Question

Two bodies of masses 8 kg are placed at the vertices A and B of an equilateral triangle ABC. A third body of mas 2 kg is placed at the centroid G of the triangle. If AG = BG = CG = 1 m, where should a fourth body of mass 4 kg be placed so that the resultant force on the 2 kg body is zero?

• A. At C
• B. At a point on the line CG such that PG = \(\frac{1}{\sqrt2}\)m
• C. At a point P on the line CG such that PG=0.5 m
• D. At a point P on the line CG such that PG=2m

Answer 1

The Correct Option is B

Step 1: Find the resultant force on the 2 kg body due to the 8 kg bodies at A and B.

Let the position of the 2 kg body be at the centroid G. The forces on the 2 kg body due to the 8 kg bodies at A and B are attractive and directed along GA and GB respectively.

Let $\vec{F}_{GA}$ be the force on the 2 kg body at G due to the 8 kg body at A.
$|\vec{F}_{GA}| = G \frac{(2 kg)(8 kg)}{(AG)^2} = \frac{16G}{AG^2} = \frac{16G}{(1 m)^2} = 16G$. The direction is along GA.

Let $\vec{F}_{GB}$ be the force on the 2 kg body at G due to the 8 kg body at B.
$|\vec{F}_{GB}| = G \frac{(2 kg)(8 kg)}{(BG)^2} = \frac{16G}{BG^2} = \frac{16G}{(1 m)^2} = 16G$. The direction is along GB.

The resultant force due to masses at A and B is $\vec{F}_{AB} = \vec{F}_{GA} + \vec{F}_{GB}$.

In an equilateral triangle, the angle AGB at the centroid is $120^\circ$.

Step 2: Using the Parallelogram law 

We can use the parallelogram law to find the magnitude of the resultant force, or vector addition.

Vectorially, $\vec{F}_{GA} = 16G \hat{GA}$ and $\vec{F}_{GB} = 16G \hat{GB}$, where $\hat{GA}$ and $\hat{GB}$ are unit vectors along GA and GB.

However, we know that for an equilateral triangle, $\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}$.

So, $\vec{GA} + \vec{GB} = - \vec{GC} = \vec{CG}$.

Thus, the resultant force due to masses at A and B is $\vec{F}_{AB} = \vec{F}_{GA} + \vec{F}_{GB} = 16G \hat{GA} + 16G \hat{GB} = 16G (\hat{GA} + \hat{GB})$.

Since AG = BG = CG = 1 m, we can consider $\vec{GA}, \vec{GB}, \vec{GC}$ as vectors from G to A, G to B, G to C.

Then $\vec{GA} + \vec{GB} + \vec{GC} = \vec{0}$.

$\vec{F}_{GA} = G \frac{2 \times 8}{1^2} \frac{\vec{GA}}{|\vec{GA}|} = 16G \frac{\vec{GA}}{1} = 16G \vec{GA}$ (This is wrong, direction should be from G to A, unit vector is $\frac{\vec{A}-\vec{G}}{|\vec{A}-\vec{G}|} = \frac{\vec{GA}}{|\vec{GA}|}$).

$\vec{F}_{GA} = 16G \frac{\vec{GA}}{|\vec{GA}|}$. $\vec{F}_{GB} = 16G \frac{\vec{GB}}{|\vec{GB}|}$.

Step 3: Finding the Direction

If we consider unit vectors $\hat{GA}$ and $\hat{GB}$.

Let's use $\vec{CG}$. We know that $\vec{GA} + \vec{GB} = - \vec{GC} = \vec{CG}$.

So, if we consider unit vectors along $\vec{GA}, \vec{GB}, \vec{CG}$.

Let's assume $\vec{GA} + \vec{GB} = k \vec{CG}$. For equilateral triangle, it is known that $\vec{GA} + \vec{GB} = \vec{CG}$.

So, the resultant force due to masses at A and B is $\vec{F}_{AB} = 16G (\hat{GA} + \hat{GB})$. This is not $16G \vec{CG}$.

Using vector sum, $\vec{F}_{AB} = \vec{F}_{GA} + \vec{F}_{GB}$. We want to balance this by a force $\vec{F}_{GP}$ due to 4 kg mass at P.

