Diploma Civil and Mechanical Engineering for Engineering Assistant · Chapter 2

Engineering Mechanics

What to remember

A body is in equilibrium when ΣFx = 0, ΣFy = 0 and ΣM = 0. Resultant of two forces comes from the parallelogram law, and three concurrent forces in equilibrium obey Lami's theorem.
Friction force F = μN, angle of friction φ satisfies tan φ = μ, and the angle of repose equals the angle of friction.
Motion equations v = u + at, s = ut + ½at², v² = u² + 2as; work = force × distance; efficiency = MA ÷ VR.

Forces and resultant

Force is a push or pull that changes the state of rest or motion of a body. SI unit: newton (N). 1 kgf = 9.81 N. A force has magnitude, direction, sense and point of application. It is a vector; mass, work and energy are scalars.
Principle of transmissibility: a force can be moved along its line of action without changing its effect on a rigid body.
Parallelogram law: two forces P and Q acting at a point at angle θ have resultant R = √(P² + Q² + 2PQ cos θ). The direction α from P is tan α = Q sin θ ÷ (P + Q cos θ).
If the forces are at 90°, R = √(P² + Q²). If they act in the same direction, R = P + Q. If opposite, R = P − Q.
Worked example 1: 30 N and 40 N at right angles give R = √(900 + 1600) = 50 N.
Worked example 2: two equal forces P with angle θ between them give R = 2P cos(θ/2). For P = 10 N and θ = 120°, R = 10 N.
Triangle law and polygon law give the resultant by drawing the forces head to tail.
Lami's theorem: if three concurrent forces keep a body in equilibrium, each is proportional to the sine of the angle between the other two: P/sin α = Q/sin β = R/sin γ.
Resolution of a force: components Fx = F cos θ and Fy = F sin θ. The resultant of many coplanar forces: R = √[(ΣFx)² + (ΣFy)²].

Force system	Description
Coplanar concurrent	Lines of action meet at one point in one plane
Coplanar parallel	Parallel lines of action in one plane
Coplanar non-concurrent	Not parallel and do not meet at one point
Collinear	Same line of action

Moments, couples and equilibrium

Moment of a force about a point = force × perpendicular distance. Unit: N·m. Example: 50 N at 0.4 m gives 20 N·m.
Varignon's theorem: the moment of the resultant about a point equals the algebraic sum of the moments of the components.
Couple: two equal, opposite, parallel forces with different lines of action. Its moment = force × distance between the lines. A couple produces rotation only, and its moment is the same about every point.
Equilibrium of coplanar forces: ΣFx = 0, ΣFy = 0 and ΣM = 0. A free body diagram shows all forces on an isolated body.
Supports: a roller gives one reaction (perpendicular to the surface); a hinge (pin) gives two reaction components; a fixed support gives two reaction components and a moment.
Types of beam: simply supported, cantilever, overhanging, fixed, continuous.
Types of load: point load, uniformly distributed load (UDL; its resultant acts at the centre of the loaded length) and uniformly varying load (resultant acts at one-third of the length from the larger end).

Friction

Friction is the force opposing relative motion between surfaces in contact. Limiting friction is the maximum friction before sliding starts. Static friction is greater than kinetic (sliding) friction, and sliding friction is greater than rolling friction.
Laws of dry friction: F = μN, independent of the area of contact and of the speed (at low speeds). μ is the coefficient of friction.
Angle of friction φ: tan φ = μ = F/N. Angle of repose α: the steepest slope at which a body just stays at rest; for a body on an inclined plane α = φ.
Worked example 3: a block of weight 500 N on a rough floor with μ = 0.3. Limiting friction = 0.3 × 500 = 150 N.
Worked example 4: a block just starts sliding at 30° incline. μ = tan 30° = 0.577.
On an inclined plane of angle α, the weight W has components W sin α (down the plane) and W cos α (normal to the plane).
Cone of friction is the cone with half angle φ around the normal. The body stays in equilibrium if the resultant reaction lies inside it.
Friction is useful in brakes, belts and walking. It is reduced by lubrication, ball bearings and streamlining.

