Four 9.0kg- spheres are located at the corners of a square of side 0.70 m. Part

Question

Four 9.0kg- spheres are located at the corners of a square of side 0.70 m.
Part A) Calculate the magnitude of the gravitational force exerted on one sphere which is in left and down corner of the square by the other three.
Part B) What is it's direction? Assume that x-axis is directed to the right.
Please solve your answer all the way through
Ok here is what I did but it says both answers are wrong
1======2
| _______|
| _______| Distance (D) of 0.70m
| _______|
3======4

Fg = (Gm1m2)/r2

F2 on 1 = GM2/D2 = (6.67 x 10-11 x 92 ) / (0.70)2
= 540.27 x 10-11 / 0.49
= 11.02 x 10-13

F3 on 1 = same as 2 on 1 = 11.02 x 10-13

F4 on 1 = GM2/R2 where R = Sqrt(D2 + D2) = D x root(2) = .98

Total Force = 2GM2/D2 + GM2/2D2
= (2.5)GM2/D2
= 2.5 x 11.02 x 10-13

= 27.56 x 10-13 N


Problem 6.9 Part B

As, F=mg sin?
27.56 x 10-13 = 9 x 9.8 x sin?
27.56 x 10-13/ 9 x 9.8 = sin?
Sin? = 0.3124 x 10-13
? = 31.24 deg

Explanation / Answer

SInce the forces acting on mass one is along +x-axis (whose unit vector is i), one is along y-axis (whose unit vector j) and thrid one along the diagnol so Total Force = GM2/D^2( i + j) + GM2/2D^2 [ cos45i + sin45j] = GM^2/D^2 [ 1.35i + 1.35j] = (6.67*10^-11) (9)^2 / (0.7)^2 [ 1.35i + 1.35j] = 14.88*10^-9 i + 14.88*10^-9 j Magnitude F = 21.0*10^-9 N And direction ? = tan^-1( Fy / Fx) = 1 so = 45 degrees

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