ICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersHow soil resists tools and supports wheels: draft, slip, tractive efficiency, weight transfer and stability.
In the exams: GATE AG Section 3 (traction asked in all 8 papers 2019–26; weight transfer and stability in 5 of 8); ICAR AICE subject 39 Unit 6; ASRB NET FMP
ICAR AIEEA-PG — Agricultural Engineering & Technology
Unit V farm machinery and power: definitions of slip, tractive efficiency, rolling resistance, cone index, weight transfer and stability, with short calculations.
How our questions for ICAR AIEEA-PG spread across the units of Mechanics of Tillage & Traction (Soil Dynamics). Difficulty: 30 easy, 17 medium, 2 hard.
| Traction theory: slip, rolling resistance and net traction | 9 | |
| Longitudinal and lateral stability | 5 | |
| Dynamic properties and strength of soil | 4 | |
| Prediction of draft | 4 | |
| Tractive efficiency | 4 | |
| Wheel and track mechanics | 4 | |
| Soil-tool mechanics | 3 | |
| Weight transfer | 3 | |
| Determination of the centre of gravity | 3 | |
| Tractor chassis mechanics | 3 | |
| Soil compaction by vehicles | 3 | |
| Cone index and cone penetrometer | 2 | |
| Traction in wetland soils | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A soil has cohesion 15 kPa and angle of internal friction 25°. What is its shear strength on a plane carrying a normal stress of 100 kPa?
C. 61.6 kPa
τ = c + σ tan φ = 15 + 100 × tan 25° = 15 + 46.6 = 61.6 kPa. Leaving out cohesion gives 46.6 kPa; using sin φ instead of tan φ gives 57.3 kPa. 105.6 kPa uses cos φ instead of tan φ.
For a cohesive soil, how does adhesion to a steel tool usually change as moisture content rises from dry to very wet?
A. It rises to a maximum at intermediate moisture and then falls as a water film lubricates the interface
Dry soil has little stickiness; adhesion rises with moisture as water bridges form, then falls when a free water film separates soil from steel. This is why sticky clays cause build-up at intermediate moisture.
A tractor covers a marked distance in 17 drive-wheel revolutions without load and in 20 revolutions when pulling a plough. What is the slip?
D. 15.0 %
s = 1 − N0/N1 = 1 − 17/20 = 15.0 %. Using (N1 − N0)/N0 = 17.6 % is the common wrong basis: slip is referred to the loaded revolutions (theoretical travel). 85.0 % is N0/N1 itself; 8.1 % divides the difference by the sum of the revolutions.
A tractor carries 20 kN on its drive wheels on dry sand (c = 0, φ = 30°). What is the maximum thrust the soil can provide?
D. 11.55 kN
H = A·c + W tan φ; with c = 0, H = 20 × tan 30° = 11.55 kN. Contact area does not matter in dry sand. Using sin φ gives 10.00 kN. 34.64 kN divides by tan φ; 17.32 kN uses cos φ.
Further reading: Micklethwaite relation, in Wong, Theory of Ground Vehicles
Why does tractive efficiency pass through a maximum as slip rises?
D. Net traction rises quickly then levels off, while the speed loss grows in proportion to slip
At zero slip, net traction and TE are zero. NT/GT rises steeply then saturates as soil strength is mobilised, while (1 − s) falls steadily; their product peaks.
A standard cone of base area 323 mm² needs a force of 450 N to penetrate soil at the standard rate. What is the cone index?
A. 1393 kPa
CI = F/A = 450 N / 323 × 10⁻⁶ m² = 1393189 Pa ≈ 1393 kPa (1.39 MPa). Dividing by mm² directly gives MPa-scale numbers in the wrong unit; using the small cone's area gives 3462 kPa.
Further reading: ASABE S313
Weight transfer from the front to the rear axle of a pulling tractor increases when
D. the hitch height is raised
ΔW = P·h/L: higher hitch or higher pull increases transfer; a longer wheelbase reduces it. Front ballast raises the front load but does not change the amount transferred.
A tractor with a rigid front axle has track width 1.6 m and CG 0.9 m above ground, midway between the wheels. What is the static side slope at which it begins to overturn?
B. 41.6°
tan θ = (t/2)/h = 0.8/0.9 = 0.889 → θ = 41.6°. Using the full track gives 60.6°, which overstates stability. 48.4° inverts the ratio; 62.7° uses sin⁻¹ instead of tan⁻¹.
When taking moments about the rear wheel contact point of a tractor on level ground, why do the front wheel rolling resistance and rear wheel thrust not appear?
D. They act horizontally at ground level, so their lines of action pass through the moment point's level and have zero arm
Both forces act at the tyre-soil interface, at ground level, so their moment about a point on the ground is zero. They do enter the horizontal force balance.
According to the widely used conclusion from Söhne's work, subsoil stress under a tyre is governed mainly by
D. the total axle load
Topsoil stress follows contact (roughly inflation) pressure; deeper stress follows total load. Lower pressure protects topsoil, but only lighter axle loads protect the subsoil.
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