Centroid and moment of inertia

Centroid is the centre of area; centre of gravity is the point where the whole weight acts. For uniform bodies they coincide.

Shape	Centroid location
Rectangle (b × d)	d/2 from base, b/2 from side
Triangle (height h)	h/3 from the base
Semicircle (radius r)	4r/3π from the diameter
Quarter circle	4r/3π from each straight edge
Solid cone (height h)	h/4 from the base
Solid hemisphere	3r/8 from the flat face

Moment of inertia (second moment of area): I = Σ a·y². Unit: mm⁴ or m⁴.

Section	I about centroidal axis	I about base
Rectangle (b wide, d deep)	bd³/12	bd³/3
Triangle (base b, height h)	bh³/36	bh³/12
Circle (diameter d)	πd⁴/64

Parallel axis theorem: I = Ig + A·h². Perpendicular axis theorem: Iz = Ix + Iy (for plane figures).
Radius of gyration: k = √(I/A).
Worked example 5: rectangle 60 mm × 100 mm: Ig = 60 × 100³ ÷ 12 = 5 × 10⁶ mm⁴. About the base, I = 5 × 10⁶ + 6000 × 50² = 20 × 10⁶ mm⁴.
The centroid of a composite figure is found by x̄ = Σ(a·x) ÷ Σa.

Kinematics and dynamics

Equations of uniform acceleration: v = u + at; s = ut + ½at²; v² = u² + 2as. For free fall use a = g = 9.81 m/s².
Worked example 6: from rest with a = 2 m/s² for 10 s: v = 20 m/s and s = 100 m.
Projectile motion (launch speed u at angle θ): time of flight T = 2u sin θ ÷ g; maximum height H = u² sin²θ ÷ 2g; range R = u² sin 2θ ÷ g. Maximum range at θ = 45°. Ranges for θ and (90° − θ) are equal.
Newton's laws: (1) a body stays at rest or in uniform motion unless a force acts (inertia); (2) F = ma, force is the rate of change of momentum; (3) action and reaction are equal and opposite.
Momentum = mv. Impulse = force × time = change of momentum. Conservation of momentum: total momentum before a collision equals total after, when no external force acts.
Worked example 7: a 5 kg body at 4 m/s sticks to a 3 kg body at rest. Common velocity = 20 ÷ 8 = 2.5 m/s.
Coefficient of restitution e = relative speed of separation ÷ relative speed of approach. e = 1 for a perfectly elastic collision and e = 0 for a perfectly inelastic one.
Work = force × displacement in the direction of force (joule). Power = work ÷ time (watt; 1 hp = 746 W). Kinetic energy = ½mv². Potential energy = mgh. Work-energy principle: net work equals the change in kinetic energy.
Worked example 8: lifting 500 N through 10 m in 20 s: work = 5000 J, power = 250 W.
Centripetal force = mv²/r, towards the centre. Simple pendulum time period T = 2π√(l/g); a spring-mass system has T = 2π√(m/k).

Simple machines and trusses

Mechanical advantage (MA) = load ÷ effort. Velocity ratio (VR) = distance moved by effort ÷ distance moved by load. Efficiency η = MA ÷ VR. For an ideal machine MA = VR and η = 100%.
Law of a machine: P = mW + C, where m is a constant slope and C is the effort to overcome friction at zero load.
A machine is reversible if η is more than 50%, and self-locking if η is 50% or less (for example a screw jack).
Examples of VR: wheel and axle = D/d; screw jack = 2πl/p (l = handle length, p = pitch); a system of n supporting ropes = n.
Worked example 9: W = 1000 N is raised by P = 100 N. MA = 10. If VR = 20, η = 10/20 = 50%.
Trusses: a frame made of straight members joined at pins. A perfect truss has m = 2j − 3 members (j = number of joints). Methods: joints and sections. Members carry only axial force: tension or compression.

