--- title: "Reactions on two cylinders in channel" subject: "Applied Mechanics" chapter: 2 key: "semester_1/applied_mechanics:two-cylinders-channel-four-contact-reactions" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/two-cylinders-channel-four-contact-reactions site: BCT Engineering --- # Reactions on two cylinders in channel Applied Mechanics · Chapter 2: Basic Concept in Statics and Static Equilibrium - Asked in: 2067 Asar, 11 marks - Video: [Engineering Mechanics - Forces - Equilibrium - Two Cylinders](https://www.youtube.com/watch?v=Is_FL6ibPsY) — Senthil nathan - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/two-cylinders-channel-four-contact-reactions - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-2 ## Question 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.] ## 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. ## Answer **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) ```text 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) ```text 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). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the distance between the centres of A and B taken as 14 cm? A. It equals the sum of the radii (5 + 9) because the smooth cylinders touch B. It equals the channel width minus the radius of A C. It equals the difference of the radii (9 - 5) D. It equals the sum of the two diameters Answer: A. Touching circles have centres separated by the sum of their radii, 5 + 9 = 14 cm. This gives the angle phi of the line of centres. (Key point: Line of centres) 2. Which equation gives N_D from the free-body diagram of cylinder A? A. N_D sin(phi) = 196.2 (vertical balance) B. N_D cos(phi) = 196.2 C. N_D = N_E D. N_E = 196.2 Answer: A. The only vertical forces on A are its weight and the vertical component of N_D (along the line of centres), so N_D sin(phi) = W_A, giving 236.3 N. (Key point: Free body of A) 3. On cylinder B, what balances the horizontal component of the push N_F from the inclined wall (N_F sin60°)? A. The horizontal component of N_D from A, N_D cos(phi) B. The floor reaction N_G C. The weight of B D. The vertical wall reaction N_E acting directly on B Answer: A. The floor reaction is vertical and smooth, so only N_D cos(phi) (= 131.7 N) balances N_F sin60°, giving N_F = 152.1 N. (Key point: Equilibrium of B horizontally) 4. Why is the floor reaction N_G (610.6 N) less than the total weight 686.7 N? A. The inclined wall's reaction N_F has an upward component of about 76.1 N that carries part of the weight B. Friction on the vertical wall carries part of the weight C. Cylinder A rests partly on the floor D. N_G is found by subtracting the weight of A only Answer: A. Vertical balance of the whole system gives N_G + N_F cos60° = 686.7 N, so N_G = 686.7 - 76.1 = 610.6 N. 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