Five free practice questions on mechanics and dynamics: unit A of paper A (4A) of the 4th Class Power Engineer certificate exam, written to the SOPEEC syllabus. They are in the exam's format: four options, one correct answer. Answer each one before opening the explanation. No account, no email.
01The questions
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A 1 800 kg rail-mounted ash cart at a generating station is pulled from rest along level track by a winch, with an acceleration of 0.40 m/s². Ignoring rolling resistance, what force must the winch cable apply?
- 0.720 kN
- 4.50 kN
- 7.06 kN
- 17.7 kN
Show the answer
A. Correct. On level track the cable only has to accelerate the cart's mass: F = ma = 1 800 kg × 0.40 m/s² = 720 N = 0.720 kN. The cart's weight is carried by the rails and does not enter.
Why not the others
B. This divides the mass by the acceleration: 1 800 ÷ 0.40 = 4 500 N. Newton's second law multiplies them: F = ma = 1 800 × 0.40 = 720 N = 0.720 kN.
C. This uses the cart's weight in place of its mass: 1 800 × 9.81 × 0.40 = 7 063 N. F = ma needs the mass in kg: 1 800 × 0.40 = 720 N = 0.720 kN.
D. This is the cart's weight, 1 800 × 9.81 = 17 658 N, the force needed to lift it, not to accelerate it along level rails. F = ma = 1 800 × 0.40 = 720 N = 0.720 kN.
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A horizontal beam pivoted at its centre carries a 600 N load 1.2 m left of the pivot and a 400 N load 1.5 m right of the pivot. Ignoring the beam's own weight, what happens?
- The right end drops, with a net moment of 120 N·m.
- The left end drops, with a net moment of 120 N·m.
- The left end drops, with a net moment of 1 320 N·m.
- The left end drops, with a net moment of 200 N·m.
Show the answer
B. Correct. Left: 600 × 1.2 = 720 N·m; right: 400 × 1.5 = 600 N·m. The left moment is larger by 720 − 600 = 120 N·m, so the beam turns with its left end going down. It is not in equilibrium (ΣM ≠ 0).
Why not the others
A. The right load has the longer arm, but its moment, 400 × 1.5 = 600 N·m, is smaller than the left one, 600 × 1.2 = 720 N·m. The left end drops under the net 120 N·m.
C. This adds the two moments (720 + 600). They act in opposite senses about the pivot, so they subtract: 720 − 600 = 120 N·m, turning the left end down.
D. This takes the difference in force, 600 − 400 = 200, and leaves out the distances. Moments compare force × distance: 720 N·m left against 600 N·m right, a net 120 N·m.
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A 400 kg drum is pushed up a ramp 6.0 m long that rises 1.2 m at a food plant, with the push parallel to the ramp. The ramp, used this way, is 80 % efficient. What push is needed?
- 100 N
- 628 N
- 785 N
- 981 N
Show the answer
D. Correct. Load = 400 × 9.81 = 3 924 N. VR = ramp length ÷ rise = 6.0 ÷ 1.2 = 5. MA = efficiency × VR = 0.80 × 5 = 4.0, so push = 3 924 ÷ 4.0 = 981 N.
Why not the others
A. This uses the drum's mass in kilograms as the load: 400 ÷ 5 ÷ 0.80 = 100. The load is its weight, 400 × 9.81 = 3 924 N, so the push is 3 924 ÷ 5 ÷ 0.80 = 981 N.
B. This multiplies the ideal push by the efficiency: 785 × 0.80 = 628 N. Friction makes a real push larger than the ideal one, so divide: 785 ÷ 0.80 = 981 N.
C. This is the push with no friction: weight ÷ VR = 3 924 ÷ 5 = 785 N. At 80 % efficiency the push must be 785 ÷ 0.80 = 981 N.
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Two horizontal tugger lines pull on the same lug of a heat-exchanger skid at a university plant: one with 2.5 kN due east, the other with 4.0 kN due north. What is the magnitude of their resultant?
- 1.50 kN
- 3.12 kN
- 4.72 kN
- 6.50 kN
Show the answer
C. Correct. Drawn head to tail the two forces form a right-angled triangle, so R = √(2.5² + 4.0²) = √22.25 = 4.72 kN, acting at tan⁻¹(4.0 ÷ 2.5) = 58° north of east.
Why not the others
A. This subtracts the forces (4.0 − 2.5), as if they acted in opposite directions. At right angles they combine by Pythagoras: √(2.5² + 4.0²) = 4.72 kN.
B. This subtracts the squares: √(4.0² − 2.5²) = √9.75 = 3.12. For two forces at right angles the resultant is the hypotenuse: √(2.5² + 4.0²) = √22.25 = 4.72 kN.
