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Homework Help: Magnitude, force and acceleration questions

  1. Dec 30, 2006 #1
    1. The problem statement, all variables and given/known data
    I find the physics book I have doesn't explain in great detail how to do these types of questions. I would like some feedback on the answers I have come up with. Thanks in advance.

    Peter and John are playing a game of tug and war on a frictionless icy surface. Peter weighs 539 N and John weighs 392 N. During tthe course of the game, John accelerates toward Peter at a rate of 3.0 m/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 exerts on Peter?
    (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 45.0 N (E), and John pulls on her with a force of 25.0 N(N). WHat is Sarah's resultant acceleration, if she weighs 294 N?

    2. Relevant equations

    magnitude of force = multiply mass by acceleration
    a=appplied force /mass
    3. The attempt at a solution

    First I calculated John and Peter's mass by dividing the Newton weight by the force of gravity which is 9.8 m/s^2

    Therefore Peter weighs 55 kg and John weighs 40 kg.

    (a) I multiplied John's mass by John's acceleration
    =40 kg X 3.0 m/s^2
    =120 N

    (b) 120 N. Based on Newton's third law that states they exert equal and opposite force on eachother.

    (c) acceleration = Fapp / mass = 120 N /55 kg = 2.19 m/s (This one I am really not sure on at all)

    (d) Sarah weighs 30 kg.
    F net = (45 N)^2(25N)^2
    = 2025 + 625
    =51.48 N

    Therefore to find Sarahs acceleration
    a= 51.48 N/30 kg = 1.71 m/s ^2

    to find the specific degrees of direction I tried to draw a free body diagram to help me viaualize this problem better.

    I came up with Tan = opposite /adjacent = 45N / 25 N = 1.8
    =60.9 degrees
    Therefore the resultant acceleration would be 1.71 m/s ^2 (N 60 degrees E)

    Any feedback would be of great help.
    Pharm 89
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 30, 2006 #2
    Why do you think that Peter's mass doesn't count? Also, why did you round down from 60.9?
  4. Dec 31, 2006 #3
    Mindscrape asks a very good question.

    To understand this problem better, you might consider a tug of war on a normal surface (one with friction). Draw FBD's of all the forces on each player (there are three, counting the tension in the rope). Which force allows the player to win (hint: it's not the forces on the rope).

    Then consider what must happen to each player if this game is played on a frictionless surface.


    p.s. I added a note to our last discussion, where I found a mistake I made. Just want to make sure you saw that.
  5. Jan 1, 2007 #4
    Happy New Year!!
    Thanks mindscrape and Dorothy for those comments.
    That makes sense that for question (a) you need to account for Peter's mass because they are asking for the force Peter exerts on John. Would it then be Peter's mass (55 kg) X 3.0m/s^2. = 165 N??

    Fo questions b, c and d do I have the right procedures or should I be rethinking how I go about all these questions.
    Pharm 89
    PS For question (d) I should have rounded the degrees to 61.
  6. Jan 1, 2007 #5
    Have you by any chance does pulley systems yet? If so, this is the same thing.

    Anyway, you have the right the procedure for the rest of the questions, but your approach to the first and most essential part is wrong. Without doing any calculations, think about this intuitively. I think the rope example is sort of crappy because rope has slack and would vary as some kind of hyperbolic trig function, so picture it as a metal rod instead. Should Peter and John have the same acceleration or different accelerations? When Peter is pushing with whatever force he pushes with to accelerate does he have to push just John, or does he also have to apply a force to get him going? Try to visualize what would happen, and then build your equations.
  7. Jan 1, 2007 #6
    Thanks Mindscrape. I havn't done pulley systems yet. Im looking at part (a) with a completely different formula now.Let me preface this formula with some general comments aboout "magnitude of the force". This is what I have learned: (1) force of gravity is a mutual attraction that the 2 objects have toward one another. Therfore, the force of gravity between the 2 objects is the same except in opposite directions because of Newton's third laws of motion.

    (2) force of gravity between object A and object B should depend on the mass of Object A and Object B. The larger the masses the larger the force of gravity between them Therefore the force of gravity between 2 objects depends on the masses of BOTH OBJECTS, NOT JUST ONE.

