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Reactions on two spheres in a channel

Short answer

Geometry gives a 60° line of centres. For sphere A: R2 sin60°=200 N → R2=230.94 N, R1=R2 cos60°=115.47 N. For sphere B: R3=200+R2 sin60°=400 N, R4=R2 cos60°=115.47 N.

Two smooth spheres of weight 200 N each are kept inside a channel as shown in the figure. The radius of each sphere is 20 cm. Determine reactions at contact surfaces. [Figure: spheres A (upper) and B (lower) in a channel of width 60 cm with floor and vertical walls; contact points labelled 1, 2, 3, 4.]

Reading of the figure: two smooth spheres A (upper) and B (lower), each W = 200 N, radius 20 cm, in a channel 60 cm wide. A touches the left wall (contact 1) and B (contact 2). B touches the floor (contact 3) and the right wall (contact 4).

Geometry

Horizontal distance between centres = 60 − 20 − 20 = 20 cm. Distance between centres = 20 + 20 = 40 cm. cos θ = 20/40, so θ = 60° (the line of centres makes 60° with the horizontal; A is up and to the left of B).

Sphere A (three forces: W, R₁ horizontal, R₂ along the line of centres)

ΣFy = 0:  R2 sin60° − 200 = 0   ->  R2 = 200/0.8660 = 230.94 N
ΣFx = 0:  R1 − R2 cos60° = 0    ->  R1 = 230.94 × 0.5 = 115.47 N

Sphere B (W, R₂ from A, R₃ vertical, R₄ horizontal)

Sphere A pushes B down and to the right with 230.94 N.

ΣFy = 0:  R3 − 200 − R2 sin60° = 0  ->  R3 = 200 + 200 = 400 N
ΣFx = 0:  R4 − R2 cos60° = 0        ->  R4 = 115.47 N

Check: the floor carries both weights, 400 N.

Answer: R₁ = 115.47 N (left wall on A), R₂ = 230.94 N (between A and B), R₃ = 400 N (floor on B), R₄ = 115.47 N (right wall on B).

✓ Answer checked by Claude Opus 5.5 (high effort)

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