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Newtons laws problem

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data

    a box of Cheerios (mass mC = 1.1 kg) and a box of Wheaties (mass mw = 3.6 kg) are accelerated across a horizontal surface by a horizontal force applied to the Cheerios box. The magnitude of the frictional force on the Cheerios box is 1.6 N, and the magnitude of the frictional force on the Wheaties box is 5.5 N. If the magnitude of is 29 N, what is the magnitude of the force on the Wheaties box from the Cheerios box?

    2. Relevant equations
    F=ma


    3. The attempt at a solution
    i set the accelerations equal and solved for F, but it was wrong. Now I have no idea.
     
  2. jcsd
  3. Oct 8, 2009 #2
    the magnitude of what is 29N? Have you set up your FBD?
     
  4. Oct 8, 2009 #3
    oops sorry. I copy and pasted the question. but basically a 29N force pushes the cheerios box to the right and it hits a Wheaties box that is right next to it, to the right.
     
  5. Oct 8, 2009 #4
    and yes I drew an FBD, and got for the cheerios box F to the right and F of friction to the left. same for the wheaties box
     
  6. Oct 8, 2009 #5
    you also have the force of the cheerios box on the wheaties box. i.e Fcw = -Fwc. Your actual effect on the Wheaties box will not be the entire force, but will be something smaller. so what will it be?

    Your step to assume that the acceleration is equal is correct.
     
  7. Oct 8, 2009 #6
    would it be 27.4 because you subtract forces?
     
  8. Oct 8, 2009 #7
    i haven't done the calculation so I don't know the actual number. but your acceleration for the wheaties box will be just the force on the wheaties box - friction on wheaties box, not the entire 29N - wheaties friction. Since force wheaties = - force wheaties on cheerios, you can set up your cheerios equation to include this same wheaties force and then solve for ma of the system.

    Or perhaps you can consider treating them as a single object with 1 friction and 1 force, from which you can get acceleration. If you do use this, what's the next step to get the actual force of cheerios on wheaties?
     
  9. Oct 8, 2009 #8
    divide by the masses added?
     
  10. Oct 8, 2009 #9
    I need to find the force on the wheaties box, not the acceleration.
     
  11. Oct 8, 2009 #10
    I had it right the whole time but used the wrong mass..Thanks for the help!!
     
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