In the previous section, we saw the magnitude of direction of the gravitational force of attraction . We also saw a solved example. In this section we will see a few more solved examples
Solved example 8.8
Find the unit of the universal gravitational constant G
Solution:
• We have: $\mathbf\small{|\vec{F}_{G}|=G\frac{m_1\; m_2}{|\vec{r}|^2}}$
• So we get: $\mathbf\small{G=\frac{|\vec{r}|^2\;|\vec{F}_{G}|}{m_1\; m_2}}$
• Thus the unit is: $\mathbf\small{N\,m^2\,kg^{-2}}$
Solved example 8.9
Two heavy bodies of masses 40 kg and 60 kg attract each other with a force of 4 × 10-5 N. If G is 6 × 10-11 N m2 kg-2, find the distance between them
Solution:
• We have: $\mathbf\small{|\vec{F}_{G}|=G\frac{m_1\; m_2}{|\vec{r}|^2}}$
• Substituting the given values, we get: $\mathbf\small{4\times 10^{-5}(N)=6\times 10^{-11}(N\,m^2\,kg^{-2})\left(\frac{40 (kg)\times 60(kg)}{|\vec{r}|^2} \right)}$
$\mathbf\small{\Rightarrow 4\times 10^{-5}=\left(\frac{1.44 \times 10^{-7}(m^2)}{|\vec{r}|^2} \right)}$
$\mathbf\small{\Rightarrow |\vec{r}|^2=\left(\frac{1.44 \times 10^{-7}(m^2)}{4\times 10^{-5}} \right)=0.0036\,(m^2)}$
$\mathbf\small{\Rightarrow |\vec{r}|=0.06\,m}$
Solved example 8.10
Two bodies A and B of masses M and 4M respectively, are placed at a distance of 2.73 m apart. Another body C of mass m is to be placed in between A and B in such a way that, the net force on C is zero. What is the distance of C from A
Solution:
1. Fig.8.14 (a) below shows the 3 bodies in a line
Let C be at a distance of x m from A
2. Fig.b shows the free body diagram of C
(i) We have: $\mathbf\small{\vec{F}_{G(A,C)}=-\left[G\frac{M\times m}{|\vec{r}_{A,C}|^3}\right]\vec{r}_{A,C}}$
$\mathbf\small{\Rightarrow \vec{F}_{G(A,C)}=-\left[G\frac{M\times m}{x^3}\right](x)(-\hat{i})}$
$\mathbf\small{\Rightarrow \vec{F}_{G(A,C)}=\left[\frac{GMm}{x^2}\right]\hat{i}}$
(ii) We have: $\mathbf\small{\vec{F}_{G(B,C)}=-\left[G\frac{m\times 4M}{|\vec{r}_{B,C}|^3}\right]\vec{r}_{B,C}}$
$\mathbf\small{\Rightarrow \vec{F}_{G(A,C)}=-\left[G\frac{m\times 4M}{(2.73-x)^3}\right](2.73-x)\hat{i}}$
$\mathbf\small{\Rightarrow \vec{F}_{G(A,C)}=-\left[\frac{4Gm M}{(2.73-x)^2}\right]\hat{i}}$
3. Net force = $\mathbf\small{\vec{F}_{G(A,C)}+\vec{F}_{G(B,C)}}$
⇒ Net force = $\mathbf\small{\left[\frac{GMm}{x^2}\right]\hat{i}+\left(-\left[\frac{4Gm M}{(2.73-x)^2}\right]\hat{i} \right)}$
4. But net force must be zero. So we get:
$\mathbf\small{\left[\frac{GMm}{x^2}\right]\hat{i}=\left[\frac{4Gm M}{(2.73-x)^2}\right]\hat{i}}$
• We have an equality of two vectors. So their magnitudes must be the same. We can write:
$\mathbf\small{\left[\frac{GMm}{x^2}\right]=\left[\frac{4Gm M}{(2.73-x)^2}\right]}$
