A grain chute that is fast on dry days and stalls on humid mornings
On a grain-handling platform, 50 kg jute sacks are slid down a straight galvanised-steel chute to a truck bed. The chute is 3.6 m long and inclined at 30° to the horizontal, and its lower end is level with the truck bed, 1.8 m below the platform. Workers give each sack a small push so that it starts at 0.5 m/s. The coefficient of kinetic friction between dry jute and the sheet is 0.25; the coefficient of static friction is 0.35. The supervisor reports that on most days the sacks arrive at the bottom too fast for one worker to control, but on humid mornings some sacks come to a stop part-way down and have to be pushed again. Management will accept a maximum delivery speed of 3.0 m/s at the lower end. Assume g = 9.81 m/s², and neglect air resistance and any rotation of the sack.
1. Calculate the speed of a dry sack at the bottom of the chute and state whether the reported behaviour is consistent with the quoted friction.
Read the worked answer
Along the chute the driving component is mg sin30° = 0.500 mg and kinetic friction is μk mg cos30° = 0.25 × 0.866 mg = 0.2165 mg. The net force is 0.2835 mg, so
a = 0.2835 × 9.81 = 2.78 m/s².
Because 0.500 > 0.2165 the sack accelerates over the whole 3.6 m:
v² = u² + 2aL = 0.5² + 2(2.78)(3.6) = 0.25 + 20.03 = 20.28, so v ≈ 4.5 m/s.
That is 1.5 times the 3.0 m/s limit, so the 'too fast' complaint is fully consistent with the data. Note what stopping would require: zero acceleration needs tan θ ≤ μk, i.e. μk ≥ tan 30° = 0.577. Dry jute on steel at 0.25 cannot produce a stall, so the humid-morning stoppages cannot be explained by the quoted dry coefficient.
2. How would you investigate and fix the humid-morning stalls without simply flattening the chute?
Read the worked answer
Diagnose the interface before changing geometry. A sack cannot stall on a 30° sheet at μk = 0.25, so either the friction has changed or the chute is not straight. Damp jute on galvanised steel commonly reaches μk of 0.5–0.7, well above the 0.577 threshold, which explains why the fault appears only on humid mornings; dents, flat spots and spilled grain acting as rollers can do the same.
Checks: pull a representative sack along the actual sheet with a spring balance, inspect the sheet for deformation, and compare a dry and a damp sack on the same day.
Remedy: keep the 30° geometry and fit a rubberised brake strip over the last 1 m, so friction is raised only over a short, controlled length; store sacks under cover overnight. Do not oil the chute - it removes what little braking exists and worsens the dry-day problem - and do not wet it, which worsens the humid-day problem.
3. If the angle were changed instead, what angle would give 3.0 m/s, and what does that imply about the design?
Read the worked answer
Requirement v = 3.0 m/s at the bottom:
9.0 = 0.25 + 2a(3.6) → a = 1.216 m/s², so sin θ − 0.25 cos θ = 1.216/9.81 = 0.1239.
Write the left side as R sin(θ − φ) with tan φ = 0.25, φ = 14.04°, R = √1.0625 = 1.031:
sin(θ − 14.04°) = 0.1202 → θ − 14.04° = 6.90° → θ ≈ 20.9°.
Check restartability: tan 20.94° = 0.383, just above μs = 0.35, so a stopped sack would barely begin moving again.
Judgement: the usable window is only about 14° to 21°, and the upper end has no safety margin - one damp morning and the chute stalls again. Tuning the angle to a single friction coefficient is fragile. A 24–25° chute with a brake strip and a short horizontal run-out at the bottom meets both the speed limit and the restart requirement far more robustly.