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Clockwise moments = anti-clockwise moments

  1. Feb 7, 2010 #1
    1. The rod shown in the diagram(mass 10g, length 100cm) is in equilibrium. Find the position of the pivot and the magnitude of the supporting force F.



    2.clockwise moments = anti-clockwise moments



    3. I'm not so sure here. I have unknown distances.
    Please see file attachment 2 - Distances 1-4.
    dtotal=100cm
    d1=30cm
    d2=20cm from 40g object to the (Weight force component) centre of rod.
    d3 and d4 are unknown

    (0.01xd1)+[0.04x(d2+(50-d4))]+(0.01xd3)=0.06xd4
    (0.01x0.30)+[0.04x(0.20+(50-d4))]+(0.01xd3)=0.06xd4
    So I have a couple of d-unknowns. Do I solve for one then plug back in to get the other?
    Can't see how this would work with [0.04x(d2+(50-d4))] this in the equation.

    Please advise?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Feb 15, 2010 #2
    Re: moments

    The first image is not for this problem - please ignore this.
    the .pdf file is correct.
    I am uploading the correct image here in a moment
     
    Last edited: Feb 15, 2010
  4. Feb 15, 2010 #3
    Re: moments

    ok.
    Ahh! So much work to get through :cry:

    Right - so i don't have unknown distances after all:

    I should start again:

    1. The rod shown in the diagram(mass 10g, length 100cm) is in equilibrium. Find the position of the pivot and the magnitude of the supporting force F.

    2.clockwise moments = anti-clockwise moments

    3.
    (weight1x30cm)+(weight2x35cm)+(weightGravityx15cm)+(weight3x35cm)
    F=(.65x0.01)+(0.35x0.04)+(0.10x0.15)+(0.35x0.06)=0.0565N

    Could someone let me know if this is the correct way to calculate this problem please?
     

    Attached Files:

  5. Feb 15, 2010 #4

    PhanthomJay

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    Re: moments

    I am not sure where you are getting the figure .35 from... When you sum moments and set their sum equal to zero for equilibrium, you can sum moments about any convenient point. Try summing moments about the right end of the beam, and solve for the distance of the pivot from the right end.....Note that your mass is in grams, which should be converted to weight units.....and the value of F you can get from looking at the net force in the y direction....
     
  6. Feb 15, 2010 #5
    Re: moments

    Hi:
    The 15cm was the distance of the weight of the beam from the center position.

    (10gx100cm)+(40gx70cm)-(Fx35cm)=0
    (0.01x1.0)+(0.04x0.7)-(Fx0.35)=0
    0.35F=0.038

    0.038/0.35=0.11N (2s.f.)

    Don't I also need to take into account the 60g?
     
    Last edited: Feb 15, 2010
  7. Feb 15, 2010 #6

    PhanthomJay

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    Re: moments

    Am I looking at the correct figure? The resultant of the 10g beam's weight acts at its center, or 50 cm from one end. The pivot force, F, acts at an unknown distance from the right end. I don't know where you get 35 cm. First find F from Newton 1 by summing vertical forces = 0. And then do the moment thing, but don't forget to include the moment from the beam's weight. Also, when summing moments about the right end, the 60 g mass has no moment arm, so its moment about the right end is 0.
     
  8. Feb 16, 2010 #7
    Re: moments

    To find the force:
    (10gx100cm)+(40gx70cm)+(10gx50cm)-x=0
    x=1+0.028+0.05=1.078N

    To find the pivot:
    Clockwise moments = Anti-clockwise moments
    1.078=Fxd
    F=1.078-d

    1.078=(1.078-d)xd
    1.078/1.078-d=d
    d=0.5 This seems wrong (algebra booboo)?

    Therefore
    Fx0.5=1.078
    F=1.078/0.5=2.158N
     
  9. Feb 16, 2010 #8

    PhanthomJay

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    Re: moments

    Nah, to find the Force, first forget about moments. Just use sum of forces in y direction = 0, per newton's first law for objects in equilibrium. Note that you have five forces involved: four of them are given and act down, and the fifth is unknown, which is the pivot force, F, acting up on the beam. Solve for it. Then sum moments to determine X.
     
  10. Feb 16, 2010 #9
    Re: moments

    You mean the sum of the forces and not the forces multiplied by distances? Just the four weights, mg?
    (10gx10N)+(40gx10N)+(10gx10N)+(60gx10N)-x=0
    10+40+10+60-x=0
    x=120N
     
  11. Feb 16, 2010 #10

    PhanthomJay

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    Re: moments

    Yes, but don't use x = 120 N, use F= 120 N. Othewise, you'll really get confused when you try to solve the distance x where F acts.

    Arre you familiar with Newton's 1st law, sum of forces =0?? That's where the 120 N comes from; 120 N acting down, therefore 120N must act up from F, which is the sole support.

    Now you can sum moments to solve for the distance x where the pivot must be located.
     
  12. Feb 16, 2010 #11
    Re: moments

    Arre you familiar with Newton's 1st law, sum of forces =0??

    I know Newtons first law to be:
    If there is no resultant force acting on an object,
    if it is at rest, it will stay at rest,
    if it is moving, it keeps moving at a constant velocity (at constant speed in a straight line).

    I'm not familiar with, sum of forces =0. I guess that from what I wrote above it means that if a body is at rest, the sum of the forces must equal 0 because it will remain in that state unless a resultant force acts upon it. It remains in equilibrium. And the same applies if it is moving.

    Taking moments about 60g
    (10gx100cm)+(40gx70cm)=(120N)(x)
    (0.01x1.0)(0.04x0.7)=120x
    0.00028=120x
    0.00028/120=0.000002333m

    Well. That doesn't seem right :yuck:

    So:
    sum of clockwise moments = sum of anti-clockwise moments
    (10gx50cm)+(40gx20cm)+(120N)(x)=(60gx50cm)
    0.005+0.008+120x=0.03
    0.013+120x=0.03
    120x=0.017
    x=0.00015m (2s.f.)

    Well, that seems a little short too :frown:
     
  13. Feb 16, 2010 #12
    Re: moments

    should i be taking moments about the pivot point?
     
  14. Feb 16, 2010 #13

    PhanthomJay

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    Re: moments

    You can take moments about any point you want , but choose a convenient one to simplify your calculations. You are very careless in your work......you forgot to include the moment from the weight of the rod's 10 gram mass,and you forgot to convert all mass units into weight units, and you multiplied when you should have added. You can take moments about the right end of the rod, or the left end of the rod, or the center of the rod, or even the center of the moon for that matter; no matter what, the location of the pivot will always be the same. Try it again, summing moments about the right end, and please don't forget what you missed last time (convert mass to weight, watch your math, don't forget the weight of the beam which acts at the center of the rod.....)........
     
  15. Feb 17, 2010 #14
    Re: moments

    Sorry.

    (0.1x100)+(0.4x70)+(0.1x50)=120x
    10+28+5=120x
    43=120x
    x=0.36 (2.s.f.)
    36.0cm

    This seems better. In this type of calculation where we are given the measurements in cm, should I change them to meter just to change them back to cm at the end. It seems a waste of effort but our school is always saying we should change to meters first.
     
  16. Feb 17, 2010 #15

    PhanthomJay

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    Re: moments

    Well, lemon, this time it was me who made the math errror, sorry,it happens to the best of us. I seldom use the metric system, so it confuses the heck out of me with all its decimal points, and because we just don't use it here in the USA. Anyway, no excuse, an objectwith a mass of 10 grams has a mass of .010 kg, which has a weight of 0.1 N; thus the force of the pivot on the beam is 1.2N, not 120 N as originally stated. So when you do your math, leaving the units of distance in cm, the result is x =36 cm, as measured from the right end of the rod. Now you could convert everything into meters, and end up with x = 0.36 m, your choice, then give the answer in cm by multiplying by 100. I find it odd that you were using x =35cm way back in the beginning before you did any calculations, which is more or less the correct answer, leading me to believe that you were given that value by a friend or book answer at the start :confused:, but anyway, your numbers look good now. Go metric!
     
  17. Feb 17, 2010 #16
    Re: moments

    Hi:
    Thanks PhantomJay.
    And no. There are no answers available to me. Only the ones I can come to here. This is a project to be handed in this Friday morning, with much still to do :(
     
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