$\mathbf\small{\Rightarrow \left[\frac{1}{x^2}\right]=\left[\frac{4}{(2.73-x)^2}\right]}$
$\mathbf\small{\Rightarrow (2.73-x)^2=4x^2}$
⇒ (2.73-x) = 2x (Taking roots on both sides)
⇒ 2.73 = 3x
⇒ x = 0.91 m
Solved example 8.11
Two bodies A and B have equal masses. The gravitational force of attraction between them is $\mathbf\small{\vec{F}_{G}}$. 25% of the mass of A is transferred to B. What is the new force of attraction
Solution:
1. Let the initial masses of the two bodies be m each
• Then we can write: $\mathbf\small{\vec{F}_{G}=-\left[\frac{Gm^2}{|\vec{r}|^3}\right]\vec{r}}$
• Where $\mathbf\small{\vec{r}}$ is the distance vector between A and B
2. Final mass of A = (m-0.25m) = 0.75m
• Final mass of B = (m+0.25m) = 1.25m
3. New force $\mathbf\small{\vec{F}_{G(new)}=-\left[\frac{G(0.75m)(1.25m)}{|\vec{r}|^3}\right]\vec{r}=-\left[\frac{0.9375Gm^2}{|\vec{r}|^3}\right]\vec{r}}$
4. Dividing (3) by (1), we get:
$\mathbf\small{\frac{\vec{F}_{G(new)}}{\vec{F}_{G}}=\left(-\left[\frac{0.9375Gm^2}{|\vec{r}|^3}\right]\vec{r} \right)\div \left(-\left[\frac{Gm^2}{|\vec{r}|^3}\right]\vec{r} \right)=0.9375}$
$\mathbf\small{\Rightarrow \vec{F}_{G(new)}=0.9375\;\vec{F}_{G}}$
Solved example 8.12
Two bodies A and B have masses m and M respectively. The gravitational force of attraction between them is $\mathbf\small{\vec{F}}$. A third body C is placed in contact with A. The body C has a mass of 2m. What is the force on B due to A? What is the total force on B ?
Solution:
Part (a):
• Initially, there are only two masses A and B
• At that time, the force is given to be $\mathbf\small{\vec{F}}$
• We can write:
$\mathbf\small{\vec{F}=\vec{F}_{G(A,B)}=-\left[\frac{GmM}{|\vec{r}|^3}\right]\vec{r}}$
• Where $\mathbf\small{\vec{r}}$ is the distance vector between A and B
Part (b):
• When C is placed near A, the force on B due to C is given by:
$\mathbf\small{\vec{F}_{G(C,B)}=-\left[\frac{G\times 2m\times M}{|\vec{r}|^3}\right]\vec{r}}$
• Taking ratios, we get:
$\mathbf\small{\frac{\vec{F}_{G(C,B)}}{\vec{F}_{G(A,B)}}=\frac{\vec{F}_{G(C,B)}}{\vec{F}}=\left(-\left[\frac{G\times 2m\times M}{|\vec{r}|^3}\right]\vec{r} \right)\div \left(-\left[\frac{G\times m\times M}{|\vec{r}|^3}\right]\vec{r} \right)}$
$\mathbf\small{\Rightarrow \frac{\vec{F}_{G(C,B)}}{\vec{F}}=2}$
$\mathbf\small{\Rightarrow \vec{F}_{G(C,B)}=2\vec{F}}$
• Thus we can write:
Total force on B = $\mathbf\small{\vec{F}_{G(A,B)}+\vec{F}_{G(C,B)}=\vec{F}+2\vec{F}=3\vec{F}}$
Solved example 8.13
In fig.8.15(a) below, three equal masses of m kg each are fixed at three corners A, B and D of the square of side a
What is the force acting on a mass 1 kg placed at the fourth corner C ?
