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 papersForces, equilibrium and motion of particles and rigid bodies: the starting point for structures, machines and tractor mechanics.
In the exams: NCEES FE Statics 9–14 and Dynamics 9–14 questions; ICAR AIEEA-PG Unit IV
ICAR AIEEA-PG - Agricultural Engineering & Technology
Unit IV names engineering mechanics alongside strength of materials and soil mechanics. Expect definitions (Varignon, Lami, D'Alembert, angle of repose, self-locking), standard centroid and moment-of-inertia results, truss determinacy and zero-force members, belt friction ratio and short numericals. Question papers are not published, so emphasis is inferred from the syllabus.
How our questions for ICAR AIEEA-PG spread across the units of Engineering Mechanics (Statics & Dynamics). Difficulty: 21 easy, 26 medium, 18 hard.
| Friction, including belt friction | 8 | |
| Impulse and momentum | 8 | |
| Centroid and centre of gravity | 7 | |
| Kinematics of particles and rigid bodies | 7 | |
| Free-body diagrams and equilibrium | 6 | |
| Analysis of trusses and frames | 6 | |
| Force systems and resultants | 5 | |
| Area and mass moment of inertia | 5 | |
| Kinetics: Newton's laws and D'Alembert's principle | 5 | |
| Work-energy principle | 4 | |
| Virtual work | 4 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
At an unloaded truss joint, three members meet; two of them are collinear. What can be said about the third member?
A. It is a zero-force member under this loading
Resolve perpendicular to the collinear pair: only the third member has a component in that direction, and no load acts, so its force is zero. Zero-force members still matter: they brace compression members and carry load when the loading changes.
Further reading: Meriam and Kraige, Engineering Mechanics: Statics / Dynamics
A square-threaded screw jack lifting a trailer holds the load without a brake when the handle is released. This is because:
D. The friction angle phi is greater than the helix angle alpha
The lowering torque is W (dm/2) tan(phi - alpha); if phi > alpha it is positive, so torque must be applied to lower the load: the screw is self-locking. A self-locking square-thread screw has efficiency below 50 per cent.
Further reading: Meriam and Kraige, Engineering Mechanics: Statics / Dynamics
A T-section toolbar has a flange 120 mm wide by 20 mm thick on top of a web 20 mm thick and 100 mm deep. How far is the centroid above the bottom of the web?
C. 82.7 mm
Flange: A = 2400 mm2 at y = 110 mm; web: A = 2000 mm2 at y = 50 mm. y-bar = (2400 x 110 + 2000 x 50) / 4400 = 82.7 mm. 60 mm is half the total depth; 80 mm averages the part centroids without weighting by area; 37.3 mm is measured from the top.
Further reading: Meriam and Kraige, Engineering Mechanics: Statics / Dynamics
The perpendicular axis theorem, Iz = Ix + Iy, can be applied to:
D. Plane areas and thin flat plates (laminae), with z normal to the plane
The theorem needs all the mass or area to lie in the xy plane, so that r^2 = x^2 + y^2 for every element. For a solid body it does not hold. For a circle it gives J = 2 I_diameter.
Further reading: Meriam and Kraige, Engineering Mechanics: Statics / Dynamics
D'Alembert's principle converts a dynamics problem into one of equilibrium by:
D. Adding an inertia force equal to -m a at the centre of mass, then applying the equations of statics
Writing sum F - m a = 0 treats -m a as an extra, fictitious inertia force. The body is then in dynamic equilibrium and moments can be taken about any point. For a rotating rigid body an inertia couple -I alpha is also added.
Further reading: Meriam and Kraige, Engineering Mechanics: Statics / Dynamics
Using a rigid-body model on level ground, a tractor with a 1.4 m track and a centre of gravity 1.0 m high turns on a 12 m radius. Assuming the tyres do not skid, at what speed does it begin to overturn outward?
D. 9.08 m/s
Tipping about the outer wheels when m v^2/r x h = m g x b/2, so v = sqrt(g r b / (2h)) = sqrt(9.81 x 12 x 0.7 / 1.0) = 9.08 m/s. Using the full track gives 12.84; 82.40 forgets the square root; 12.97 m/s swaps h and b/2 (sqrt(9.81 x 12 x 1.0 / 0.7)). Real tractors overturn at lower speeds because of tyre deflection, slopes and bumps.
Further reading: Goering et al., Off-Road Vehicle Engineering Principles
A 3500 kg tractor reversing at 1.5 m/s couples automatically to a stationary 1500 kg trailer and they move off together. What is their common speed just after coupling?
D. 1.05 m/s
Momentum is conserved: 3500 x 1.5 = 5000 v, v = 1.05 m/s. 1.25 m/s comes from wrongly conserving kinetic energy, which is not conserved in a coupling (plastic) impact; 0.45 m/s uses the trailer's mass in the numerator. 0.75 m/s simply halves the tractor's speed.
Further reading: Meriam and Kraige, Engineering Mechanics: Statics / Dynamics
In a collision between two bodies, the coefficient of restitution e = 0 means the impact is:
C. Perfectly plastic: the bodies move together after impact
e = relative separation speed / relative approach speed. e = 0 means no separation: the bodies stay together, and the kinetic-energy loss is greatest. e = 1 is perfectly elastic. Momentum is conserved in both.
Further reading: Meriam and Kraige, Engineering Mechanics: Statics / Dynamics
Grain leaves a conveyor at 3 kg/s moving horizontally at 4 m/s, strikes a deflector and leaves it moving vertically downward at 4 m/s. What is the magnitude of the force the grain exerts on the deflector?
B. 16.97 N
F = m-dot x change in velocity. The velocity change has components -4 (horizontal) and -4 (vertical), magnitude 4 sqrt 2 = 5.66 m/s. F = 3 x 5.66 = 16.97 N. 12 N counts only one component; zero wrongly assumes that an unchanged speed means no force. 24.00 N adds the two components arithmetically (3 x 8).
Further reading: Meriam and Kraige, Engineering Mechanics: Statics / Dynamics
A frictionless screw jack has a lead of 10 mm and a handle radius of 0.25 m. What handle force holds a 15 kN load?
C. 95.5 N
One turn: effort moves 2 pi x 0.25 = 1.571 m, load moves 0.010 m. P x 1.571 = 15000 x 0.010, P = 95.5 N. Leaving out 2 pi gives 600 N; leaving the lead in millimetres gives a figure 1000 times too large. Real jacks need more because of friction. 191.0 N takes 0.25 m as the handle diameter (pi x 0.25).
Further reading: Meriam and Kraige, Engineering Mechanics: Statics / Dynamics
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