A uniform sphere with mass 64.8 kg is held with its center at the origin, and a

Question

A uniform sphere with mass 64.8 kg is held with its center at the origin, and a second uniform sphere with mass 81.6 kg is held with its center at the point x = 0, y = 3 m. (a) What are the magnitude and direction of the net gravitational force due to these objects on a third uniform sphere with mass 0.485 kg placed at the point x = 4 m, y = 0? magnitude N direction Ý° measured counterclockwise from the positive x-axis (b) 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?

x = m

y=. M

Explanation / Answer

F (1 on 3) = 6.67*10^-11*(64.8)*(0.485)(-x)/(4^2) = -1.31*10^-10 N

F (2 on 3) = 6.67*10^-11*(81.6)*(0.485)(-x)/(5^2) = 1.056*10^-10 N

Similarly

F (2 on 3)x = 1.056*10^-10 N *(4/5)(-x) = - 8.45*10^-11 N (x)

  
F (2 on 3)y = 1.056*10^-10 N *(3/5)(y) = 6.336*10^-11 N (y)

Fx = F(1 0n 3)x + F(2 on 3)x = -2.155 *10^-10 N (x)

Fy = F(1 0n 3)y + F(2 on 3)y = 6.336*10^-11 N (y)

Magnitude = sqrt(Fx^2+ Fy^2) = 2.25*10^-10 N

Direction = tan inverse(Fy/Fx ) = theta = 16.38 degree (Alonf -x axis)

b. for 2nd case we ve

{(3-y)/y}^2 = (81.6)/(64.8)

hence

3-y = 1.12y

y = 1.42 m

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