Solution:
1. Any square will lie on a plane. So this is a 2D problem. We will need the x and y axes only. There is no need for the z-axis
• Assume that, the origin of the coordinate axes is at A. This is shown in fig.b
2. For this square:
(i) C will be equidistant from B and D
C will be at a distance of (2)a from A
3. Now we can begin the calculations:
• The distance vectors are shown in fig.8.16(a) below:
• The free body diagram of the 1 kg mass at C is shown in fig.16(b)
4. Force due to mass at B
(i) $\mathbf\small{\vec{F}_{G(B,C)}=-\left[G\frac{m\times 1}{|\vec{r}_{B,C}|^3}\right]\vec{r}_{B,C}}$
$\mathbf\small{\Rightarrow \vec{F}_{G(B,C)}=-\left[\frac{Gm}{a^3}\right]\vec{r}_{B,C}}$
(ii) So our next task is to write $\mathbf\small{\vec{r}_{B,C}}$ in component form
• To reach C from B:
♦ We have no horizontal distance to travel
♦ We travel only a distance BC vertically up
• So the components of $\mathbf\small{\vec{r}_{B,C}}$ are $\mathbf\small{(0)\hat{i}}$ and $\mathbf\small{a\hat{j}}$
• We can write: $\mathbf\small{\vec{r}_{B,C}=a\hat{j}}$
(iii) So the result in (i) becomes: $\mathbf\small{\vec{F}_{G(B,C)}=-\left[\frac{Gm}{a^3}\right]a\hat{j}}$
5. Force due to mass at D
(i) $\mathbf\small{\vec{F}_{G(D,C)}=-\left[G\frac{m\times 1}{|\vec{r}_{D,C}|^3}\right]\vec{r}_{D,C}}$
$\mathbf\small{\Rightarrow \vec{F}_{G(D,C)}=-\left[\frac{Gm}{a^3}\right]\vec{r}_{D,C}}$
(ii) So our next task is to write $\mathbf\small{\vec{r}_{D,C}}$ in component form
• To reach C from D:
♦ We have no vertical distance to travel
♦ We travel only a distance DC horizontally to the right
• So the components of $\mathbf\small{\vec{r}_{D,C}}$ are $\mathbf\small{(a)\hat{i}}$ and $\mathbf\small{0\hat{j}}$
• We can write: $\mathbf\small{\vec{r}_{D,C}=a\hat{i}}$
(iii) So the result in (i) becomes: $\mathbf\small{\vec{F}_{G(D,C)}=-\left[\frac{Gm}{a^3}\right]a\hat{i}}$
6. Force due to mass at A:
(i) $\mathbf\small{\vec{F}_{G(A,C)}=-\left[G\frac{m\times 1}{|\vec{r}_{A,C}|^3}\right]\vec{r}_{A,C}}$
$\mathbf\small{\Rightarrow \vec{F}_{G(A,C)}=-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]\vec{r}_{A,C}}$
(ii) So our next task is to write $\mathbf\small{\vec{r}_{A,C}}$ in component form
• To reach C from A:
♦ First we travel a distance AB horizontally to the right
♦ Then we travel a distance BC vertically up
• So the components of $\mathbf\small{\vec{r}_{A,C}}$ are $\mathbf\small{a\hat{i}}$ and $\mathbf\small{a\hat{j}}$
• We can write: $\mathbf\small{\vec{r}_{A,C}=a\hat{i}+a\hat{j}}$
(iii) So the result in (i) becomes:
$\mathbf\small{\vec{F}_{G(A,C)}=-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right](a\hat{i}+a\hat{j})}$
$\mathbf\small{\Rightarrow \vec{F}_{G(A,C)}=\left(-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]a\hat{i}\right)+\left(-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]a\hat{j}\right)}$
7. When we add the results in (4), (5) and (6), we get the net force experienced by the mass at C
So we can write:
The net force experienced by the mass at C
= $\mathbf\small{\vec{F}_{G(B,C)}+\vec{F}_{G(D,C)}+\vec{F}_{G(A,C)}}$
= $\mathbf\small{\left(-\left[\frac{Gm}{a^3}\right]a\hat{j} \right)+\left(-\left[\frac{Gm}{a^3}\right]a\hat{i} \right)+\left(-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]a\hat{i}\right)+\left(-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]a\hat{j}\right)}$
(i) First we will add the horizontal terms:
$\mathbf\small{\left(-\left[\frac{Gm}{a^3}\right]a\hat{i} \right)+\left(-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]a\hat{i}\right)}$
= $\mathbf\small{-\left[\frac{Gm}{a^3}\right]\left(a\hat{i}+\left[\frac{1}{2\sqrt{2}}\right]a\hat{i}\right)}$
= $\mathbf\small{-\left[\frac{Gm}{a^2}\right]\left(1+\frac{1}{2\sqrt{2}}\right)\hat{i}}$
(ii) Next we add the vertical terms:
$\mathbf\small{\left(-\left[\frac{Gm}{a^3}\right]a\hat{j} \right)+\left(-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]a\hat{j}\right)}$
= $\mathbf\small{-\left[\frac{Gm}{a^3}\right]\left(a\hat{j}+\left[\frac{1}{2\sqrt{2}}\right]a\hat{j}\right)}$
= $\mathbf\small{-\left[\frac{Gm}{a^2}\right]\left(1+\frac{1}{2\sqrt{2}}\right)\hat{j}}$
8. We can write:
■ The resultant force on C has two components:
• The horizontal component is: $\mathbf\small{-\left[\frac{Gm}{a^2}\right]\left(1+\frac{1}{2\sqrt{2}}\right)\hat{i}}$
• The vertical component is: $\mathbf\small{-\left[\frac{Gm}{a^2}\right]\left(1+\frac{1}{2\sqrt{2}}\right)\hat{j}}$
9. Note that, the magnitudes of both components are the same
So we can write:
■ The resultant force on C has two components:
• The horizontal component is: $\mathbf\small{-p\hat{i}}$
• The vertical component is: $\mathbf\small{-p\hat{j}}$
Where $\mathbf\small{p=\left[\frac{Gm}{a^2}\right]\left(1+\frac{1}{2\sqrt{2}}\right)}$
This is shown in fig.8.17(a) below:
10. We have two vectors
♦ The magnitudes of the two vectors are equal
♦ The two vectors are perpendicular to each other
• So the magnitude of the resultant vector will be given by:
$\mathbf\small{\sqrt{p^2+p^2}=\sqrt{2p^2}=\sqrt{2}\;p}$ (Details here)
= $\mathbf\small{\sqrt{2}\left[\frac{Gm}{a^2}\right]\left(1+\frac{1}{2\sqrt{2}}\right)}$
= $\mathbf\small{\left[\frac{Gm}{a^2}\right]\left(\sqrt{2}+\frac{1}{2}\right)}$
• Also, the resultant will be making an angle of 45o with the horizontal. This is shown in fig.8.17(b)
• In (6) we wrote: $\mathbf\small{\vec{r}_{A,C}=a\hat{i}+a\hat{j}}$
• We can use trigonometry and write:
$\mathbf\small{\vec{r}_{A,C}=(a\cos45)\hat{i}+(a\sin45)\hat{j}}$
• The reader is advised to write all steps using this alternate method and prove that, the resultant force on the mass at C is the same as that obtained above
Solved example 8.8
Find the unit of the universal gravitational constant G
Solution:
• We have: $\mathbf\small{|\vec{F}_{G}|=G\frac{m_1\; m_2}{|\vec{r}|^2}}$
• So we get: $\mathbf\small{G=\frac{|\vec{r}|^2\;|\vec{F}_{G}|}{m_1\; m_2}}$
• Thus the unit is: $\mathbf\small{N\,m^2\,kg^{-2}}$
Solved example 8.9
Two heavy bodies of masses 40 kg and 60 kg attract each other with a force of 4 × 10-5 N. If G is 6 × 10-11 N m2 kg-2, find the distance between them
Solution:
• We have: $\mathbf\small{|\vec{F}_{G}|=G\frac{m_1\; m_2}{|\vec{r}|^2}}$
• Substituting the given values, we get: $\mathbf\small{4\times 10^{-5}(N)=6\times 10^{-11}(N\,m^2\,kg^{-2})\left(\frac{40 (kg)\times 60(kg)}{|\vec{r}|^2} \right)}$
$\mathbf\small{\Rightarrow 4\times 10^{-5}=\left(\frac{1.44 \times 10^{-7}(m^2)}{|\vec{r}|^2} \right)}$
$\mathbf\small{\Rightarrow |\vec{r}|^2=\left(\frac{1.44 \times 10^{-7}(m^2)}{4\times 10^{-5}} \right)=0.0036\,(m^2)}$
$\mathbf\small{\Rightarrow |\vec{r}|=0.06\,m}$
Solved example 8.10
Two bodies A and B of masses M and 4M respectively, are placed at a distance of 2.73 m apart. Another body C of mass m is to be placed in between A and B in such a way that, the net force on C is zero. What is the distance of C from A
Solution:
1. Fig.8.14 (a) below shows the 3 bodies in a line
Let C be at a distance of x m from A
 |
| Fig.8.14 |
(i) We have: $\mathbf\small{\vec{F}_{G(A,C)}=-\left[G\frac{M\times m}{|\vec{r}_{A,C}|^3}\right]\vec{r}_{A,C}}$
$\mathbf\small{\Rightarrow \vec{F}_{G(A,C)}=-\left[G\frac{M\times m}{x^3}\right](x)(-\hat{i})}$
$\mathbf\small{\Rightarrow \vec{F}_{G(A,C)}=\left[\frac{GMm}{x^2}\right]\hat{i}}$
(ii) We have: $\mathbf\small{\vec{F}_{G(B,C)}=-\left[G\frac{m\times 4M}{|\vec{r}_{B,C}|^3}\right]\vec{r}_{B,C}}$
$\mathbf\small{\Rightarrow \vec{F}_{G(A,C)}=-\left[G\frac{m\times 4M}{(2.73-x)^3}\right](2.73-x)\hat{i}}$
$\mathbf\small{\Rightarrow \vec{F}_{G(A,C)}=-\left[\frac{4Gm M}{(2.73-x)^2}\right]\hat{i}}$
3. Net force = $\mathbf\small{\vec{F}_{G(A,C)}+\vec{F}_{G(B,C)}}$
⇒ Net force = $\mathbf\small{\left[\frac{GMm}{x^2}\right]\hat{i}+\left(-\left[\frac{4Gm M}{(2.73-x)^2}\right]\hat{i} \right)}$
4. But net force must be zero. So we get:
$\mathbf\small{\left[\frac{GMm}{x^2}\right]\hat{i}=\left[\frac{4Gm M}{(2.73-x)^2}\right]\hat{i}}$
• We have an equality of two vectors. So their magnitudes must be the same. We can write:
$\mathbf\small{\left[\frac{GMm}{x^2}\right]=\left[\frac{4Gm M}{(2.73-x)^2}\right]}$
$\mathbf\small{\Rightarrow \left[\frac{1}{x^2}\right]=\left[\frac{4}{(2.73-x)^2}\right]}$
$\mathbf\small{\Rightarrow (2.73-x)^2=4x^2}$
⇒ (2.73-x) = 2x (Taking roots on both sides)
⇒ 2.73 = 3x
⇒ x = 0.91 m
Solved example 8.11
Two bodies A and B have equal masses. The gravitational force of attraction between them is $\mathbf\small{\vec{F}_{G}}$. 25% of the mass of A is transferred to B. What is the new force of attraction
Solution:
1. Let the initial masses of the two bodies be m each
• Then we can write: $\mathbf\small{\vec{F}_{G}=-\left[\frac{Gm^2}{|\vec{r}|^3}\right]\vec{r}}$
• Where $\mathbf\small{\vec{r}}$ is the distance vector between A and B
2. Final mass of A = (m-0.25m) = 0.75m
• Final mass of B = (m+0.25m) = 1.25m
3. New force $\mathbf\small{\vec{F}_{G(new)}=-\left[\frac{G(0.75m)(1.25m)}{|\vec{r}|^3}\right]\vec{r}=-\left[\frac{0.9375Gm^2}{|\vec{r}|^3}\right]\vec{r}}$
4. Dividing (3) by (1), we get:
$\mathbf\small{\frac{\vec{F}_{G(new)}}{\vec{F}_{G}}=\left(-\left[\frac{0.9375Gm^2}{|\vec{r}|^3}\right]\vec{r} \right)\div \left(-\left[\frac{Gm^2}{|\vec{r}|^3}\right]\vec{r} \right)=0.9375}$
$\mathbf\small{\Rightarrow \vec{F}_{G(new)}=0.9375\;\vec{F}_{G}}$
Solved example 8.12
Two bodies A and B have masses m and M respectively. The gravitational force of attraction between them is $\mathbf\small{\vec{F}}$. A third body C is placed in contact with A. The body C has a mass of 2m. What is the force on B due to A? What is the total force on B ?
