# Chapter 13 - Newton's Theory of Gravity - Exercises and Problems - Page 354: 36

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.

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