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Reactions on two cylinders in channel

Short answer

Using smooth-contact geometry, the angle between cylinder centres is 56.1°. Equilibrium of A gives N_D = 236.3 N and N_E = 131.7 N. Equilibrium of B gives N_F = 152.1 N and N_G = 610.6 N.

Two cylinders A and B rest in a channel as shown in figure below. 'A' has a diameter of 10 cm and weight 20 kg. 'B' has 18 cm diameter and weight 50 kg. The channel is 18 cm wide at the bottom with one side vertical and other side at 120° as shown. Determine the reactions at four contact points. [Figure: cylinder B on the channel floor between a vertical wall and an inclined wall at 60° to the floor, cylinder A resting on B and against the vertical wall; contact points labelled including D, E, F, G.]

Reading of the figure: B (d = 18 cm, r = 9 cm, 50 kg) sits on the floor and touches the inclined wall at F. A (d = 10 cm, r = 5 cm, 20 kg) rests on B (contact D) and touches the vertical wall at E. The inclined wall makes 120° with the floor inside the channel (60° as marked outside). B touches the floor at G. The channel is 18 cm wide at the bottom, between the foot of the inclined wall and the vertical wall. All contacts are smooth. Take g = 9.81 m/s², so W_A = 196.2 N and W_B = 490.5 N.

Geometry

Origin at the foot of the inclined wall, x to the right, y up. The vertical wall is x = 18.

  • Centre of B: y_B = 9. Its distance from the inclined wall (normal n = (sin 60°, cos 60°)) must be 9: x_B sin 60° + 9 cos 60° = 9, so x_B = 5.196 cm.
  • Centre of A: x_A = 18 − 5 = 13 cm.
  • AB = 5 + 9 = 14 cm. Horizontal offset = 13 − 5.196 = 7.804 cm, so vertical offset = √(14² − 7.804²) = 11.62 cm.
  • The line of centres makes angle φ with the horizontal: cos φ = 7.804/14 = 0.5574, sin φ = 0.8302 (φ = 56.1°).

Cylinder A (weight 196.2 N; N_D along the line of centres; N_E horizontal)

ΣFy = 0:  N_D sinφ − 196.2 = 0   ->  N_D = 196.2/0.8302 = 236.3 N
ΣFx = 0:  N_D cosφ − N_E = 0     ->  N_E = 236.3 × 0.5574 = 131.7 N

Cylinder B (weight 490.5 N; N_D from A acts down-left; N_F perpendicular to the inclined wall; N_G vertical)

ΣFx = 0:  N_F sin60° − N_D cosφ = 0   ->  N_F = 131.7/0.8660 = 152.1 N
ΣFy = 0:  N_G + N_F cos60° − 490.5 − N_D sinφ = 0
          N_G = 490.5 + 196.2 − 0.5 × 152.1 = 610.6 N

Check: ΣFy for the whole system: N_G + N_F cos 60° = 610.6 + 76.1 = 686.7 N = 196.2 + 490.5 ✓.

Answer: N_F = 152.1 N (about 15.5 kgf, normal to the inclined wall), N_G = 610.6 N (62.2 kgf, floor), N_D = 236.3 N (24.1 kgf, between A and B), N_E = 131.7 N (13.4 kgf, vertical wall).

✓ Answer checked by Claude Opus 5.5 (high effort)

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