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Force of gravity on one object caused by four objects

  1. Apr 11, 2013 #1
    1. A square of edge length s is formed by four spheres of masses, m1, m2, m3, and m4. What is the x-component and the y-component of the net gravitational force from them on a central sphere of mass m5. State your answers in terms of the given variables. (Use any variable or symbol stated above along with the following as necessary: and G for the gravitational constant.)

    In the problem y is given positive direction up, x is positive right.
    m1 - +y,-x (top left corner); m2 - +y,+x (top right); m3 - -y,-x (bot left); m4 - -y,+x (bot right)




    2. The force of gravity between two objects equation: Fgrav = (G*M*m)/r^2



    3. I would assume due to it being a square, the radius between the center and the corner squared (r^2) would equal (s^2)/2 and the angle we're using is 45 degrees. That being said it should be the Fgrav * sin(45) for y component and Fgrav * cos(45) for x comp. so it should look like:
    Fx = (2Gm5cos(45))/(s^2) * (m2+m4-m1-m3)
    Fy = (2Gm5sin(45))/(s^2) * (m1+m2-m3-m4)

    I've worked it and reworked it thinking perhaps I missed something in the simplification process but no good. WebAssign (the computer program) says that it is wrong. Please help, my two roommates, one neighbor, and I all worked independently and got the same answer. The only other thing I thought of was maybe the radius of the spheres but that information is not given.
     
  2. jcsd
  3. Apr 12, 2013 #2

    haruspex

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    Your answer looks correct. How smart is WebAssign at figuring out whether two algebraic expressions are equivalent? Are you actually entering "sin(45)"? Might it expect radians? Would putting √2 instead of 2 sin(45) help?
     
  4. Apr 12, 2013 #3

    Simon Bridge

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    Those coordinates don't make any sense.

    You mean that m1 is at coordinate (x,y)=(-1,1)s/2 etc so that the square has sides length s and is centered on the origin?
    Why "assume"? Didn't you work it out? i.e. using a sketch of the situation ... the distance from center to any corner forms the hypotenuse of a 1-1-√2 triangle where the other sides have length s/2 ... therefore the distance is s/√2 and r2=s2/2 ... no "assume"ing needed.

    OK - so your trouble is with directions.

    You don't need to resolve the forces into x and y components -
    If you wanted to "brute force" it so the numbers come out automagically without much thought, then use: $$\vec{F}=\frac{Gm_1m_2}{|\vec{r}|^3}\vec{r}$$

    But since the geometry is simple you know how it will come out...
    just write out the magnitude of the force multiplied by a unit vector pointing in the direction of the force.

    note: cos45 = sin45 = 1/√2
    the unit vector pointing to the mass at position (1,1)s/2 is ##\frac{1}{\sqrt{2}}(\hat{\imath}+\hat{\jmath})##

    you have another shortcut in that two of the forces are opposite direction to the other two.
     
  5. Apr 12, 2013 #4
    Resolved it

    Turns out that it was the sin(45) and cos(45). The program I'm guessing didn't register it as 45 degrees.

    By substituting 2sin(45) and 2cos(45) for √2 everything worked out.

    Thanks for the advice guys!
     
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