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Marble Static Equilibium

  1. Sep 8, 2016 #1
    1. The problem statement, all variables and given/known data
    Two uniform, 75.0-g marbles 2.00 cm in diameter are stacked as shown in Fig. P11.75 in a container that is 3.00 cm wide. (a) Find the force that the container exerts on the marbles at the points of contact A, B, and C. (b) What force does each marble exert on the other?

    2. Relevant equations
    T = F*l = 0
    F = ma = 0

    3. The attempt at a solution

    I just cannot figure out the geometry of this problem, at all, which I need to even begin the question. How do I know where the point of contact of the two marbles is?
     

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  3. Sep 8, 2016 #2

    kuruman

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    The points of contact are labeled in the figure as A, B and C. How far do you think A and B are from the bottom left corner of the container? What about point C?
     
  4. Sep 8, 2016 #3
    A is 1 cm above, B is 1 cm to the right, and C I don't know. I know that Force B would be equal to 0.075*2*g, and all I can say about force C and force A are that they are equal and opposite to eachother
     
    Last edited: Sep 8, 2016
  5. Sep 8, 2016 #4

    kuruman

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    What can you say about the torques generated by all the external forces acting on the two marbles? Can you find expressions about some origin, say point A?
     
  6. Sep 8, 2016 #5
    Okay, well if I look at the torque of the first marble about point A, we have the torque from the force of the wall at A being zero, then we have the torque of the Force B being 0.15*g*1, and the torque of the second marble acting on the first, which I don't know how to find
     
  7. Sep 8, 2016 #6

    kuruman

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    Your last message is incomplete. Anyway, point B is better for calculating torques because two of them are zero about that point. I think you were going to say that you don't know how to find the lever arm of point C about B. You need to figure out the sine and cosine of the angle formed by the center-to-center distance and the horizontal. It's simple trig.
     
  8. Sep 8, 2016 #7
    My last message is complete I just forgot to put the period. Which two torques are zero about point B? And why are we looking for the lever arm of point C when point C is not in contact with the bottom left marble.
     
  9. Sep 8, 2016 #8

    kuruman

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    The torques that are zero about point B are the ones generated by the normal force at B and the weight of the bottom marble. Point C is not in contact with the bottom marble but it is in contact with the top marble. If you consider as your system the two marbles together then you don't have to worry about the contact force between the two marbles which is what you are asked to find in part (b). If you try to do this problem with a free body diagram (FBD) of only one marble, you will end up with too many unknowns and you will have to draw another FBD for the second marble. Therefore, you might as well draw the FBD for both marbles together in the first place and be done with. :wink:
     
  10. Sep 8, 2016 #9
    I see, okay. But why is it valid to treat them both as one system together since they are not attached to eachother?

    Okay, so if I treat them as a system together, then the lever arm at point C would simply be the vertical distance to point C, but we aren't given how high above the bottom of the jar point C is, how would I find that?
     
  11. Sep 8, 2016 #10

    kuruman

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    It doesn't matter that the two marbles are not attached to one another. The beauty of the FBD idea (thanks to Sir Isaac Newton) is that you can consider any number of objects as your "system" as long as you account correctly for the external forces (those outside the system) that act on the components of this system.

    Anyway, start at point B and draw a line to the center of the bottom marble then draw a line to the center of the top marble then to point C. Draw a right triangle that has horizontal and vertical right sides and the center-to-center line segment as its hypotenuse. Do the trig on this right triangle to find the overall vertical displacement that takes you to point C. As I wrote earlier, you need to figure out the sine and cosine of the angle formed by the center-to-center distance and the horizontal.
     
  12. Sep 8, 2016 #11
    How do you know what the center to center distance is? Clearly the horizontal distance of the centers would be 1, but how do you know the distance from one center to another?
     
  13. Sep 8, 2016 #12

    Nidum

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  14. Sep 8, 2016 #13

    kuruman

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    What is the distance from F to D? Hint: Point F is at the center of the circle and point D is on the circumference. How about from D to G? Add the two. (Thanks for the drawing, Nidum)
     
    Last edited: Sep 8, 2016
  15. Sep 8, 2016 #14
    Why did you draw a straight line between the two centers, isn't it possible to have the two centers not lie on a line through both their radiuses, I'd draw what I mean, but paint won't allow me to, something along these lines
     

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  16. Sep 8, 2016 #15

    kuruman

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    Imagine a plane that is tangent to the two marbles at the point of contact. The line connecting the center of one marble to the point of contact is perpendicular to that plane and so is the line connecting the center of the other marble to the point of contact. Since these two lines are perpendicular to the same plane, they must be parallel; since they pass though the same point (the point of contact) they form a straight line between the two centers, just like Nidum drew. Your drawing shows two line segments that do not go through the point of contact and are not a straight line from one center to the other. The center to center straight line must go through the point of contact.
     
  17. Sep 8, 2016 #16
    I see, that makes perfect sense thanks for the explanation. Okay, so now I form the triangle with horizontal 1 and hypotenuse 2, solving for the height gives me √(22 - 12 = √3. So a total vertical distance of √3 + 1. So the torque on the system from point C woould be Fc * (√3 + 1).

    So i get that Fa = Fc*(√3 + 1), I must have done something wrong since I previously got Fa = Fc when I balance the forces in the x direction.
     
  18. Sep 8, 2016 #17

    kuruman

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    It is still true that FA = FC. The contradiction arises because you forgot the torque generated by the weight of the top marble. Also, use symbols instead of a mix of symbols and numbers. If you use 1 for the radius, you might get in trouble when you plug in to find numerical values. Use R for radius, then plug in at the very end. When you eventually find the forces they should be a number times mg, the weight of one marble.
     
  19. Sep 8, 2016 #18
    Right, okay, so the center of gravity in between the two balls should be at a horizontal distance of 1.5 from the corner, which is 0.5 from point B, so the torque due to grabity would be 0.5*0.15g, so in the end we have 0.075g + Fa = (√3 + 1)Fa, which gives me 0.075g = √3Fa, and in the end, We get Fa = 0.42N = Fc. And Fb = 1.4715.

    b) For the force of each marble on the other, we have that for the forces in the x direction of the bottom marble, Fm * cosθ = Fa

    From the triangle we formed before, cosθ would be 1/2, so we get Fm = Fa*2 = 0.84N.

    Is that right?
     
  20. Sep 8, 2016 #19

    kuruman

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    That's right.
     
  21. Sep 8, 2016 #20
    Great thanks for the help!
     
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