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Newton's 3rd law

  1. Nov 16, 2006 #1
    Hi, I'm having problems solving this problem:

    A small box of mass m1 is sitting on a board of mass m2 with length L . The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box is us. The coefficient of kinetic friction between the board and the box is us.

    Throughout the problem, use for the magnitude of the acceleration due to gravity.

    Find Fmin., the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).

    Express your answer in terms of some or all of the variables us ,m1 ,m2 ,g ,L and . Do not include Force of friction in your answer.
     
  2. jcsd
  3. Nov 16, 2006 #2
    This is what i've done:

    I know that acceleration of board must be larger than the acceleration of the box.

    I've found the acceleration of the box to be a=Ff/m1 which then equals
     
  4. Nov 16, 2006 #3
    a= usg

    now i need to find the acceleration of the board.

    I've found that Fx=F-Ff

    now i need to find the acceleration of the board.

    I changed the Ff into usm2g

    F-usm2g=max

    then i changed m into m1 and m2 because the board has both masses

    f-usm2g=m1m2a

    then for a I got:

    a=f-usm2g/m1m2

    however this is wrong. if anyone can help me with this that would be great thanks
     
  5. Nov 16, 2006 #4

    PhanthomJay

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    You are mixing up your free body diagrams. If you are going to isolate the board, do not include the mass of the block. If you are going to isolate the system, do not include the friction force betwen the block and board.
     
  6. Nov 16, 2006 #5
    So for the board, would i just include a normal force, weight force, and the force applied
     
  7. Nov 16, 2006 #6

    PhanthomJay

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    No, if you're just looking at the board in a FBD, you have the applied force acting right, and the friction force between box and board acting left. You have the weight and normal force also, but you don't need them here to solve. The masss is just m2. Alternatively, look at the block and board system as a FBD. then you just have the applied force to consider, but the mass is (m1 + m2.)
     
  8. Nov 16, 2006 #7
    okay so would this make the horizontal component of the FBD for the board:

    max=Fapplied,x - Ffriction,x

    (m1+m2)a=Fapplied,x-usm2g,x

    a=Fapplied,x-usm2g,x/(m1+m2)
     
  9. Nov 16, 2006 #8

    PhanthomJay

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    Ah, there you go again mixing up your FBD's. For the board, m2a= Fapplied - usm1g
    For the box, m1a = usm1g
    and check your work with a FBD of the system, (m1 +m2)a = Fapplied

    Now you've got more than enough equations to solve for F. Note that for applied forces greater than F, the block will start to slide. This is sometimes not easily understood.
     
  10. Nov 16, 2006 #9
    okay, i'm starting to see how it's done and how i've mixed up my FBD. I think I just need to practice some more problems.

    I rearanged the m1a=usm1g to a=usm1g/m1

    i then substituted this value for a into:

    Fapplied=m2a+usm1g

    Thanks so much for your help, I really appreciated it!!! Now I'm going to try some more practice problems while it's fresh in my mind. thanks again!
     
  11. Nov 16, 2006 #10
    okay, i'm starting to see how it's done and how i've mixed up my FBD. I think I just need to practice some more problems.

    I rearanged the m1a=usm1g to a=usm1g/m1

    i then substituted this value for a into:

    Fapplied=m2a+usm1g

    Thanks so much for your help, I really appreciated it!!! Now I'm going to try some more practice problems while it's fresh in my mind. thanks again!
     
  12. Nov 16, 2006 #11
    okay, i'm starting to see how it's done and how i've mixed up my FBD. I think I just need to practice some more problems.

    I rearanged the m1a=usm1g to a=usm1g/m1

    i then substituted this value for a into:

    Fapplied=m2a+usm1g

    Thanks so much for your help, I really appreciated it!!! Now I'm going to try some more practice problems while it's fresh in my mind. thanks again!
     
  13. Nov 16, 2006 #12
    okay, i'm starting to see how it's done and how i've mixed up my FBD. I think I just need to practice some more problems.

    I rearanged the m1a=usm1g to a=usm1g/m1

    i then substituted this value for a into:

    Fapplied=m2a+usm1g

    Thanks so much for your help, I really appreciated it!!! Now I'm going to try some more practice problems while it's fresh in my mind. thanks again!
     
  14. Nov 16, 2006 #13
    okay, i'm starting to see how it's done and how i've mixed up my FBD. I think I just need to practice some more problems.

    I rearanged the m1a=usm1g to a=usm1g/m1

    i then substituted this value for a into:

    Fapplied=m2a+usm1g

    Thanks so much for your help, I really appreciated it!!! Now I'm going to try some more practice problems while it's fresh in my mind. thanks again!
     
  15. Nov 16, 2006 #14
    oh wow i'm sorry i didn't mean to put that many up, i don't know what happened
     
  16. Nov 16, 2006 #15

    PhanthomJay

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    [tex]you're welcome^5[/tex]!
     
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