Solution:
Part (a):
• Initially, there are only two masses A and B
• At that time, the force is given to be $\mathbf\small{\vec{F}}$
• We can write:
$\mathbf\small{\vec{F}=\vec{F}_{G(A,B)}=-\left[\frac{GmM}{|\vec{r}|^3}\right]\vec{r}}$
• Where $\mathbf\small{\vec{r}}$ is the distance vector between A and B
Part (b):
• When C is placed near A, the force on B due to C is given by:
$\mathbf\small{\vec{F}_{G(C,B)}=-\left[\frac{G\times 2m\times M}{|\vec{r}|^3}\right]\vec{r}}$
• Taking ratios, we get:
$\mathbf\small{\frac{\vec{F}_{G(C,B)}}{\vec{F}_{G(A,B)}}=\frac{\vec{F}_{G(C,B)}}{\vec{F}}=\left(-\left[\frac{G\times 2m\times M}{|\vec{r}|^3}\right]\vec{r} \right)\div \left(-\left[\frac{G\times m\times M}{|\vec{r}|^3}\right]\vec{r} \right)}$
$\mathbf\small{\Rightarrow \frac{\vec{F}_{G(C,B)}}{\vec{F}}=2}$
$\mathbf\small{\Rightarrow \vec{F}_{G(C,B)}=2\vec{F}}$
• Thus we can write:
Total force on B = $\mathbf\small{\vec{F}_{G(A,B)}+\vec{F}_{G(C,B)}=\vec{F}+2\vec{F}=3\vec{F}}$
Solved example 8.13
In fig.8.15(a) below, three equal masses of m kg each are fixed at three corners A, B and D of the square of side a
 |
| Fig.8.15 |
Solution:
1. Any square will lie on a plane. So this is a 2D problem. We will need the x and y axes only. There is no need for the z-axis
• Assume that, the origin of the coordinate axes is at A. This is shown in fig.b
2. For this square:
(i) C will be equidistant from B and D
C will be at a distance of (2)a from A
3. Now we can begin the calculations:
• The distance vectors are shown in fig.8.16(a) below:
 |
| Fig.8.16 |
4. Force due to mass at B
(i) $\mathbf\small{\vec{F}_{G(B,C)}=-\left[G\frac{m\times 1}{|\vec{r}_{B,C}|^3}\right]\vec{r}_{B,C}}$
$\mathbf\small{\Rightarrow \vec{F}_{G(B,C)}=-\left[\frac{Gm}{a^3}\right]\vec{r}_{B,C}}$
(ii) So our next task is to write $\mathbf\small{\vec{r}_{B,C}}$ in component form
• To reach C from B:
♦ We have no horizontal distance to travel
♦ We travel only a distance BC vertically up
• So the components of $\mathbf\small{\vec{r}_{B,C}}$ are $\mathbf\small{(0)\hat{i}}$ and $\mathbf\small{a\hat{j}}$
• We can write: $\mathbf\small{\vec{r}_{B,C}=a\hat{j}}$
(iii) So the result in (i) becomes: $\mathbf\small{\vec{F}_{G(B,C)}=-\left[\frac{Gm}{a^3}\right]a\hat{j}}$
5. Force due to mass at D
(i) $\mathbf\small{\vec{F}_{G(D,C)}=-\left[G\frac{m\times 1}{|\vec{r}_{D,C}|^3}\right]\vec{r}_{D,C}}$
$\mathbf\small{\Rightarrow \vec{F}_{G(D,C)}=-\left[\frac{Gm}{a^3}\right]\vec{r}_{D,C}}$
(ii) So our next task is to write $\mathbf\small{\vec{r}_{D,C}}$ in component form
• To reach C from D:
♦ We have no vertical distance to travel
♦ We travel only a distance DC horizontally to the right
• So the components of $\mathbf\small{\vec{r}_{D,C}}$ are $\mathbf\small{(a)\hat{i}}$ and $\mathbf\small{0\hat{j}}$
• We can write: $\mathbf\small{\vec{r}_{D,C}=a\hat{i}}$
(iii) So the result in (i) becomes: $\mathbf\small{\vec{F}_{G(D,C)}=-\left[\frac{Gm}{a^3}\right]a\hat{i}}$
6. Force due to mass at A:
(i) $\mathbf\small{\vec{F}_{G(A,C)}=-\left[G\frac{m\times 1}{|\vec{r}_{A,C}|^3}\right]\vec{r}_{A,C}}$
$\mathbf\small{\Rightarrow \vec{F}_{G(A,C)}=-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]\vec{r}_{A,C}}$
(ii) So our next task is to write $\mathbf\small{\vec{r}_{A,C}}$ in component form
• To reach C from A:
♦ First we travel a distance AB horizontally to the right
♦ Then we travel a distance BC vertically up
