# Force of gravity on one object caused by four objects

1. Apr 11, 2013

### drdude24

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. Apr 12, 2013

### haruspex

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?

3. Apr 12, 2013

### Simon Bridge

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.

4. Apr 12, 2013

### drdude24

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.