Three spherical balls of masses 1kg, 2kg and 3kg are placed at the corners of an equilateral triangle of side 1m. The magnitude of the gravitational force exerted by 2kg and 3kg masses on 1kg mass.

We are tasked to calculate the gravitational force exerted by two masses (2 kg and 3 kg) on a 1 kg mass, placed at the corners of an equilateral triangle with a side length of 1 meter.

Given Data:

• Side of the equilateral triangle: a = 1 m
• Masses at the corners: 1 kg, 2 kg, and 3 kg
• Gravitational constant: G

Step-by-Step Solution:

1. Force between 1 kg and 2 kg:

   The formula for gravitational force between two masses is: 
   F = (G * m₁ * m₂) / r² 
   Here, m₁ = 1 kg, m₂ = 2 kg, and r = 1 m. 
   Substituting values: 
   F₁ = (G * 1 * 2) / (1²) = 2G

2. Force between 1 kg and 3 kg:

   Using the same formula with m₁ = 1 kg, m₂ = 3 kg, and r = 1 m: 
   F₂ = (G * 1 * 3) / (1²) = 3G

3. Resultant Force on the 1 kg mass:

   The forces F₁ and F₂ act at an angle of 60° (since the triangle is equilateral). 
   The resultant force F_R is given by: 
   FR = √(F₁² + F₂² + 2 * F₁ * F₂ * cosθ) 
   Substituting values: 
   FR = √( (2G)² + (3G)² + 2 * 2G * 3G * cos60° ) 
   Since cos60° = 1/2, the equation becomes: 
   FR = √(4G² + 9G² + 6G²) 
FR = √(19G²) 
FR = G√19

Final Answer:

The magnitude of the resultant gravitational force exerted by the 2 kg and 3 kg masses on the 1 kg mass is:

F_R = G√19