NCEES FE Other Disciplines · USA · NCEES
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
NCEES FE Other Disciplines
Statics (9-14 questions): resultants, equilibrium of rigid bodies, trusses and frames, centroids, area moments of inertia, static friction. Dynamics (9-14 questions): particle and rigid-body kinematics, Newton's second law, work-energy, impulse-momentum. Almost all are short numerical problems solved with the NCEES Reference Handbook, so speed with FBDs and unit handling decides the score.
How our questions for NCEES FE Other Disciplines spread across the units of Engineering Mechanics (Statics & Dynamics). Difficulty: 19 easy, 32 medium, 19 hard.
| Kinetics: Newton's laws and D'Alembert's principle | 8 | |
| Work-energy principle | 8 | |
| Free-body diagrams and equilibrium | 7 | |
| Analysis of trusses and frames | 7 | |
| Area and mass moment of inertia | 7 | |
| Impulse and momentum | 7 | |
| Force systems and resultants | 6 | |
| Friction, including belt friction | 6 | |
| Kinematics of particles and rigid bodies | 6 | |
| Centroid and centre of gravity | 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
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 front loader raises an 800 kg round bale through 2.5 m in 4 s at steady speed. What is the average useful power delivered to the bale?
B. 4.91 kW
Work = m g h = 800 x 9.81 x 2.5 = 19620 J; power = 19.62 kJ / 4 s = 4.91 kW. Using mass instead of weight gives 0.50 kW; 19.62 is the energy in kJ, not a power. The hydraulic pump must supply more because of losses. 78.48 kW multiplies the energy by the time.
Further reading: Meriam and Kraige, Engineering Mechanics: Statics / Dynamics
A trailer starts from rest and rolls freely down a ramp, dropping 2 m in height. Ignoring friction and wheel rotational inertia, what is its speed at the bottom?
D. 6.26 m/s
Energy conservation: m g h = 0.5 m v^2, v = sqrt(2 g h) = sqrt(2 x 9.81 x 2) = 6.26 m/s. The result is independent of mass and of the ramp shape. Omitting the square root gives 39.24. 4.43 m/s drops the 2 (sqrt(g h)); 19.62 is g h itself.
Further reading: Meriam and Kraige, Engineering Mechanics: Statics / Dynamics
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
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
An ideal hydraulic jack has a pump piston of 20 mm diameter and a lifting ram of 80 mm diameter. What force on the pump piston supports 16 kN on the ram?
A. 1.0 kN
Equal pressure (or virtual work with equal displaced volume): F1 / A1 = F2 / A2, so F1 = 16 x (20/80)^2 = 16 / 16 = 1.0 kN. Using the diameter ratio instead of the area ratio gives 4 kN. 0.25 kN cubes the diameter ratio; 256 kN multiplies by the area ratio instead of dividing.
Further reading: Meriam and Kraige, Engineering Mechanics: Statics / Dynamics
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