Exam traps

Mass is a scalar and weight (force) is a vector. 1 kgf is 9.81 N.
Moment is force times the perpendicular distance, not any distance.
Friction does not depend on the area of contact. Static friction exceeds kinetic friction.
The angle of repose equals the angle of friction, and tan φ = μ (not sin or cos).
The centroid of a triangle is at h/3 from the base, not h/2 or 2h/3 from the base.
I of a rectangle is bd³/12 about the centroid but bd³/3 about the base.
Maximum range of a projectile is at 45° (not 30° or 60°), neglecting air resistance.
Efficiency is MA ÷ VR, and a machine with η ≤ 50% is self-locking.

One-liners

1. Resultant of perpendicular forces P and Q is √(P² + Q²).
2. Lami's theorem applies to three concurrent forces in equilibrium.
3. Moment of a couple = force × distance between the forces.
4. Friction F = μN, with tan φ = μ.
5. Centroid of a triangle is at h/3 from the base.
6. Centroid of a semicircle is at 4r/3π from the diameter.
7. Parallel axis theorem: I = Ig + Ah².
8. v² = u² + 2as.
9. Range of a projectile R = u² sin 2θ ÷ g.
10. Impulse equals change of momentum.
11. For a perfectly elastic collision, e = 1.
12. A perfect truss satisfies m = 2j − 3.

