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Force balancing problem

  1. Jan 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Two identical small charged spheres hang in equilibrium with equal masses as shown in the figure. The length of the strings are equal (0.21 m) and the angle (shown in the figure) with the vertical is identical (7 degrees). The acceleration of gravity is 9.8 m/s^2 and the value of Coulomb's constant is 8.98755e9. Find the magnitude of the charge on each sphere. Answer in units of C.

    *The masses are 0.02 kg.

    2. Relevant equations


    3. The attempt at a solution

    I know I have to find F(electrical) first and it is equal to the X-component, but I am completely lost otherwise. Thanks.
  2. jcsd
  3. Jan 9, 2009 #2


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    You know the horizontal force from the angle don't you?

    You know the distance of the separation and since the charges are equal then ... 1 equation - 1 unknown.
  4. Jan 9, 2009 #3
    So the horizontal force of the triangle is 0.21*sin 7 = .0255925621 and the distance of separation is .0255925621(2)=0.0511851242, right?

    The first number would be F and the second number would be r in Coulomb's law equation?
  5. Jan 9, 2009 #4


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    I think you should begin with a force diagram. Show ALL the forces on one of the spheres. Separate them into horizontal and vertical. List them here if you need more help after that.
  6. Jan 9, 2009 #5
    Okay, there's the force of gravity pointing downward, the force of electricity pointing horizontally, and the force of tension in a diagonal position as the hypotenuse of the triangle.
  7. Jan 9, 2009 #6


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    Maybe you want to check the units for force?

    Your first number is only units of meters isn't it?
  8. Jan 9, 2009 #7
    Would I have to multiply the first number by the product of mass and gravity (force and tension)? Is my second number correct?
  9. Jan 9, 2009 #8


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    Force is Newtons is kg m / s2 ... so multiplying by mass and gravity sounds like a plan ... but why would you multiply that force by the length of the cord?

    The second number is OK, but you could lose the precision.
  10. Jan 9, 2009 #9
    "You could lose the precision."

    Typical forum-crawling physics elitist, eh? Haha, just kidding... maybe, but not really.

    I would multiply those two quantities because the force of tension is evenly distributed on the string, right?
  11. Jan 9, 2009 #10


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    I'm thinking like this for the left sphere:
    Horizontal Forces: -Fe + Ft sin 7 = 0
    Vertical Forces: (similar)

    where Fe is the electric force, Ft the tension force in the string.
  12. Jan 9, 2009 #11
    I'm still not getting the right answers... let me write this out.

    length*sin(theta)*mass*gravity = K*Q^2/R^2


    I am getting 3.82*10^-8 for Q and it is incorrect.
  13. Jan 9, 2009 #12
    Wait... so I would just solve for Q in:

  14. Jan 9, 2009 #13


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    No, not correct. The electric force is NOT equal to zero.
    You have Horizontal Forces: -Fe + Ft sin 7 = 0, not Fe = 0.
    It is quite a complex problem and I have been holding back on you - I don't want to spoil your frustration with this problem because only such frustration can make you learn physics. This is a wonderful problem and your teacher will have great difficulty finding another one this hard if I spoil it for you.

    I will offer some hints. You cannot find the tension force, Ft (at least not until you have found the charge). So you can't solve the equation you have for the horizontal forces because it has two unknowns. You can't find Fe because you don't know the charge and you can't find Ft.

    Do not neglect the vertical part of the problem. It appears to be useless at first, but go through the routine of writing that the sum of the vertical forces is zero - because there is no acceleration so F = ma = 0. Ask yourself what parts you can find numbers for. It will turn out you CAN calculate something with one of the two equations, and that is the breakthrough you need to finish it. Good luck!
  15. Jan 9, 2009 #14


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    As I already pointed out length times mass times gravity is not just force. Those are the units of N-m.

    Lose the length.
  16. Jan 9, 2009 #15
    Thanks for all of the help so far. So it would just be sin(theta)*mass*gravity = K*Q^2/R^2?

    This is the last problem on my Internet Homework and I can't afford to guess again or enter an answer I am uncertain of.
  17. Jan 9, 2009 #16


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    Almost. Except Sinθ is on the Tension in the string.

    But happily you know that m*g = cosθ * T ... so that T = mg/cosθ

    This yields then sinθ * T = k*q2/r2 = mg* tanθ
  18. Jan 10, 2009 #17
    Yes, I've already used that and it yields the incorrect answer stated above. What did you get when you plugged and chugged?
  19. Jan 10, 2009 #18


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    I'm not seeing where you were using tan7°.

    I haven't calculated it out. Just be careful of units and powers of 10.

    You already had the value for r from the 2*sin7° *(.21) so ... looks to me like you don't have all that much to do.
  20. Jan 10, 2009 #19
    Alright. Let's run through this together.

    We can agree that R=0.051185.



    So what am I doing wrong?
  21. Jan 10, 2009 #20


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    I got more or less the same number. It might be an idea to run it through again being careful about the digits - 9.81 or better instead of 9.8 for instance. A neat way to reduce the calcs and chances of rounding inaccuracy is to use symbols until the very end. Using that trick that sin over cos is tan, you end up with
    kqq/rr = mg tan(7)
    so q = sqrt( mgr^2 tan(7)/k )
    Try running that through in one go with the calculator and rounding only the final answer.
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