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

force = 2.73 x 10^ -15 N

angle = 153.435

Step-by-step 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(m_1m_3)/(d^2+(2d)^2)`

and

`F_(12)=G(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(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(m_1m_3)/(d^2+(2d)^2)\cos \theta+G(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(m_1m_3)/(d^2+(2d)^2)*\frac{2}{\sqrt[]{5}}\; \hat{i}`
`\Rightarrow\boxed{F_x=G(m_1m_3)/(5d^2)*\frac{2}{\sqrt[]{5}}\; \hat{i}}`

and

`\boxed{F_y=G(m_1m_3)/(5d^2)*\frac{1}{\sqrt[]{5}}+G(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*10^(-11))((2.2*10^(-4))^2)/(5(0.036)^2)*\frac{2}{\sqrt[]{5}}\; \hat{i}`
`\boxed{F_x=4.46*20^(-16)\; \; \hat{i}}`

and

`F_y=(6.67*10^(-11))((2.2*10^(-4))^2)/(5(0.036)^2)*\frac{1}{\sqrt[]{5}}+(6.67*10^(-11))((2.2*10^(-4))^2)/((0.036)^2)\hat{j}`
`\Rightarrow\boxed{F_y=2.7*10^(-15)\; \hat{j}}`

Therefore, the net gravitational force is

`F_{\text{tot}}=4.46*20^(-16)\; \; \hat{i}+2.7*10^(-15)\; \hat{j}`

the magnitude of this net force is

`|F_{\text{tot}}|=\sqrt[]{(4.46*20^(-16))^2+(2.7*10^(-15))^2}`
`\boxed{|F_{\text{tot}}|=2.73*10^(-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.