D. This adds the forces as scalars (2.5 + 4.0). That holds only if both pull the same way. At right angles: R = √(2.5² + 4.0²) = 4.72 kN.
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On a graph of velocity against time, the slope of the line and the area under it represent
- acceleration and distance travelled
- distance travelled and acceleration
- velocity and acceleration
- acceleration and average velocity
Show the answer
A. Correct. Slope = change in velocity ÷ time = acceleration (a horizontal line means zero acceleration). Area = velocity × time = distance travelled (displacement, for motion in one direction).
Why not the others
B. This swaps them. The slope is rise ÷ run = change in velocity ÷ time, which is acceleration; the area is velocity × time, which is distance.
C. Velocity is read straight off the vertical axis. The slope gives acceleration and the area under the line gives the distance travelled.
D. The slope is acceleration, but the area is velocity × time, which is a distance in metres. Average velocity would be that area divided by the time.
02Answer key
| Question | Answer | Learning outcome |
|---|---|---|
| Q1 | A | Define mass, force, acceleration, velocity, and weight |
| Q2 | B | Determine the direction and calculate the magnitude of the moment of a force |
| Q3 | D | Define the term simple machine and apply to calculations of mechanical advantage, velocity ratio and efficiency of simple machines |
| Q4 | C | Define scalar and vector quantities as they apply to drawing vector diagrams |
| Q5 | A | Draw graphs of velocity as a function of time |
03What unit A covers
The mechanics behind plant work: SI units, mass and weight, moments and equilibrium, simple machines such as pulleys, screw jacks and ramps, vectors and resultant forces, speed, velocity and acceleration, work, power and energy, gauge and absolute pressure, friction, stress, strain and factor of safety, and belt, gear and chain drives.
The SOPEEC 4th Class syllabus names this unit “Elementary Mechanics and Dynamics” (ABSA numbers it A1) and lists nine objectives under it. SOPEEC does not publish how many exam questions come from each unit, and neither do the ABSA, Technical Safety BC, TSASK or TSSA pages we checked. The right-hand column is how many questions our bank holds on each objective, Standard and Advanced together, not a count from the exam.
| Syllabus objective | In our bank |
|---|---|
| A-1 Perform basic calculations and define basic terms used in the study of mechanics. | 7 |
| A-2 Perform calculations using forces and moments, and determine whether or not a system is in equilibrium. | 8 |
| A-3 Define simple machines and perform calculations relating to mechanical advantage, velocity ratio, and efficiency. | 7 |
| A-4 Define and identify scalar and vector quantities, and solve simple vector problems graphically. | 6 |
| A-5 Define speed, velocity, distance, displacement, and acceleration, and solve simple linear problems involving these terms. | 10 |
| A-6 Define force, work, pressure, power, and, energy and perform calculations involving the relationships between these mechanical terms. | 9 |
| A-7 Describe friction and solve problems involving friction on a horizontal plane. | 7 |
| A-8 Explain the following: a. The physical properties of materials and how these properties affect the materials behaviour when external forces are applied b. Stress and the deformation of bodies caused by externally applied forces as well as the internal forces that resist these deformations, including tensile, compressive, and shear stresses plus factor of safety | 9 |
| A-9 Explain the common examples of power transmission systems, including speed changes, transmitted power, and efficiency. | 8 |
04What the questions turn on
Absolute uses the local barometer
Absolute pressure is gauge pressure plus the atmospheric pressure actually present, so a plant at altitude adds its own barometer reading, not the standard 101.3 kPa. A sealed vessel moved to a lower barometer keeps its absolute pressure, so its gauge reads higher.
Moments use the perpendicular distance
A moment is the force times the perpendicular distance to the force's line of action. A push at an angle to a handle turns less than the same push at right angles, and a beam is in equilibrium only when the moments balance.
Efficiency is MA divided by VR
Mechanical advantage is load divided by effort. Velocity ratio comes from the distances moved, such as the number of rope falls carrying the load or 2πR over the pitch of a screw jack. Efficiency is MA divided by VR.
Drive ratios multiply stage by stage
On a belt drive, diameter times speed is the same on both pulleys; on gears and chains, teeth times speed. Through a jackshaft, work each stage in turn. An idler gear reverses rotation but leaves the overall ratio unchanged.
05The other units
Paper 4A has twelve units, A to L. Each has its own page of free questions:
The 4th Class Power Engineering practice exam (4A) page describes the full question bank. How these questions are written and checked: how our questions are made. Found a mistake? Tell us and we will fix it.