- 2070 Chaitra · 8 marks
Reactions on three cylinders in a ditch
Short answer
Using cylinder radii and ditch width to find contact angles, equilibrium on each cylinder gives: C-left wall 4.00 kN horizontal, C-B contact 5.00 kN, B-A contact 13.33 kN, B-right wall 14.67 kN, A-left wall 10.67 kN, A-floor 10.00 kN vertical.
Determine the reactions at the contact points, if three cylinders are piled in a rectangular ditch as shown in figure. Given that the weight of the cylinders are: WA = 2 kN, WB = 5 kN, WC = 3 kN. [Figure: rectangular ditch of width 18 cm; cylinder A (rA = 4 cm) and cylinder B (rB = 6 cm) on the floor side by side, cylinder C (rC = 4 cm) resting on top against the left wall.]
Reading of the figure: a ditch 18 cm wide. A (r = 4 cm, W = 2 kN) rests on the floor against the left wall. B (r = 6 cm, W = 5 kN) rests on A and touches the right wall. C (r = 4 cm, W = 3 kN) rests on B and touches the left wall. All contacts are smooth. (Floor level and wall heights are not needed.)
Geometry
Take the floor as y = 0 and the left wall as x = 0.
- A centre: (4, 4). B centre: x = 18 − 6 = 12.
- AB = r_A + r_B = 10 cm, horizontal offset 12 − 4 = 8 cm, so vertical offset = √(10² − 8²) = 6 cm. B centre = (12, 10).
- C centre: x = 4. BC = 10 cm, horizontal offset 8 cm, vertical 6 cm, so C is 6 cm above B: (4, 16).
- Both lines of centres have slope 6/8: cos θ = 0.8, sin θ = 0.6 (θ = 36.87° with the horizontal).
Cylinder C (W = 3 kN; wall reaction R₁ horizontal; N_BC along BC)
N_BC acts on C up and to the left, along (−0.8, 0.6).
y: 0.6 N_BC − 3 = 0 -> N_BC = 5 kN
x: R1 − 0.8 N_BC = 0 -> R1 (left wall on C) = 4 kN
Cylinder B (W = 5 kN; N_BC from C acts on B along (0.8, −0.6); N_AB from A along (0.8, 0.6); right wall R₄ horizontal, to the left)
y: 0.6 N_AB − 5 − 0.6 × 5 = 0 -> N_AB = 8/0.6 = 13.33 kN
x: 0.8 × 5 + 0.8 × 13.33 − R4 = 0 -> R4 = 4 + 10.67 = 14.67 kN
Cylinder A (W = 2 kN; N_AB from B acts on A along (−0.8, −0.6); floor R₃ up; left wall R₂ to the right)
y: R3 − 2 − 0.6 × 13.33 = 0 -> R3 = 10 kN
x: R2 − 0.8 × 13.33 = 0 -> R2 = 10.67 kN
Checks: R₃ = 2 + 5 + 3 = 10 kN ✓. Left walls R₁ + R₂ = 4 + 10.67 = 14.67 kN = R₄ ✓.
Answer:
| Contact | Reaction |
|---|---|
| C with left wall | 4.00 kN (horizontal) |
| C with B | 5.00 kN (along line of centres) |
| B with A | 13.33 kN (along line of centres) |
| B with right wall | 14.67 kN (horizontal) |
| A with left wall | 10.67 kN (horizontal) |
| A with floor | 10.00 kN (vertical) |
✓ Answer checked by Claude Opus 5.5 (high effort)
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