We need $\vec{F}_{AB} + \vec{F}_{GP} = \vec{0}$, so $\vec{F}_{GP} = - \vec{F}_{AB} = - (\vec{F}_{GA} + \vec{F}_{GB})$.

We assumed $\vec{GA} + \vec{GB} = \vec{CG}$. If we assume unit vectors along $\vec{GA}$ and $\vec{GB}$ are approximately in the direction of $\vec{GA}$ and $\vec{GB}$.

Let's consider magnitude of resultant of $\vec{F}_{GA}$ and $\vec{F}_{GB}$. Angle between $\vec{F}_{GA}$ and $\vec{F}_{GB}$ is $120^\circ$.

$|\vec{F}_{AB}| = \sqrt{|\vec{F}_{GA}|^2 + |\vec{F}_{GB}|^2 + 2 |\vec{F}_{GA}| |\vec{F}_{GB}| \cos 120^\circ} = \sqrt{(16G)^2 + (16G)^2 + 2 (16G) (16G) (-\frac{1}{2})} = \sqrt{2 (16G)^2 - (16G)^2} = \sqrt{(16G)^2} = 16G$.

The direction of resultant force is along the angle bisector of $\angle AGB$. The angle bisector of $\angle AGB$ is along GC. So $\vec{F}_{AB}$ is along GC direction.

So $\vec{F}_{AB} = 16G \hat{GC}$.

We need to place 4 kg mass at P such that force $\vec{F}_{GP}$ is equal and opposite to $\vec{F}_{AB}$.

$\vec{F}_{GP} = G \frac{(2 kg)(4 kg)}{(GP)^2} \hat{GP} = \frac{8G}{(GP)^2} \hat{GP}$.

We need $\vec{F}_{GP} = - \vec{F}_{AB} = - 16G \hat{GC} = 16G \hat{CG}$.

So $\frac{8G}{(GP)^2} \hat{GP} = 16G \hat{CG}$.

$\frac{8}{(GP)^2} \hat{GP} = 16 \hat{CG}$.

$\hat{GP} = 2 (GP)^2 \hat{CG}$.

For directions to be same, $\hat{GP} = \hat{CG}$. So P is on the line CG, such that $\vec{GP}$ is in the direction of $\vec{CG}$. This is not possible to balance $\vec{F}_{AB}$ which is in the direction of $\vec{GC}$.

$\hat{GP}$ must be in the direction of $\hat{CG}$.

$\frac{8}{(GP)^2} = 16$.

$(GP)^2 = \frac{8}{16} = \frac{1}{2}$.

$GP = \frac{1}{\sqrt{2}} m$.

And $\hat{GP} = \hat{CG}$. So P is on the line CG in the direction of C from G, such that $PG = \frac{1}{\sqrt{2}} m$.

Final Answer: The fourth body of mass 4 kg be placed at a point on the line CG such that PG = \(\frac{1}{\sqrt2}\)m

Answer 2

Let the masses at vertices A and B be $m_A = m_B = 8$ kg. 

The mass at the centroid G is $m_G = 2$ kg. 

We want to find the position of a fourth mass, $m_P = 4$ kg, such that the net force on $m_G$ is zero.

The gravitational force between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = G\frac{m_1m_2}{r^2}$.

The forces on $m_G$ due to $m_A$ and $m_B$ are equal in magnitude and directed along AG and BG, respectively. Since the triangle is equilateral, these forces form an angle of 120° with each other. 

Their resultant force acts along GC with a magnitude of $\frac{Gm_Am_G}{AG^2} = \frac{G(8)(2)}{(1)^2}=16G$.

For the net force on $m_G$ to be zero, the fourth mass must be placed along the line CG such that its gravitational force on $m_G$ equals the resultant force due to $m_A$ and $m_B$, but in the opposite direction. Let P be the position of the fourth mass and PG = x.

$\frac{Gm_Pm_G}{PG^2} = \frac{G(4)(2)}{x^2}=\frac{8G}{x^2}$. 

Setting the magnitudes equal gives:

$\frac{8G}{x^2} = 16G$

$x^2 = \frac{8}{16} = \frac{1}{2}$

$x = PG = \frac{1}{\sqrt{2}}$

The correct answer is (B) at a point P on the line CG such that PG = $\frac{1}{\sqrt{2}}$ m.