Three identical particles A, B and C of mass 100 kg each are placed in a straight line with AB = BC = 13 m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13 m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately

Question

Three masses $200\,\text{kg}$, $300\,\text{kg}$ and $400\,\text{kg}$ are placed at the vertices of an equilateral triangle of side $20\,\text{m}$. They are rearranged on the vertices of a bigger triangle of side $25\,\text{m}$ with the same centre. The work done in this process is ___ J.
(Gravitational constant $G = 6.7 \times 10^{-11}\,\text{N m}^2\text{kg}^{-2}$)

Answer 1

To solve this problem, we need to calculate the gravitational force on a particle P due to three particles A, B, and C aligned in a straight line, with P positioned along the perpendicular bisector of line AC.

Step 1: Understand the configuration and placement of particles

Particles A, B, and C have masses of 100 kg each, placed in a straight line with distances AB = BC = 13 m.
The particle P, with the same mass of 100 kg, is placed at a distance of 13 m from point B along the perpendicular bisector of AC.

Step 2: Use the formula for gravitational force

The gravitational force between two masses is given by the formula:

F = \frac{G \cdot m_1 \cdot m_2}{r^2}

where G is the gravitational constant, m_1 and m_2 are the masses, and r is the distance between the masses.

Step 3: Calculate the net gravitational force on particle P

Position symmetry and perpendicular bisector: The forces due to A and C on P will cancel along their line of symmetry due to equal distances and masses, yet they will add in the perpendicular direction.

The distance from P to both A and C combine with the distances given, forming a right triangle with sides along AB, BC, and perpendicular:

Using Pythagoras, the distance PA = PC = \sqrt{13^2 + 13^2} = \sqrt{2 \times 169} = 13\sqrt{2} m.

The horizontal components of the forces due to A and C will cancel each other out, so we only need to calculate the force in the vertical (perpendicular) direction:

The total force from A and C on P vertically:

Force due to A on P vertically: F_{AP} = \frac{G \cdot 100 \cdot 100}{(13\sqrt{2})^2} \cdot \sin(45^\circ)
Force due to C on P vertically: F_{CP} = \frac{G \cdot 100 \cdot 100}{(13\sqrt{2})^2} \cdot \sin(45^\circ)

Adding the vertical components:

F_{AP\_ver} + F_{CP\_ver} = 2 \cdot \frac{G \cdot 100 \cdot 100}{169} \cdot \frac{1}{2} = \frac{G \cdot 100 \cdot 100}{169}

Since forces due to particles A and C are symmetric, their vertical component equals the single perpendicular force difference, achieving double and calculated.

Focusing on B, the gravitational force is simple:

F_{BP} = \frac{G \cdot 100 \cdot 100}{13^2}

Step 4: Total force acting on P:

Total perpendicular force from A and C effectively becomes zero as they balance out within perpendicular distance contributing vertically toward particle P: 50G per A-side + C-side.
Total force towards B naturally becomes 50G.

Sum the contributions:

F_{total} = 50G + 50G = 100G.

Conclusion: Therefore, the gravitational force F on the fourth particle P is approximately 100G. Thus, the correct answer is 100G.

Answer 2