Question

You are sitting at your dinner table and notice three peas have fallen off of your plate as shown. Each pea has a mass of 0.22g and d=3.6cm. A. Find the magnitude and direction (CCW from the +axis) of the net gravitational force on the pea labeled M1 B. Find the direction (CCW from the +axis) of the net gravitational force on the pea labeled m1

Answer 1

Answer:

force = 2.73 x 10^ -15 N

angle = 153.435

Explanation:

Let us redraw the diagram

Let us call

F13 = force on m1 due to m3

F12 = force on m1 due to m2

Now along the x-axis, we have

$F_x=F_{13}\sin \theta\; \hat{i}$

and along the y-axis, we have

$F_y=F_{13}\cos \theta+F_{12}\; \hat{j}$

Now,

$F_{13}=G\frac{m_1m_3}{d^2+(2d)^2}$

and

$F_{12}=G\frac{m_1m_2}{d^2}$

Therefore, the force along the x-axis becomes

$\begin{gathered} F_x=F_{13}\sin \theta\; \hat{i} \\ \Rightarrow F_x=G\frac{m_1m_3}{d^2+(2d)^2}\sin \theta\; \hat{i} \end{gathered}$

and the force along the y-axis becomes

$\begin{gathered} F_y=F_{13}\cos \theta+F_{12}\; \hat{j} \\ \Rightarrow F_y=G\frac{m_1m_3}{d^2+(2d)^2}\cos \theta+G\frac{m_1m_2}{d^2}\hat{\; j} \end{gathered}$

Since

$\sin \theta=\frac{2d}{\sqrt[]{d^2+(2d)^2}}=\frac{2d}{d\sqrt[]{5}}=\frac{2}{\sqrt[]{5}}$

and

$\cos \theta=\frac{d}{\sqrt[]{d^2+(2d)^2}}=\frac{d}{d\sqrt[]{5}}=\frac{1}{\sqrt[]{5}}$

The above equations become

$F_x=G\frac{m_1m_3}{d^2+(2d)^2}\times\frac{2}{\sqrt[]{5}}\; \hat{i}$$\Rightarrow\boxed{F_x=G\frac{m_1m_3}{5d^2}\times\frac{2}{\sqrt[]{5}}\; \hat{i}}$

and

$\boxed{F_y=G\frac{m_1m_3}{5d^2}\times\frac{1}{\sqrt[]{5}}+G\frac{m_1m_2}{d^2}\hat{j}}$

To find the numerical value, we now put

G = 6.67 x 10^-11

m_1 = m_2 = m_3 = 2.2 x 10^-4 kg

d = 0.036 m

into the above equations and get

$F_x=(6.67\times10^{-11})\frac{(2.2\times10^{-4})^2}{5(0.036)^2}\times\frac{2}{\sqrt[]{5}}\; \hat{i}$$\boxed{F_x=4.46\times20^{-16}\; \; \hat{i}}$

and

$F_y=(6.67\times10^{-11})\frac{(2.2\times10^{-4})^2}{5(0.036)^2}\times\frac{1}{\sqrt[]{5}}+(6.67\times10^{-11})\frac{(2.2\times10^{-4})^2}{(0.036)^2}\hat{j}$$\Rightarrow\boxed{F_y=2.7\times10^{-15}\; \hat{j}}$

Therefore, the net gravitational force is

$F_{\text{tot}}=4.46\times20^{-16}\; \; \hat{i}+2.7\times10^{-15}\; \hat{j}$

the magnitude of this net force is

$|F_{\text{tot}}|=\sqrt[]{(4.46\times20^{-16})^2+(2.7\times10^{-15})^2}$$\boxed{|F_{\text{tot}}|=2.73\times10^{-15}}$

Finally, we need to find the direction of this force.

To specify, the direction, we need to find the angle this force makes counterclockwise with respect the x-axis.

The angle is given by

$\cos \theta=\frac{1}{\sqrt[]{5}}$$\Rightarrow\theta=63.435^o$

adding additional 90 degrees gives the angle with respect to the positive x-axis.

$\theta+90^o=153.435^o$

Hence, the force is directed at about 153 degrees counterclockwise with respect to the x-axis.