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Newton's law

  1. Oct 31, 2007 #1
    Here is the question but before I begin, I've noticed that their weight are expressed in N.
    Do I have to do a conversion? or is that correct? I was just confused..

    Peter and John are playing a game of tug-of-war on a frictionless, icy surface. Peter
    weighs 539N and John weighs 392N. During the course of the game, John accelerates
    toward Peter at a rate of 3.0m/s[tex]^{2}[/tex].

    a) What is the magnitude of the force that Peter exerts on John?
    F[tex]_{N}[/tex] = F[tex]_{P}[/tex] - F[tex]_{J}[/tex]
    F[tex]_{N}[/tex] = 539 - 392
    F[tex]_{N}[/tex] = 147N
    b) What is the magnitude of the force that John exerts on Peter?
    F[tex]_{N}[/tex] = -147
    c) What is the magnitude of Peter's acceleration toward John?
    d) Sarah decides to join the game as well. Now Peter pulls on Sarah with a force of
    45.0N[E] and John pulls on her with a force of 25.0N[N] What is Sarah's resultant
    acceleration, if she weighs 294N?
     
    Last edited: Oct 31, 2007
  2. jcsd
  3. Oct 31, 2007 #2
    Nice conceptual question! :approve:
    First things first: If there is any tug used for this tug-of-war, then that tug is of negligible mass. Note that this ensures same tension throughout the tug.

    "I've noticed that their weight are expressed in N.
    Do I have to do a conversion? or is that correct? I was just confused.."

    Go through your textbook. Mass has a unit kg, weight is a force and hence N is correct. [In fact, the common answer -- when somebody asks you, what is your weight? -- 50kg, is scientifically erroneous!]

    Secondly, I will suggest you to write any statement in Physics, whenever you have enough reasons to support your answer!
    "F = F - F
    F = 539 - 392
    F = 147N"

    With what logic did you write that? [Was it, just because answer required should be in units of N.. so why not subtract/add these two?]

    Furthermore, note that magnitude of anything is always positive!

    Okay, I will start with giving you hints.

    Assuming there has been used a light rope for the tug-of-war, the tension in the rope would be same.
    For each person you can apply F = m*a.
    And, note that the acceleration given is relative acceleration and NOT the absolute one (with respect to inertial refernce frame).

    Tell me, if this much hint is sufficient or not. What could you make out from what I have told you. Attack the first part, we shall go on to other parts later on.
     
  4. Oct 31, 2007 #3

    Doc Al

    User Avatar

    Staff: Mentor

    N stands for Newtons, the standard unit for force. Given their weights, you can figure out their masses.
    That's not right. Their weights represent the force of gravity (the earth) pulling them down. What you need to figure out are the horizontal forces that Peter and John exert.

    Hint: What are their masses? Hint 2: Use Newton's 2nd law.
     
  5. Oct 31, 2007 #4
    Thanks guys in advance.
    I'll begin with question a) & b)

    a) Since their weight is expressed in N, I have to find their mass by "m=N/g" right?
    then for Peter I would get;

    F[tex]_{p}[/tex]=539/9.8
    F[tex]_{p}[/tex]=55kg

    and with given unit a=3.0m/s (the surface is frictionless ice, a is equal for both??)
    F[tex]_{net}[/tex]=55 x 3
    F[tex]_{net}[/tex]=165N

    Therefore, the magnitude of the force that Peter exerts on John is 165N.

    b)

    F[tex]_{j}[/tex]=392/9.8
    F[tex]_{j}[/tex]=40kg

    F[tex]_{net}[/tex]=40 x 3
    F[tex]_{net}[/tex]=120N

    Therefore, the magnitude of the force that John exerts on Peter is 120N.

    I gave it a shot, is there anything wrong in the calculation?
     
    Last edited: Oct 31, 2007
  6. Oct 31, 2007 #5
    oops! I think I've made a mistake on a) with the acceleration unit.

    so would this work?
    F[tex]_{p}[/tex] = -F[tex]_{j}[/tex]
    ma=-(ma)
    55(a)=-(120)
    a=-2.2m/s

    so then I would go back to a)
    F[tex]_{net}[/tex]=55 x -2.2
    F[tex]_{net}[/tex]=-121N

    121N would be my new answer

    Is it correct?
     
