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Three identical spheres, each of mass m, are kept inside a box arranged so that two spheres rest on the bottom and the third rests on top of the two, all in contact. The box accelerates vertically upward with acceleration g/4. Friction is negligible everywhere. Which of the following statements are correct?

1. The normal force from the spheres on the bottom of the box is (9/4)mg
2. The normal force from the spheres on the bottom of the box is (15/4)mg
3. The normal force between spheres A and B (the two bottom spheres) is 2*sqrt(3)*m*g
4. The normal force between spheres A and B (the two bottom spheres) is 7*m*g/(4*sqrt(3))

Correct answer: The normal force from the spheres on the bottom of the box is (15/4)mg

Solution

With upward acceleration g/4, the effective gravity is g_eff = g + g/4 = 5g/4. Total normal force from box bottom = 3m * (5g/4) = 15mg/4. The geometry (equilateral triangle of centres) gives the angle between the line of centres and vertical as 30 deg. Top sphere equilibrium: 2N*cos(30 deg) = m*5g/4 -> N = 5mg/(4*sqrt(3)). Force between bottom spheres: F = N*sin(30 deg) = 5mg/(4*sqrt(3)) * 1/2 = 5mg/(8*sqrt(3)) = 5*sqrt(3)*mg/24.

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