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Homework Help: 400N child on swing

  1. Nov 12, 2007 #1


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    [SOLVED] 400N child on swing

    1. The problem statement, all variables and given/known data
    A 400N child is in a swing attatched to ropes 2.0m long

    Find the gravitational potential energy of the child earth system relative to the child's lowest position when

    a) the ropes are horizontal

    b) rope makes angle 30 deg with vertical

    c) child is at bottom of circular arc

    2. Relevant equations
    [tex] U_g= mgy[/tex]

    3. The attempt at a solution

    First of all for this question in general would I be finding the Wnet of the system and thus the change in potential energy of the system? Is that what they want when they say "find the gravitational potential energy of the child earth system?

    When they say a "400N child"
    does that mean I can consider that the force of the mass and the gravity already multiplied and thus I can just use that in place of mg

    a) I think that the distance y for a would be 2m b/c the ropes are horizontal and thus since the length of the ropes are 2m then it would be logical that the distance traveled would be 2m.

    Would I assume that the child is a distance 0.5m above the ground?
    I think I would.. assuming they want the net force

    Wnet= [tex]\Delta U_g[/tex]

    Ui= mgyi = 400N (0.5m)= 200J ====> does that seem sort of large??

    I would like to see if my thinking is correct before I go further..

    Could someone Please help me ?

    Thanks :smile:
  2. jcsd
  3. Nov 12, 2007 #2
    Where did the 0.5 come from?
  4. Nov 12, 2007 #3


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    well the book assumed heights on another question so...
    but seriously the child wouldn't be sitting on the floor right?
  5. Nov 12, 2007 #4
    Ah. Well, you don't need to assume anything for this. It asks for everything relative to the child, right? This means it doesn't matter how high off the ground the child is, only how high he/she gets from his/her lowest point.

    You know how high that is, so just use the Energy = Force * Height equation.
  6. Nov 12, 2007 #5


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    but how would I find the height for part b?

    b) gives the angle that it has with the vertical but I don't know how I'd find the height traveled by it.

    c) child=> at bottom of arc so I think that means it doesn't move and thus the height would be 0?
    wouldn't that give potential E= 0? I know that that can't be though...

  7. Nov 12, 2007 #6
    My interpretation is that the "child's lowest position" is when the "child is at bottom of circular arc" so I'd take that as datum and measure all other heights above it.

    This leads to the GPE in part c being trivially easy.

    For part b the rope is straight, at 30 degress to vertical and the child at the end of it. How much higher is the child compared to when the rope is vertical?
  8. Nov 12, 2007 #7


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    first..what is "datum"??

    c) isn't it...0m ? since the child doesn't move...

    this gets me confused since I know that the equation is

    Ug= mgy

    but if y= 0...that would make the rest of the equation = 0 too...

    b) I seriously don't know how I'll get the difference in height of the child compared with before with a 0 deg angle...

  9. Nov 12, 2007 #8
    Datum is Latin for "given" (and you thought this was the Physics forum!) and is used here in the sense it is used when mapmaking and surveying to indicate level zero. See http://en.wikipedia.org/wiki/Datum for more than you ever wanted to know.


    "but if y= 0...that would make the rest of the equation = 0 too...". Exactly. That's what I meant by trivially easy; it doesn't get much easier!


    It's so easy I don't know how to explain. I guess you are somehow making it harder than it really is. Have to go now. Will need a diagram if you don't figure it out before tomorrow.
  10. Nov 12, 2007 #9


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    Last edited by a moderator: May 3, 2017
  11. Nov 12, 2007 #10
    Thanks for the diagram. Yes you have to use the angle.

    Draw a horizontal from the mid point of the grey seat. Now you have a right angled triangle. The hypotenuse is the full length of the rope, lets call it L. The vertical side is the full length of the rope minus h, that's (L - h)

    The length of the vertical side is Lcosθ
    L is the length of the rope
    θ is the angle of the rope with the vertical


    Now I really am going!
  12. Nov 12, 2007 #11


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    Hm...Is this what you mean??
    I wasn't sure what you mean by vertical...b/c you state vertical twice ..once as L-h then 2nd as Lcos theta...

    http://img100.imageshack.us/img100/1567/88164147ft0.th.jpg [Broken]

    What I got from your explanation was that...

    L-h ==> the line that was drawn from the 2 different points of the swing...

    Lcos theta = L??? do I plug that into L in the L-h ??
    Last edited by a moderator: May 3, 2017
  13. Nov 12, 2007 #12

    this is what he meant by a right triangle.

    Use [tex]cos(\theta) = adjacent/hypotenuse[/tex] to find x.

    Then, to find how high the child has risen, subtract x from 2m.
  14. Nov 12, 2007 #13


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    Thank You both I get how to do the problem now :smile:
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