Practice questions

The SI unit of force is
1. joule
2. pascal
3. watt
4. newton
Answer
D. newton
One newton gives 1 kg an acceleration of 1 m/s².
Which of the following is a scalar quantity?
1. Force
2. Work
3. Velocity
4. Momentum
Answer
B. Work
Work has magnitude only. The others are vectors.
Lami's theorem is applicable to
1. any number of non-concurrent forces
2. two parallel forces
3. three concurrent forces in equilibrium
4. a couple
Answer
C. three concurrent forces in equilibrium
Each force is proportional to the sine of the angle between the other two.
A couple consists of two forces that are
1. equal and perpendicular
2. equal, opposite and parallel with different lines of action
3. unequal and parallel
4. equal and opposite on the same line
Answer
B. equal, opposite and parallel with different lines of action
This is the definition of a couple.
The condition for equilibrium of coplanar forces is
1. ΣFx = 0 only
2. ΣM = 0 only
3. ΣFx = ΣFy
4. ΣFx = 0, ΣFy = 0 and ΣM = 0
Answer
D. ΣFx = 0, ΣFy = 0 and ΣM = 0
All three conditions must hold.
The angle of friction φ is related to the coefficient of friction μ by
1. sin φ = μ
2. cos φ = μ
3. tan φ = μ
4. φ = μ
Answer
C. tan φ = μ
tan φ = F/N = μ.
Limiting friction is
1. the maximum friction force just before sliding starts
2. zero friction
3. the friction force at rest with no applied force
4. the friction force during sliding
Answer
A. the maximum friction force just before sliding starts
Beyond this value the body slides.
The centroid of a triangle of height h lies at what distance from its base?
1. 2h/3
2. h/4
3. h/2
4. h/3
Answer
D. h/3
The medians meet at one-third of the height from the base.
The moment of inertia of a rectangle about its centroidal axis parallel to the base is
1. bd³/12
2. b³d/12
3. bd²/6
4. bd³/3
Answer
A. bd³/12
For a rectangle I = bd³/12.
The first equation of motion is
1. s = ut + ½at²
2. v = u + at
3. s = vt
4. v² = u² + 2as
Answer
B. v = u + at
This relates final velocity, initial velocity, acceleration and time.
The maximum range of a projectile for a given speed is obtained at an angle of
1. 30°
2. 90°
3. 45°
4. 60°
Answer
C. 45°
R = u² sin 2θ ÷ g is maximum when sin 2θ = 1.
The coefficient of restitution for a perfectly elastic collision is
1. 0
2. 1
3. 0.5
4. infinite
Answer
B. 1
For a perfectly elastic collision e = 1.
Impulse is equal to
1. change in kinetic energy
2. force × distance
3. mass × acceleration
4. change in momentum
Answer
D. change in momentum
Force × time = change of momentum.
The efficiency of a machine is
1. VR ÷ MA
2. load × effort
3. MA ÷ VR
4. effort ÷ load
Answer
C. MA ÷ VR
η = MA/VR.
A perfect truss with j joints has how many members?
1. 2j − 3
2. j − 2
3. 2j + 3
4. 3j − 2
Answer
A. 2j − 3
The relation m = 2j − 3 gives a perfect (just rigid) truss.
Two forces of 30 N and 40 N act at right angles. The resultant is
1. 10 N
2. 50 N
3. 35 N
4. 70 N
Answer
B. 50 N
R = √(30² + 40²) = 50 N.
Two forces of 6 N and 8 N act at right angles. The resultant is
1. 10 N
2. 14 N
3. 2 N
4. 7 N
Answer
A. 10 N
R = √(36 + 64) = 10 N.
The moment of a 50 N force acting at a perpendicular distance of 0.4 m is
1. 2 N·m
2. 200 N·m
3. 125 N·m
4. 20 N·m
Answer
D. 20 N·m
M = 50 × 0.4 = 20 N·m.
Two forces of 20 N each, 0.5 m apart, form a couple. The moment of the couple is
1. 20 N·m
2. 40 N·m
3. 10 N·m
4. 5 N·m
Answer
C. 10 N·m
Moment = 20 × 0.5 = 10 N·m.
A block of weight 400 N rests on a floor with μ = 0.25. The limiting friction force is
1. 400 N
2. 100 N
3. 160 N
4. 1600 N
Answer
B. 100 N
F = μW = 0.25 × 400 = 100 N.
A body just begins to slide on a rough plane at an inclination of 45°. The coefficient of friction is
1. 0.5
2. 0.707
3. 1.414
4. 1
Answer
D. 1
μ = tan 45° = 1.
A body with initial velocity 10 m/s and acceleration 2 m/s² for 5 s reaches a velocity of
1. 20 m/s
2. 25 m/s
3. 30 m/s
4. 15 m/s
Answer
A. 20 m/s
v = u + at = 10 + 2 × 5 = 20 m/s.
A body starts from rest with acceleration 3 m/s². The distance travelled in 4 s is
1. 12 m
2. 36 m
3. 24 m
4. 48 m
Answer
C. 24 m
s = ½at² = 0.5 × 3 × 16 = 24 m.
A ball is thrown vertically upward at 19.62 m/s (g = 9.81 m/s²). The total time before it returns to the ground is
1. 2 s
2. 4 s
3. 1 s
4. 8 s
Answer
B. 4 s
T = 2u/g = 2 × 19.62 ÷ 9.81 = 4 s.
A 5 kg body moving at 4 m/s sticks to a 3 kg body at rest. The common velocity is
1. 1.5 m/s
