Posted by **myra** on Sunday, May 27, 2012 at 5:57pm.

Four 7.5 kg spheres are located at the corners of a square of side 0.60 m. Calculate the magnitude and direction of the gravitational force on one sphere due to the other three.

- physics -
**Tiffany**, Wednesday, March 25, 2015 at 11:48pm
R=.6M

mass=7.5 kg

Going clockwise label the masses in a square. M1 top left corner, M2 top right corner, M3 bottom right corner, M4 bottom left corner.

Start with M1 and M2 use Newton's universal gravity formula;

F=G((M1*M2)/r^2)

Where G always equals 6.67*10^-11

Back to M1 and M2

equation:

Fg=6.67*10^-11((7.5kg*7.5kg)/0.6m^2)

Fg=1.04*10^-8 @ 0°

Same exact thing for M1 and M4 but instead of 0° it's 270°

For M1 and M3 it's a diagonal radius so to solve for the Diagonal radius make a right triangle and use the equation a^2+b^2=c^2 where c^2 is what we are finding.

Equation:

.6^2+.6^2=c^2

.72=c^2

Square root.72 To get the radius of .84

Now solve M1 and M3

Equation:

Fg=6.67*10^-11((7.5kg*7.5kg)/0.84m^2)

Fg=5.31*10^-9 @ 315°

Now you have to add the vectors to get the vector of C^2 by making a right triangle connecting M1,M2,M3 making M2 where the 90° is

From M1 to M2 we already know it's 1.04*10^-8 but she still need to find the cos to find the other part of the line.

Equation:

Cos(45=x/5.31*10^-9

X=3.75*10^-9

Y (aka M2 to M3) equals the same because 45 makes sin and cos the same

After finding that add

(1.04*10^-8)+(3.75*10^-9)

Getting; 1.41*10^-8 Remember both lines M1 to M2 and M2 to M3 are equal so they are both 1.41*10^-8

With that said to finalize this problem find C^2 now

Equation:

(1.41*10^-8)^2 + (1.41*10^-8)^2 =C^2

4.00*10^-16=CA

square root 4.00*10^-16 and you should get 2.0*10^-8

ANSWER: 2.0*10^-8 @ 45°

- physics -
**Julia **, Wednesday, November 18, 2015 at 7:50pm
I had been looking online for the answer to this question for 3 hours and this is the only process that I found was right THANK YOU

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