ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 · 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 AICE-JRF/SRF (PhD) — Farm Machinery and Power Engineering
Unit 6 soil dynamics in tillage and traction at PG depth: Mohr-Coulomb and soil-metal friction, soil failure and critical depth, Reece's equation, Bekker and Janosi-Hanamoto models, mobility numbers, compaction models.
How our questions for ICAR AICE-JRF/SRF spread across the units of Mechanics of Tillage & Traction (Soil Dynamics). Difficulty: 2 easy, 26 medium, 25 hard.
| Traction theory: slip, rolling resistance and net traction | 7 | |
| Soil-tool mechanics | 6 | |
| Soil compaction by vehicles | 5 | |
| Dynamic properties and strength of soil | 4 | |
| Tractive efficiency | 4 | |
| Wheel and track mechanics | 4 | |
| Cone index and cone penetrometer | 4 | |
| Longitudinal and lateral stability | 4 | |
| Tractor chassis mechanics | 4 | |
| Traction in wetland soils | 4 | |
| Weight transfer | 3 | |
| Determination of the centre of gravity | 3 | |
| Prediction of draft | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
In moist plastic soil, the draft of a tine recorded against time is smooth and continuous, while in dry friable soil it shows a repeating saw-tooth pattern. What does the saw-tooth indicate?
D. Successive shear failure of soil blocks ahead of the tool
In brittle soil, force builds as a block is compressed, drops when it shears away, and builds again: a saw-tooth. Flow failure in plastic soil gives a smoother trace.
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 longer track contact length, at the same load and width, develops a given thrust at lower slip because
D. shear displacement accumulates along the contact length, mobilising soil shear strength sooner
Shear displacement j at a point grows with its distance from the front of the contact (j = s·x). A long contact mobilises more of the soil's shear strength at a given slip, so thrust builds at lower slip.
A penetrometer using the 130 mm² cone records 260 N at 10 cm depth. A colleague mistakenly uses the 323 mm² base area. By what factor does the colleague underestimate the cone index?
A. 2.48
Correct CI = 260/130×10⁻⁶ = 2.0 MPa; with the wrong area 260/323×10⁻⁶ = 0.805 MPa. Ratio = 323/130 = 2.48. CI scales with 1/area, not with the square or square root. 0.40 inverts the ratio; 6.17 squares it; 1.58 takes its square root.
A tractor (track 1.4 m, CG height 0.8 m, CG on the centreline) turns on level ground on a 5 m radius. Above what speed will it begin to overturn outwards?
D. 23.6 km/h
Overturn when v²/(g r) = (t/2)/h → v = √(g r t/(2h)) = √(9.81 × 5 × 1.4/(2 × 0.8)) = 6.55 m/s = 23.6 km/h. Using full track in place of half gives 33.4 km/h; quoting m/s as km/h is a unit slip. 154.5 km/h forgets the square root (42.9 × 3.6).
A tractor of weight 30 kN, wheelbase 2.2 m, has a level front reaction of 9.0 kN. With the front raised 20° (rear wheels on the ground) the front reads 7.5 kN. The axle centres are level on level ground at 0.7 m height. How high is the CG above the ground?
B. 1.00 m
y above the axle line = (Wf − Wf′)L/(W tan θ) = (9.0 − 7.5) × 2.2/(30 × tan 20°) = 3.30/10.919 = 0.302 m. Height above ground = 0.7 + 0.302 = 1.00 m. Stopping at y forgets the axle height; using sin θ or cos θ uses the wrong trigonometric function. 0.30 m stops at y; 1.02 m uses sin θ; 0.82 m uses cos θ.
On a 2WD tractor pulling a load on level ground in steady motion, the horizontal force balance gives
C. rear wheel net thrust = drawbar pull + front wheel rolling resistance
The rear tyres must overcome both the implement and the front wheels' rolling resistance. Gross thrust is larger still, since it also covers the rear wheels' own motion resistance.
In a three-point linkage, what is the virtual hitch point?
C. The point where the lines of the upper and lower links, extended, intersect in side view
The linkage acts as if the implement were hitched at the instantaneous centre where the extended link lines meet; its position determines how the implement's forces are transferred to the tractor.
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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