Question

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 has mass m2 = 5.1 kg and lies on the y-axis at x2 = 0, y2 = 2.8 m. Particle 3 has mass m3 = 3.7 kg and lies at the origin. What is the direction of the net gravitational force on particle 3, expressed as an angle counterclockwise from the +x-axis?

Answer 1

`F=18.58* 10^(-11)\ N`

`\theta=30.276^(\circ)`

Step-by-step explanation:

Given:

mass of first particle,
`m_1=6.7\ kg`

mass of second particle,
`m_2=5.1\ kg`

mass of third particle,
`m_3=3.7\ kg`

coordinate position of first particle in meters,
`(x_1,y_1)\equiv(4.2,0)`

coordinate position of second particle in meters,
`(x_2,y_2)\equiv(0,2.8)`

coordinate position of third particle in meters,
`(x_3,y_3)\equiv(0,0)`

Now, gravitational force on particle 3 due to particle 1:

`F_(31)=G(m_1.m_3)/(r_(31)^2)`

`F_(31)=6.67* 10^(-11) * (6.7* 3.7)/(4.2^2)`

`F_(31)=9.37* 10^(-11)\ N`

towards positive Y axis.

gravitational force on particle 3 due to particle 2:

`F_(32)=G(m_2.m_3)/(r_(21)^2)`

`F_(32)=6.67* 10^(-11) * (5.1* 3.7)/(2.8^2)`

`F_(32)=16.05* 10^(-11)\ N`

towards positive X axis.

Now the net force

`F=\sqrt{F_(31)\ ^2+F_(32)\ ^2}`

`F=\sqrt{(10^(-11))^2(9.37^2+16.05^2)}`

`F=18.58* 10^(-11)\ N`

For angle in counterclockwise direction from the +x-axis

`tan\theta=(9.37* 10^(-11))/(16.05* 10^(-11))`

`\theta=30.276^(\circ)`