2. 4 m/s
3. 1.6 m/s
4. 2.5 m/s
Answer
D. 2.5 m/s
Momentum 20 kg·m/s ÷ total mass 8 kg = 2.5 m/s.
The kinetic energy of a 4 kg body moving at 5 m/s is
1. 50 J
2. 100 J
3. 10 J
4. 20 J
Answer
A. 50 J
KE = ½mv² = 0.5 × 4 × 25 = 50 J.
The potential energy of a 10 kg mass at 5 m height (g = 9.81 m/s²) is
1. 50 J
2. 981 J
3. 490.5 J
4. 98.1 J
Answer
C. 490.5 J
PE = mgh = 10 × 9.81 × 5 = 490.5 J.
A force of 500 N lifts a load through 10 m in 20 s. The power is
1. 100 W
2. 250 W
3. 5000 W
4. 25 W
Answer
B. 250 W
Work = 5000 J, so power = 5000 ÷ 20 = 250 W.
A machine has an effort of 100 N lifting 1000 N, and a velocity ratio of 20. Its efficiency is
1. 20%
2. 200%
3. 50%
4. 10%
Answer
C. 50%
MA = 10, so η = 10 ÷ 20 = 50%.
The centroid of a triangle of height 12 cm is at what distance from the base?
1. 4 cm
2. 6 cm
3. 8 cm
4. 3 cm
Answer
A. 4 cm
h/3 = 12/3 = 4 cm.
For a rectangle 60 mm × 100 mm (deep), the moment of inertia about the centroidal axis parallel to the 60 mm side is
1. 1.8 × 10⁶ mm⁴
2. 2 × 10⁷ mm⁴
3. 2 × 10⁶ mm⁴
4. 5 × 10⁶ mm⁴
Answer
D. 5 × 10⁶ mm⁴
I = bd³/12 = 60 × 10⁶ ÷ 12 = 5 × 10⁶ mm⁴.
A simply supported beam of 6 m carries a UDL of 10 kN/m over the full span. Each reaction is
1. 30 kN
2. 10 kN
3. 15 kN
4. 60 kN
Answer
A. 30 kN
Total load = 60 kN, shared equally, so each reaction is 30 kN.
A truss has 5 joints. For it to be a perfect truss the number of members must be
1. 5
2. 8
3. 10
4. 7
Answer
D. 7
m = 2j − 3 = 7.
The moment of inertia of a rectangle about its base is bd³/3. This follows from which theorem?
1. Perpendicular axis theorem
2. Lami's theorem
3. Parallel axis theorem
4. Varignon's theorem
Answer
C. Parallel axis theorem
Ibase = bd³/12 + bd × (d/2)² = bd³/3.
A projectile is launched at 30° and another at 60° with equal speed. Their ranges are
1. in the ratio 2:1
2. equal
3. in the ratio 1:2
4. in the ratio 1:3
Answer
B. equal
sin 60° = sin 120°, so complementary angles give equal ranges.
A screw jack is self-locking when its efficiency is
1. more than 75%
2. exactly 100%
3. more than 50%
4. 50% or less
Answer
D. 50% or less
When η ≤ 50% the machine will not run back on its own.
Which statements are correct? 1. Friction force depends on the area of contact. 2. Static friction is greater than kinetic friction.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
Dry friction is independent of the contact area.
Which statements are correct? 1. The moment of a couple is the same about any point. 2. A couple produces translation of a body.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
A couple produces pure rotation, with no resultant force.
Which statements are correct? 1. The angle of repose equals the angle of friction. 2. The centroid of a semicircle is at 4r/3π from its diameter.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are standard results.
Which statements are correct? 1. Impulse is the change of momentum. 2. In a perfectly inelastic collision the bodies separate with equal speeds.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
In a perfectly inelastic collision (e = 0) the bodies stick together.
Which statements are correct? 1. A scalar has magnitude only. 2. Weight is a scalar quantity.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Weight is a force, so it is a vector.
Match: (a) Triangle (b) Semicircle (c) Hemisphere (d) Cone. The centroid or centre of gravity distances from the base are respectively
1. 2h/3, 4r/3π, h/4, 3r/8
2. h/2, r/2, r/4, h/3
3. h/3, 4r/3π, 3r/8, h/4
4. h/4, 3r/8, 4r/3π, h/3
Answer
C. h/3, 4r/3π, 3r/8, h/4
These are the standard locations for the four shapes.
The resultant of two equal forces of 10 N each is 10 N. The angle between them is
1. 60°
2. 90°
3. 180°
4. 120°
Answer
D. 120°
R = 2P cos(θ/2) = 10 gives cos(θ/2) = 0.5, so θ/2 = 60° and θ = 120°.
A rectangle of base 60 mm and depth 100 mm has area 6000 mm². Its moment of inertia about its base is
1. 6 × 10⁶ mm⁴
2. 2 × 10⁷ mm⁴
3. 5 × 10⁶ mm⁴
4. 1.5 × 10⁷ mm⁴
Answer
B. 2 × 10⁷ mm⁴
I = 5 × 10⁶ + 6000 × 50² = 2 × 10⁷ mm⁴.
The resultant of a triangular load varying from zero to w per unit length over a span L acts at what distance from the zero end?
1. L/3
2. L/2
3. L/4
4. 2L/3
Answer
D. 2L/3
The resultant of a triangular load passes through its centroid, which is 2L/3 from the zero end.

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