Answer
The net gravitational force on the 20.0-kg mass is $6.09\times 10^{-7}~N$ in the +y-direction.
Work Step by Step
We can find the distance $R$ from the center of the 20-kg mass to the center of each of the other masses.
$R = \sqrt{(5.0~cm)^2+(20.0~cm)^2}$
$R = 20.6~cm$
We can find the gravitational force that one 10.0-kg mass exerts on the 20.0-kg mass.
$F = \frac{G~(10.0~kg)(20.0~kg)}{R^2}$
$F = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(10.0~kg)(20.0~kg)}{(0.206~m)^2}$
$F = 3.14\times 10^{-7}~N$
We can find the angle $\theta$ from the +y-axis that each 10.0-kg mass attracts the 20.0 kg mass.
$tan(\theta) = \frac{5.0~cm}{20.0~cm}$
$\theta = arctan(\frac{5.0~cm}{20.0~cm})$
$\theta = 14.0^{\circ}$
By symmetry, the x-component of the net gravitational force on the 20.0-kg mass is zero. Therefore, the net gravitational force on the 20.0-kg mass is double the y-component of the gravitational force exerted by one 10.0-kg mass.
$F_{net} = 2~F_y$
$F_{net} = 2~F~cos(\theta)$
$F_{net} = (2)(3.14\times 10^{-7}~N)~cos(14.0^{\circ})$
$F_{net} = 6.09\times 10^{-7}~N$
The net gravitational force on the 20.0-kg mass is $6.09\times 10^{-7}~N$ in the +y-direction.
