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Magnitude of the force (Tug-o-war)

  1. Jul 29, 2007 #1
    Peter and john are playing a game of tug-o-war on a frictionless, icy surface. peter wieghs 539 N and john wieghs 392 N. during the course of the game, john accelerates toward peter at a rate of 3.0m/s^2.

    A) what is the magnitude of the force that peter exerts on john?

    B)What is the magnitude of the force that john exerst on peter?

    C)What is the magnitude of peters acceleration toward john?

    d) Sarah decides to join the game as well. now peter pulls on sarah with a force of 5.0 N [E], and john pills on her with a force of 25.0 N [N]. What is sarahs resultant acceleration, if she wieghs 294 N.


    ok so i think i can do D, just with the rules of a right triangle and using f=ma but how do i do abc i am soo confused on the magnitude of the force...please help???
     
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  3. Jul 29, 2007 #2
    They give you the persons weight, so I think you can get their mass from that... they also give you an acceleration. Somehow with an acceleration and a mass you can find a force
     
  4. Jul 29, 2007 #3

    mgb_phys

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    I assumed a tug of war is in a straight line so there are no triangleds to worry about.
    Draw a diagram with the people as blocks, then think about what forces are going in what direction, and which way people will move.
    Remember that if you pull on one of a rope with a force that same force must act at the other end of the rope. After that it's just f=ma.
     
  5. Jul 30, 2007 #4
    i still dont understand i am sorry i have mass but the magnitude of force ? would that be the persons mass multiplied by gravity? or would it be the persons wieght multipleid by the accelration im giving? cuz isnt that c. where i use the accelration im givin? i am very confused
     
  6. Jul 30, 2007 #5

    mgb_phys

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    For the first part of the question you have to find the mass of the two people, from knowing that weight = mass * g ( g=9.8 m/s^2)

    Then it's just f=ma for the given force and the mass you have just calculated. Post back here when you have tried this!
     
  7. Jul 30, 2007 #6
    a)539/9.8=m ,m=55 Fn = ma = (55)(0)?? = 0N??

    b)392/9.8=m m=40 Fn=ma = (40)(3) = 120N??

    c)=120N same as last question???
     
  8. Jul 31, 2007 #7

    mgb_phys

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    I haven't checked your figures but think about a and b.
    If one person exerts a force on one end of a rope shouldn't the same force be applied at the other end.

    c, asks you for an accelration not a force. What force do you have 'left over' to accelrate the person and how much would that force accelrate someone with that mass?

    You shouldn't try and do these classical physics problems just from the formula - you should understand what is happening what the forces are and where they balance, draw a force diagram.

    ps. But when you get on to quantum theory you shouldn't try and understand what is happening - just believe in the equations!.
     
  9. Aug 2, 2007 #8
    ok can u just explain to me wat i should do, i have found that peter wieghs 55kg and john wieghs 40kg john is moving toward peter P--<--0--<--J The magnitude of force of peter would be Fn=m(a) =55(-3)(-3 is the accerlation that john is moving toward peter at) a) -165 N on john Fn=m(a) 40(3) (3 is the accerlation john is moving toward peter at) b)=120N

    c)??? please help more...im sorry
     
  10. Aug 2, 2007 #9

    mgb_phys

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    You're nearly there:
    They are on icy surfaces so there is nothing to keep peter stationary when he pulls on the rope, ther fore they both accelerate toward each other.
    Because both are moving you have to use the total mass of both of them to work out the force using f=ma.

    Now if peter pulls john with a certain force that same force must pull back, you can't "lose" force in a rope.

    It says John is accelarating toward peter at 3.0m/s^2, but they are both moving so this is the same as saying that peter is accelarating toward John!

    It might be easier to picture if you didn't think of one pulling the other, think of a motorised winch or a spring between them pulling them both!
     
  11. Aug 2, 2007 #10
    still a little confused so lets just start on a) wat is the magnitude of the force that peter exerts on john? ok so i know john's mass is 55 kg now wat do i do? wat should i know ? Fn=ma wat is my accelration and is Fn the magnitude of the force?
     
  12. Aug 2, 2007 #11

    mgb_phys

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    The two people are being pulled together, so the total mass is m = ( 539+392)/9.8
    They accelerate toward each other at a= 3.0m/s^2, put this into F = m a

    The trick to part 'a' is realising that neither of them is fixed so they are both pulled together and 3.0 is the 'closing' acceleration between them
    That pretty much answers b,c as well. For part 'd' I would have to draw it!
     
  13. Aug 3, 2007 #12
    hahaha so wait u want me to put down on paper the exact same thing for a b and c oh dont worry about d i figured that one out first day but ya ok so heres the identical answers for a b c

    a)95 kg is the total mass f = m a = 95 (3) = 285N Peter exerts a force of 285 N towards john

    b)95 kg is the total mass f = m a = 95 (3) = 285N John exerts a force of 285 N towards peter

    c)95 kg is the total mass f = m a = 95 (3) = 285N Peters magnitude of his acceleration toward john is 285N


    ?????i dont think thats all right please give me solid help i am soo beyond confused over this simple problem
     
  14. Aug 3, 2007 #13

    PhanthomJay

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    The confusion may be in the wording of the problem itself. While John's and peter's acceleration with respect to each other are the same, their accelerations with respect to the ground are not. Newton 2 is applied using the latter. I am assuming that the given acceleration of john toward peter is 3m/s/s with respect to the ground. Apply newton 2 to a Free body diagram of John to calculate the rope tension acting on John. Use newton 3 to calculate the rope tension acting on peter. Then do a FBD of peter to calculate his acceleration with respect to the ground. Let's forget Sarah for the moment, that's a whole different problem.
     
  15. Aug 3, 2007 #14
    ohh so this is a tension problem? i do not know how since i have not learned tension yet well i have but not in the unit so how would they excpect me to answer this problem without knowing about tension? ok im going to give u my best possible answer P-->--0--<--J Can u explain to me what i know, what i have what equastions i can use and can someone please give me a free body diagram
     
  16. Aug 3, 2007 #15

    PhanthomJay

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    the tension in the rope is same as the force that peter exerts on John. And using newton's 3rd law, that must also be the force that John exerts on peter. Draw an FBD of John alone. Only the tension force acts on John in the horizontal direction. Calculate it from Newton 2. Then look at an FBD of peter and calculate Peter's acceleration.
     
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