    This is the formula I have:
    Fg = G(Ma) (Mb) / r^2

    Ma = mass of Object A in kg (Peter = 55 kg)
    Mb = mass of object B (John = 40 kg )
    r= distance between 2 objects in metres
    Fg = magnitude of force of gravity between objects A and B.
    G -gravatational constant = 9.8 m/s^2

    Therfore I would have to find the distance between Peter and John
    found the distance based on knowing John accelerates toward Peter at a rate of 3.0 m/s and used the distance formula.
    for the Time - i based on it 1 second and velocity = 3.0 m/s
    d= v1(t) + 1/2g(t)^2
    = 3.0 m/s ^2(1s) + 1/2 (9.8 m/s ^2) (1s^2
    = 34.9 m
    Then plugged this info into the formula fo rthe magnitude of the force of gravity
    Fg = (9.8m/s^2) (55kg) (40kg) / (34.9 m)^2
    = 17.8 N

    I would assume that for part B - the force that John exerts on Peter would be the same, the only difference is it is in the opposite direction.

    Am I getting any closer to understanding this problem or I am confusing things more.
    Thanks for the help.
    Pharm 89
  8. Jan 1, 2007 #7
    The formula you used is generally used for celestial bodies, and the G is not the 9.81m/s^2 "g" you are used to, rather it is 6.something*10^-11 units of some sort.

    I guess I unintentionally confused you thoroughly. You have to have played tug of war at some point in your life. The first time you went through the problem you put in some numbers that probably looked good, and produced a reasonable numerical answer, but what you need to check if what you put into the formula makes sense physically. Basically what you found was the force John would have been exerting on the ice if he was merely running on the ice. You can't do this for the rope scenario because John is part of a system.

    Would it be easier playing tug of war against a mouse or an elephant?

    You know that you would rather play against somebody lighter than somebody heavier.

    Now, here is your check for the second part. If you are playing tug of war with somebody, what happens when you start to win or (the more common case among physicists) lose? Are they going at a different speed than you are?

    I think it would be tough to have a winner if all people went at different speeds. Also, in response to your question for part B, the pushing and pulling forces have to be the same, unless you are using a quantum rope, because otherwise different parts of rope would be going in different directions and such.

    I remember when I first took high school physics that I hadn't even formally done high school algebra, which might have turned out to be an advantage. It took me a while to realize, but physics is so much different from math because math only gives you the tools to solve the physics. Math has a lot of formulas that you just systematically carry out, whereas physics is about taking situations and applying the situation to the math. Fortunately, with mechanics at least, you have already know the answer. You don't realize it, but your everyday experiences are a cheat sheet, use it!

    If you still don't get your problem, look up an example of a pulley and make sure you really nail the pulley problem. Then go back to this, and see what is the same, you should find that it is all very similar.
    Last edited: Jan 1, 2007
  9. Jan 1, 2007 #8
    I think this is all getting a bit more complicated than this problem deserves. Most pulley problems (in my experience), have a force which directly opposes the pull of the rope (often gravity). That's not true in this problem. And Newton's law of gravity is way more than you need.

    My suggestion is to keep it simple. There is only one force at work here, which is supplied by the rope. Since there is no friction, gravity plays no role --- the answer to this problem, including the tension on the rope, would be the same if this were conducted in deep space as on a frictionless surface. It would be the same on mars as on the moon, as long as friction stays out of it. Well, I am getting poetic, but I hope you take my point :smile:

    Anywhere you are, a given force will accelerate a given mass the same amount.

  10. Jan 2, 2007 #9
    Well, it would be similar to a simple two mass pulley problem because the system is accelerated - just like the tug of war. One is due to the force some guy applies that accelerates the system, the other is due to the force of gravity accelerating the system.
  11. Jan 3, 2007 #10
    That's true. Although in this case the two people are both accelerating towards each other, probably at different rates.
  12. Jan 3, 2007 #11
    Actually, this is embarrasing. I hope he doesn't give up on the Physics Forums... but, I think he did fine. I just went back and re-read his original post, and I must have misread his solution. He did just fine. We've been torturing this poor guy for nothing.

    Wowzers. Really sorry, Pharm89. Please don't give up on us.

    Last edited: Jan 3, 2007
  13. Jan 3, 2007 #12
    Sweet, so I am actually understanding this physics problem. Was I right on (a) with my very orginal post with 120 N or on my second attempt in which I incorporated Peter's weight which gave me 165 N.