• So the components of $\mathbf\small{\vec{r}_{A,C}}$ are $\mathbf\small{a\hat{i}}$ and $\mathbf\small{a\hat{j}}$
• We can write: $\mathbf\small{\vec{r}_{A,C}=a\hat{i}+a\hat{j}}$
(iii) So the result in (i) becomes:
$\mathbf\small{\vec{F}_{G(A,C)}=-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right](a\hat{i}+a\hat{j})}$
$\mathbf\small{\Rightarrow \vec{F}_{G(A,C)}=\left(-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]a\hat{i}\right)+\left(-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]a\hat{j}\right)}$
7. When we add the results in (4), (5) and (6), we get the net force experienced by the mass at C
So we can write:
The net force experienced by the mass at C
= $\mathbf\small{\vec{F}_{G(B,C)}+\vec{F}_{G(D,C)}+\vec{F}_{G(A,C)}}$
= $\mathbf\small{\left(-\left[\frac{Gm}{a^3}\right]a\hat{j} \right)+\left(-\left[\frac{Gm}{a^3}\right]a\hat{i} \right)+\left(-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]a\hat{i}\right)+\left(-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]a\hat{j}\right)}$
(i) First we will add the horizontal terms:
$\mathbf\small{\left(-\left[\frac{Gm}{a^3}\right]a\hat{i} \right)+\left(-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]a\hat{i}\right)}$
= $\mathbf\small{-\left[\frac{Gm}{a^3}\right]\left(a\hat{i}+\left[\frac{1}{2\sqrt{2}}\right]a\hat{i}\right)}$
= $\mathbf\small{-\left[\frac{Gm}{a^2}\right]\left(1+\frac{1}{2\sqrt{2}}\right)\hat{i}}$
(ii) Next we add the vertical terms:
$\mathbf\small{\left(-\left[\frac{Gm}{a^3}\right]a\hat{j} \right)+\left(-\left[\frac{Gm}{(\sqrt{2}\,a)^3}\right]a\hat{j}\right)}$
= $\mathbf\small{-\left[\frac{Gm}{a^3}\right]\left(a\hat{j}+\left[\frac{1}{2\sqrt{2}}\right]a\hat{j}\right)}$
= $\mathbf\small{-\left[\frac{Gm}{a^2}\right]\left(1+\frac{1}{2\sqrt{2}}\right)\hat{j}}$
8. We can write:
■ The resultant force on C has two components:
• The horizontal component is: $\mathbf\small{-\left[\frac{Gm}{a^2}\right]\left(1+\frac{1}{2\sqrt{2}}\right)\hat{i}}$
• The vertical component is: $\mathbf\small{-\left[\frac{Gm}{a^2}\right]\left(1+\frac{1}{2\sqrt{2}}\right)\hat{j}}$
9. Note that, the magnitudes of both components are the same
So we can write:
■ The resultant force on C has two components:
• The horizontal component is: $\mathbf\small{-p\hat{i}}$
• The vertical component is: $\mathbf\small{-p\hat{j}}$
Where $\mathbf\small{p=\left[\frac{Gm}{a^2}\right]\left(1+\frac{1}{2\sqrt{2}}\right)}$
This is shown in fig.8.17(a) below:
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| Fig.8.17 |
♦ The magnitudes of the two vectors are equal
♦ The two vectors are perpendicular to each other
• So the magnitude of the resultant vector will be given by:
$\mathbf\small{\sqrt{p^2+p^2}=\sqrt{2p^2}=\sqrt{2}\;p}$ (Details here)
= $\mathbf\small{\sqrt{2}\left[\frac{Gm}{a^2}\right]\left(1+\frac{1}{2\sqrt{2}}\right)}$
= $\mathbf\small{\left[\frac{Gm}{a^2}\right]\left(\sqrt{2}+\frac{1}{2}\right)}$
• Also, the resultant will be making an angle of 45o with the horizontal. This is shown in fig.8.17(b)
Another method:
• We can use trigonometry and write:
$\mathbf\small{\vec{r}_{A,C}=(a\cos45)\hat{i}+(a\sin45)\hat{j}}$
• The reader is advised to write all steps using this alternate method and prove that, the resultant force on the mass at C is the same as that obtained above
In the next section, we will see gravitational force of attraction between extended objects
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