A uniform sphere with mass 70.0 kg is held with its center at the origin, and a

Question

A uniform sphere with mass 70.0 kg is held with its center at the origin, and a second uniform sphere with mass 90.0 kg is held with its center at the point x = 0, y = 3.00 m. A third uniform sphere with mass 0.400 kg placed at the pointx = 4.00 m, y = 0.

What is the magnitude of the net gravitational force due to 70.0 kg and90.0 kg spheres on 0.400 kg sphere.

What is the direction of the net gravitational force due to 70.0 kg and90.0 kg spheres on 0.400 kgsphere.

Where, other than infinitely far away, could the third sphere be placed such that the net gravitational force acting on it from the other two spheres is equal to zero?

Explanation / Answer

cos theta = 4/5

sin theta =3/5

r13 = 4

r23 = sqrt[(4 - 0 )^2 + (0 - 3)^2 ] = 5

F1 = Gm1m3/r13^2

F1 = (6.67 x 10^-11 * 70 * 0.400 ) /16 = 1.16725 x 10^-10 N

F2 = Gm2m3/r23^2

F2 = 9.6048 x 10^-11

F1x = -1.16725 x 10^-10 N

F1y = 0

F2x = -F2cos theta = -(9.6048 x 10^-11) *(0.800) = -7.68384 x 10^-11 N

F2y = F2 sintheta = (9.6048 x 10^-11 ) * (0.600) = 5.76288 x 10^-11

Fx = F1x + F2x = -1.935634 x 10^-10 N

Fy = F1y + F2y = 5.76288 x 10^-11

F = sqrt ( Fx^2 + Fy^2)

F = 2.0196 x 10^-10 N

part b )

theta = tan^-1 (Fy/Fx) = -71.43 degree

part c )

Fnet = 0 if F1 = F2

Gm1m3/y^2 = Gm2m3/(3-y)^2

m2/(3-y)^2 - m1/y^2

y = 3 * sqrt(m1) / [ sqrt(m1) + sqrt(m2) ]

y = 1.41 m

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