  7. Oct 31, 2007 #6
    Nopes.. acceleration wont be same for them! Why would it be? [I told you, please reason out every step of yours!]
    Instead, since the rope is light, tension HAS to be same.
    Furthermore, I pointed it out.. (but, I guess you didnt take notice of it, or didnt understand it!)
    "John accelerates toward Peter at a rate of 3.0m/s."
    .. this acceleration is not of any individual w.r.t. the fixed ground. Instead, it is relative acceleration of John w.r.t. Peter (or, vice-versa).
    So, if acceleration of John is a1 and that of Peter it is a2, then
    |a1 - a2| = 3.0 m/s^2
    And as they must accelerate towards each other (why?), a1 and a2 must be in opposite direction. Taking one direction to be positive, and other negative we get,
    a1 + a2 = 3.

    Think of another equation from Newton's Law(s). And solve!

    P.S.: Letters in bold denote vector quantities.
     
    Last edited: Oct 31, 2007
  8. Oct 31, 2007 #7
    I'm sorry, I'm so confused, I don't even know what to think T_T
    so my last thread(quoted) is not correct as well?
     
  9. Oct 31, 2007 #8
    Yes, that is also not correct.
     
  10. Oct 31, 2007 #9
    See, tension in the rope is constant. So, both Peter and John would be pulled by same force, T, (not acceleartion). Now, since they have different masses, their accelerations would be different. Assume, their accelerations are a1 and a2. Now, as the accelerations would be in opposite directions (why?), using relative velocity (see my second post in this thread), a1 + a2 = 3.
    Did you get two equations? Read this post carefully, and you will be able to work out two equations!
     
  11. Oct 31, 2007 #10
    this is serious..
    I've been through this course without any serious problem so far
    but I'm definitely lost here.

    I should read the stuff all over again

    sry and thanks for your time saket
     
  12. Oct 31, 2007 #11

    Doc Al

    User Avatar

    Staff: Mentor

    keep it simple

    Despite the way the sentence is worded, I would assume that the acceleration is with respect to the ground, not with respect to each other. The "toward Peter" is just there to show direction.

    So you should be able to apply Newton's 2nd law to find the net force on John, which happens to be the force that Peter exerts on him (since that's the only horizontal force acting on John).
     
  13. Nov 1, 2007 #12
    It may be! It can be solved assuming either and only the answer given in the text could confirm what is exactly meant.
    I see your point "Doc Al", but I chose that because the question seemed to be checking student's understanding of vectors as well.
    Anyways, let the text-book decide what it wants to say! :rofl:
     
  14. Nov 3, 2007 #13
    Ok, I'm back with some new answers that I've come up with;

    First of all, their mass from N to kg.

    Peter's mass:
    539N = m(9.8m/s[tex]^{2}[/tex])
    m = 55kg

    John's mass:
    392N = m(9.8m/s[tex]^{2}[/tex])
    m = 40kg

    a) Peter exerting on John.
    (I still have no clue how to find the a of Peter before knowing that both of their force is equal.
    However, I know that F1=-F2 so I've used John's given unit to solve this.)
    F = (40kg)(3.0m/s[tex]^{2}[/tex])
    F = 120 N
    Since F[tex]_{1}[/tex] = -F[tex]_{2}[/tex],
    the magnitude of force that Peter exerting on John is also 120N.

    b) John exerting on Peter.
    Since F[tex]_{1}[/tex] = -F[tex]_{2}[/tex],
    the magnitude of force that John exerting on Peter is 120N.

    c) a = 120N / 55kg
    a = 2.2 m/s[tex]^{2}[/tex]

    d) Fnet = Fpeter + Fjohn
    Fnet = (45.0)[tex]^{2}[/tex] + (25.0)[tex]^{2}[/tex]
    Fnet = 51.50 N

    ma = 51.50 N
    a = 51.50 N / 30 kg
    a = 1.72m/s[tex]^{2}[/tex]

    tanθ = opp/adj
    tanθ = 25.0 / 45.0
    θ = 29 degrees

    Therefore, Sarah's acceleration is 1.72m/s[tex]^{2}[/tex] E29N.

    Is this correct?
     
    Last edited: Nov 3, 2007
  15. Nov 3, 2007 #14

    Doc Al

    User Avatar

    Staff: Mentor

    Looks perfect to me.
     
  16. Nov 3, 2007 #15
    Finally!
    Thanks Doc Al, as always :)
     
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