Question

Sphere Awith mass$80kg$is located at the origin of an xycoordinate system; sphere Bwith mass$\mathbf{60}\mathbf{}\mathit{k}\mathit{g}\mathbf{}$is located at coordinates$\left(0.25m,0\right)$; sphere Cwith massis located in the first quadrant$\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathit{m}$from Aand$\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{}\mathit{m}$from B. In unit-vector notation, what is the gravitational force on Cdue to AandB?

Short answer

The magnitude of the gravitational force on the sphere C is.

$F=-4.4\times {10}^{-8}N\hat{j}$

Step-by-step solution

Listing the given quantities

The mass of sphere A is$m_A=80kg$

The mass of sphere B is$m_B=60kg$

The mass of sphere C is$m_C=0.20kg$

The coordinates of sphere A are$m_A\left(0,0\right)$

The coordinates of sphere B are$m_B\left(0.25m,0\right)$

The distance between sphere A and sphere C is $r_{AC}=0.20m$

The distance between sphere B and sphere C is$r_{BC}=0.15m$ .

Understanding the concept of force

We can find the magnitude of force on sphere C due to sphere A and B by using the formula for gravitational force.Then using the cosine law, we can find the angle made by force ${\mathit{F}}_{\mathbf{B}}$and ${\mathit{F}}_{\mathbf{A}}$. From this, we can find the magnitude of the net gravitational force on the sphere C due to A and B.

Formulae:

$$\begin{gathered} \overrightarrow{F}=\frac{GMm}{r^2}\hat{r} \\ \overrightarrow{F_{1,net}}=\sum _{i=1}^n\overrightarrow{F_{1i}} \end{gathered}$$

Step 3: Calculations ofthe magnitude of force

The magnitude of force on sphere C due to sphere A is given by

$\begin{array}{rcl}{F}_{AC}& =& \frac{G{m}_A{m}_C}{r_{AC}^2}\\ & =& \frac{6.67\times {10}^{-11}\left(80\right)\left(0.20\right)}{{0.2}^2}\\ & =& 2.7\times {10}^{-8}N\\ & & \end{array}$

The magnitude of force on sphere C due to sphere B is given by

$\begin{array}{rcl}{F}_{BC}& =& \frac{G{m}_B{m}_C}{r_{BC}^2}\\ & =& \frac{6.67\times {10}^{-11}\left(60\right)\left(0.20\right)}{{0.15}^2}\\ & =& 3.6\times {10}^{-8}N\\ & & \end{array}$

Now, by using cosine rules, we can calculate the angles made by forces$F_A$and$F_B$as

$\begin{array}{rcl}{\theta }_A& =& \pi +\left(\frac{r_{AC}^2+r_{AB}^2-r_{BC}^2}{2r_{AC}{r}_{AB}}\right)\\ & =& \pi +\left(0.80\right)\\ & =& 217°\\ & & \end{array}$

In a similar way, the angle made by force$F_B$with x axis can be calculated as

$\begin{array}{rcl}{\theta }_B& =& \left(\frac{r_{BC}^2+r_{AB}^2-r_{AC}^2}{2r_{BC}{r}_{AB}}\right)\\ & =& -53°\\ & & \end{array}$
Hence, the net force acting on the sphere C can be written as

$F=F_{AC}\left(cos{\theta }_{A}i+sin{\theta }_{A}j\right)+F_{BC}\left(cos{\theta }_{B}i+sin{\theta }_{B}j\right)$

Solving this equation, we get

$F=-4.4\times {10}^{-8}N\hat{j}$

The magnitude of the gravitational force on the sphere C is$F=-4.4\times {10}^{-8}N\hat{j}$.