Answer:
[tex]F=18.58\times 10^{-11}\ N[/tex]
[tex]\theta=30.276^{\circ}[/tex]
Explanation:
Given:
mass of first particle, [tex]m_1=6.7\ kg[/tex]
mass of second particle, [tex]m_2=5.1\ kg[/tex]
mass of third particle, [tex]m_3=3.7\ kg[/tex]
coordinate position of first particle in meters, [tex](x_1,y_1)\equiv(4.2,0)[/tex]
coordinate position of second particle in meters, [tex](x_2,y_2)\equiv(0,2.8)[/tex]
coordinate position of third particle in meters, [tex](x_3,y_3)\equiv(0,0)[/tex]
Now, gravitational force on particle 3 due to particle 1:
[tex]F_{31}=G\frac{m_1.m_3}{r_{31}^2}[/tex]
[tex]F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}[/tex]
[tex]F_{31}=9.37\times 10^{-11}\ N[/tex]
towards positive Y axis.
gravitational force on particle 3 due to particle 2:
[tex]F_{32}=G\frac{m_2.m_3}{r_{21}^2}[/tex]
[tex]F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}[/tex]
[tex]F_{32}=16.05\times 10^{-11}\ N [/tex]
towards positive X axis.
Now the net force
[tex]F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}[/tex]
[tex]F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}[/tex]
[tex]F=18.58\times 10^{-11}\ N[/tex]
For angle in counterclockwise direction from the +x-axis
[tex]tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}[/tex]
[tex]\theta=30.276^{\circ}[/tex]
