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Two particles, each of mass m, are fixed at points P and Q with PQ = L. R is the midpoint of PQ. A third particle of mass m is placed at point S on the perpendicular bisector of PQ at a distance L from R. Find the magnitude of the net gravitational force exerted on the particle at S by the two particles at P and Q.

1. Gm²/L²
2. 16Gm²/5L²
3. 16Gm²/(5*sqrt(5)*L²)
4. 4*sqrt(2)*Gm²/(5L²)

Correct answer: 16Gm²/(5*sqrt(5)*L²)

Solution

Distance from P (or Q) to S: d = sqrt((L/2)² + L²) = sqrt(L²/4 + L²) = sqrt(5L²/4) = (L sqrt5)/2. Force from each = Gm²/d² = Gm²/(5L²/4) = 4Gm²/(5L²). By symmetry horizontal components cancel; vertical (along RS) components add: each contributes a factor cos(theta) = (vertical leg L)/d = L/((L sqrt5)/2) = 2/sqrt5. Net = 2 * 4Gm²/(5L²) * (2/sqrt5) = 16Gm²/(5 sqrt5 L²).

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