    Thanks for all the help.
    Pharm 89:smile:
  14. Jan 3, 2007 #13
    I think you were right in your first post.
  15. Jan 4, 2007 #14
    No, it isn't right, sorry. :(

    Picture that John is on the left and Peter is on the right. Peter runs to the right at 3m/s^2. In order to do that Peter pushes against the ground, and the ground pushes back with the equal force, a miniature third law problem.

    Equation for Peter: [tex]\Sigma_{(F_{Pnet})x}=F_P-T_{J on P}=m_p a_x[/tex]

    Equation for Paul: [tex]\Sigma_{(F_{Jnet})x}=T_{J on P}=m_J a_x[/tex]

    by Newton's Third Law: [tex]T_{J on P} - T_{P on J}= 0[/tex]
    or equivalently
    [tex]T_{J on P}=T_{P on J}[/tex]

    I'm going to rename the tension of the rope as T_R. Now solve for Peter's equation with the third law constraint.

    [tex]m_p a_x = F_P - T_R[/tex]

    Sub John eq into Peter Eq through tension constraint

    [tex]m_p a_x = F_P - m_J a_x[/tex]


    [tex]F_p = a_x (m_p + m_J)[/tex]

    Solve for F_p, you can do that.

    Now you have to figure out what the rope's tension is, which is T_R. Accelerations are equal because the rope isn't elastic. They have to be equal or else physics wouldn't work.
    Last edited: Jan 4, 2007
  16. Jan 4, 2007 #15
    But there is no force other than the rope. Nobody is pushing on the ground, because the ground is frictionless.

    The only thing that F_P could be in this problem is the force that P is exerting on the rope. But that has to be equal to T_R, by the third law, so this equation says that ma = 0, which we know is not true, from the problem statement.

    In a tug of war on a surface with friction, then you would be correct: F_P would be the friction exerted against the ground by P's shoes, and it would oppose the rope. No such frictional force exists in this case.

    I would suggest to you that the accelerations can't be equal, because the same force is accelerating two different masses and there is no force opposing this. Even if there were friction, the accelerations would differ, because friction is proportional to mg, and the more massive player would have a greater force of friction resisting the pull of the rope. Finally, consider that people DO win tug of wars (I always lose, I have personally had this experience), so accelerations HAVE to differ, or no one would win.

    Consider these two cases:

    Equal masses on a frictionless surface:

    The game will end in a draw, as both players accelerate at the same speed to the center. By the third law, A pulls on B with the same force that B pulls on A.


    Unequal masses on a frictionless surface:

    The advantage goes to the more massive player. He will accelerate less, and so cover less distance.

    Another way to view this situation is to consider the center of mass. In both cases, the players will meet at the center of mass, which does not move, as we know. In the first case, the center of mass is in the middle of the rope: result, draw.

    In the second case, the center of mass is closer to the heavier player. Result: advantage to the player with the greater mass.

    But none of this involves any force opposing the pull of the rope.

    Last edited: Jan 4, 2007
  17. Jan 4, 2007 #16
    I don't know, maybe I am misinterpreting the problem. What I do know is that the problem is ambiguous, and if two masses are tied together and there is either a tension pulling one one of the masses (e.g. the force on the mass to the right) that what I posted will be true. In other words, if the rope is always taught then there is no possible way for one person to have more acceleration than another without the rope either being untaught or stretched.

    If, on the other hand, this is merely about how one of the two people pull on the rope to make the other person come towards them then what you say would be true; that the accelerations don't have to be equivalent because the rope will slack like a catenary.

    Regardless, a typical tug of war game would follow my analysis with some extra terms for friction.
  18. Jan 4, 2007 #17


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    The bottom line here is that Pharm was 100% correct in his original post, as very nicely explained by Dorothy in subsequent replies.

    The accelerations of each persons are different. The rope doesn't go slack in between the 2, it's the part of the rope that is left laying on the ice behind each of them that is slack (The rope length between the 2 'shortens' as each moves toward each other, requiring at least one of them to pull on the rope left hand over right hand, left over right, until they meet).

    BTW, all calculated and given accelerations in this problem are with respect to the ground, not with respect to each other. That should have been clarified in the original problem statement.
  19. Jan 4, 2007 #18
  20. Jan 4, 2007 #19


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    Last edited: Jan 4, 2007
  21. Jan 5, 2007 #20
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