Question

You are sitting at your dinner table and noticed three peas have fallen off of your plate as shown. Each Pea has a mass of 0.22 g and D= 3.6 cm. I got part A that is 2.73×10-15th but part B I need help finding the direction (CCW from the +x axis) of the net gravitational force on the pea labeled M1

Answer 1

In order to find the direction of the net gravitational force, let's draw the forces acting on m1:

The direction F13 is pointing is given by:

`\begin{gathered} \tan (\theta)=(2d)/(d) \\ \tan (\theta)=2 \\ \theta=\tan ^(-1)(2) \\ \theta=63.435\degree \end{gathered}`

Let's find the forces F12 and F13, and decompose the second one in its horizontal and vertical components using this angle:

`\begin{gathered} F12=(G\cdot M\cdot m)/(d^2)=(6.674\cdot10^(-11)\cdot0.22\cdot10^(-3)\cdot0.22\cdot10^(-3))/((3.6\cdot10^(-2))^2)=2.49\cdot10^(-15) \\ F13=\frac{G\cdot M\cdot m}{(\sqrt[]{d^2+(2d)^2_{}})^2_{}}=(6.674\cdot10^(-11)\cdot0.22\cdot10^(-3)\cdot0.22\cdot10^(-3))/(5\cdot(3.6\cdot10^(-2))^2)=4.98\cdot10^(-16) \\ F13_x=F13\cdot\sin (\theta)=4.98\cdot10^(-16)\cdot\sin (63.435\degree)=4.45\cdot10^(-16) \\ F13_y=F13\cdot\cos (\theta)=4.98\cdot10^(-16)\cdot\cos (63.435\degree)=2.23\cdot10^(-16) \end{gathered}`

Adding the vertical forces, we have:

`F_y=F13_y+F12=2.23\cdot10^(-16)+24.9\cdot10^(-16)=27.13\cdot10^(-16)=2.713\cdot10^(-15)`

Finally, the direction of the net force is given by:

(Let's use a negative Fy, since it's pointing down)

`\begin{gathered} \theta=\tan ^(-1)((F_y)/(F_x))=\tan ^(-1)((F_y)/(F13_x)) \\ =\tan ^(-1)((-2.713\cdot10^(-15))/(0.445\cdot10^(-15))) \\ =\tan ^(-1)(-6.0966) \\ =-80.68\degree \end{gathered}`