the forces pointing down.

    Only 1 force is point down which is the weight, mg.
  7. Oct 25, 2010 #6
    oh i understand what u are saying. im really sorry but i still dont get why you have to add
    Fn with the vertical force . please don't give up on me.
  8. Oct 25, 2010 #7


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    As is, without splitting anything up. Do you agree that the forces acting are the normal force (up), the weight (down) and the force on the handle (at 30° to the horizontal)?

    Now if we take the force on the handle acting at 30° to the horizontal, you can split this up into two forces, a vertical force (acting upwards) and a horizontal force. The horizontal force causes the cart to move forward.

    Does the cart move up or down? It does not, meaning that the vertical forces point up = vertical forces pointing down.
  9. Oct 25, 2010 #8
    i am so sorry. but if the cart doesnt move down or up then doesnt that mean you have to use horizontal force?

    would the diagram look like this?
    ↑vertical force
    ↓ mg
  10. Oct 25, 2010 #9


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    Horizontally, there is that horizontal force and friction. In order to find the value of the friction, you need the normal force. That is why we are considering vertical motion at the moment.
  11. Oct 25, 2010 #10
    oh okay thanks you so much!!!!!!!!!:) o:)
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