Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Gravitational Force homework Question

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    AS seen in the figure, two spheres of mass m and a third sphere of mass M form a equilateral triangle. The net gravitational force on the central sphere from the three other spheres is zero. (a) What is M in terms of m? (b) If we double the value of m4, what then is the magnitude of the net gravitational force on the central sphere?

    My apologies I am on one of my universities computers that has all the programs like locked up and it's a mac, and I'm not that familiar with macs to begin with, so my only option at the moment is to describe the picture...

    Picture a equilateral triangle... at one vertex picture a sphere that is larger than the rest with a mass M. At the other two vertexes of the triangle picture a sphere that is smaller than the one of mass M that each of masses m. At the center of the triangle picture a sphere of mass m4...

    Imagine a equilateral triangle with this orientation
    Where the sphere of mass M is at the top most vertex of the triangle (with the point of view of being the bottom of being the side of the triangle closests towards the bottom of the screen).

    Answer: (a) M = m (b) 0

    2. Relevant equations

    3. The attempt at a solution

    I agree with the answer the worksheet provides me for (b) 0
    I however disagree with the answer the worksheet provides me for (a). I got M = sqrt(2)m. I'm not exactly sure if I am wrong or the answer on the sheet is right (there have been worksheets in the past that my professor has made that have had wrong answers)...

    Can anyone confirm that the answer the answer sheet is right or if I am?
  2. jcsd
  3. Sep 25, 2011 #2
    How did you get M = sqrt(2)m ?
  4. Sep 25, 2011 #3
    Well here's what I did, let me know if you can't follow my work
    attached is my work and the question

    Attached Files:

  5. Sep 25, 2011 #4
    Ok so I believe that this is more correct
    M = m sqrt(3)
    and here's my work
    Last edited: Sep 25, 2011
  6. Sep 25, 2011 #5
    see attached
    note that number 5 is the second question in my work below
    can anyone confirm that m sqrt(3) is correct or wrong or if it it's really just m all by itself

    Attached Files:

  7. Sep 25, 2011 #6
    I can't read question 5 on that picture. So I can't see what you have done.
  8. Sep 25, 2011 #7
    here we go
    I accidentally included a random pi/3 next to sphere 3 in the file below sorry

    Attached Files:

  9. Sep 25, 2011 #8
    Looking at your diagram, your angles are incorrect for an equilateral triangle. Everything else is correct.
  10. Sep 25, 2011 #9
    What would be the correct angles, I thought that the angles within a equilateral triangle were pi/3, I drew the center of the sphere to be at the origin of my coordinate system and took directly upwards to be the positive y direction, and directly towards the right from this point to be the positive x direction.

    If we draw three lines from the center of a equilateral triangle to each vertex three more equilateral triangles would be formed that were smaller and within the larger one. When I choose my coordinate axes I thought about the angle that was formed from each force vector and the angle from my coordinate axes. I don't see which angles I got specifically wrong and what they should be and thought that what I was doing was correct but I guess not. Also thanks for your help.
  11. Sep 25, 2011 #10
    The way you've drawn your diagram, if you draw three lines from the center to each vertex, each of the angles between two lines would be pi/3.

    If you add them all together, they would all add up to pi. This is not correct.

    You can find the correct angle between two lines by finding the value which those angles must add up to, and then dividing it by the number of angles.
  12. Sep 25, 2011 #11
    ah thanks
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook