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Newton's Second law problem

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the Net force on the system
    see thumbnail for given.

    Normal force action on object 1 - 150N
    Normal force action on object 2 - 346N
    Force of friction action on object #1 - 15N
    Force of friction action on object #2 - 34.6N
    M1 moves to the left

    2. Relevant equations

    F = ma
    FR = μmg

    3. The attempt at a solution

    [itex]\sum[/itex]Fx = (.1)(20)(10)cos(30) + (.1)(40)(10)sin(30)
    = 37.32N

    The answer should be 63.8N im not sure what i am forgetting
     

    Attached Files:

  2. jcsd
  3. Oct 2, 2011 #2
    Are the masses at rest or moving?
     
  4. Oct 2, 2011 #3
    i believe they are moving to the left
     
  5. Oct 2, 2011 #4
    Then there is motion and coefficient of KINETIC friction must be used.
     
  6. Oct 2, 2011 #5
    If i use (.2)(200cos(30)) + (.2)(400sin(30)) only equals 74N and the answer is 63.8
     
  7. Oct 2, 2011 #6
    Correcting the coefficient of friction is the FIRST step.
    Now you have to explain how you got the above expression.
     
  8. Oct 2, 2011 #7
    well that's were i get a bit lost. I need to find the net force, and i figured that if i used the x component of box 1 and box 2 so i got 200cos30 for box one, and 400sin30 for box two
     
  9. Oct 2, 2011 #8
    Now i suggest you draw a diagram for M1 only with all the forces acting on M1.
     
  10. Oct 2, 2011 #9
    O right, the applied force. Now would i add that to what i had gotten before? so 100cos(30) ?
     
  11. Oct 2, 2011 #10
    Why don't you use F(net) = Ma for M1 along the horizontal?
     
  12. Oct 2, 2011 #11
    so wouldnt that be Fx = Facos(30) - T
     
  13. Oct 2, 2011 #12
    Fa = ?
     
  14. Oct 2, 2011 #13
    force applied = 100N
     
  15. Oct 2, 2011 #14
    What about the friction on M1?
     
  16. Oct 2, 2011 #15
    .2(200cos(30)?
     
  17. Oct 2, 2011 #16
    Not correct.
    M1 is in vertical equilibrium. hence find the normal reaction on M1. THEN the friction force on M1.
     
  18. Oct 2, 2011 #17
    so it would be .2(100cos(30))?
     
  19. Oct 2, 2011 #18
    use equilibrium in the vertical direction for M1.In the vertical dir there are THREE forces.
     
  20. Oct 2, 2011 #19
    The way i would approach this problem is to first figure out how to find the force pulling at that certain angle using the coefficient of static friction since static means the forces are all in equilibrium and you can make the sum equal to zero. SO, you know that just before a netforce higher than zero is provided, the coefficient of static friction times the normal force is the friction.
     
  21. Oct 2, 2011 #20
    Normal force = mg - Force applied y
    (20*10) - 100sin(30) = 150N

    Fg = mg
    Fg = 200N
     
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