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

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.