Lesson 10: Mechanics — Forces & Kinematics | A Level Mathematics

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1. Forces and Equilibrium M1

Force: A vector quantity that can cause a body to accelerate, decelerate, or change direction. Measured in Newtons (N). Forces have both magnitude and direction and must be combined as vectors.

Types of Forces

Force	Symbol	Description
Weight	W = mg	Gravitational force pulling body downward. Always acts vertically downward through the centre of mass. g = 9.8 m/s² (use this unless told otherwise).
Normal Reaction	R or N	Perpendicular contact force from a surface on a body. Always acts perpendicularly away from the surface.
Tension	T	Pulling force in a string, cable, or rod. Acts along the string, away from the object.
Friction	F	Force opposing relative motion between surfaces. Acts along the surface, opposing motion (or tendency of motion).
Thrust/Compression	—	Pushing force in a rod. Acts along the rod, toward the object.
Applied Force	P or F	External force applied to a body by a person, engine, etc.

Resolving Forces

Resolution of Forces:
Any force F at angle θ to the horizontal has components:
Horizontal component: F cos θ Vertical component: F sin θ
For equilibrium: ΣFₓ = 0 AND ΣFᵧ = 0 (net force in every direction = 0)

📐 Worked Example 1 — Resolving Forces on an Inclined Plane

A block of mass 5 kg rests on a smooth inclined plane at 30° to the horizontal. A force P acts parallel to the plane and up the slope to keep the block in equilibrium. Find P and the normal reaction R.

Forces on block: Weight W=5×9.8=49N (down), Normal Reaction R (perpendicular to plane), Force P (up the slope).

Resolve along the plane (↑ positive):
P − W sin30° = 0 → P = 49×0.5 = 24.5 N

Resolve perpendicular to plane (away from surface positive):
R − W cos30° = 0 → R = 49×(√3/2) = 42.4 N

Friction

Friction Laws

F ≤ μR (F = friction force, R = normal reaction, μ = coefficient of friction)

F = μR (when object is on the point of sliding or sliding)

Angle of friction: tan λ = μ where λ is the angle between resultant contact force and normal

📐 Worked Example 2 — Friction on a Horizontal Surface

A block of mass 8 kg rests on a rough horizontal surface. A horizontal force of 30 N is applied. The block is on the point of sliding. Find μ and the friction force. Then find the acceleration if the applied force is increased to 50 N.

Vertical equilibrium: R=W=8×9.8=78.4 N
On point of sliding: F=μR=30 N → μ=30/78.4=0.383

With P=50N: Friction F=μR=0.383×78.4=30N (kinetic friction)
Net force = 50−30=20N. By F=ma: 20=8×a → a=2.5 m/s²

Equilibrium of a Particle Under Multiple Forces

📐 Worked Example 3 — Three-Force Equilibrium (Lami's Theorem)

Three concurrent forces keep a particle in equilibrium: 20 N at 0°, T N at 120°, and S N at 240°. Find T and S.

Resolve horizontally (→ positive):
20 + T cos120° + S cos240° = 0
20 − T/2 − S/2 = 0 → T + S = 40 ...(1)

Resolve vertically (↑ positive):
T sin120° + S sin240° = 0
T(√3/2) − S(√3/2) = 0 → T = S ...(2)

From (1) and (2): 2T=40 → T=S=20N

2. Moments and Equilibrium of a Rigid Body M1

Moment of a force about a point = Force × perpendicular distance from the point to the line of action of the force. Measured in Newton-metres (Nm). Anticlockwise moments are positive by convention.

Moment and Equilibrium Conditions

Moment = F × d (F = force, d = perpendicular distance)

For equilibrium of a rigid body:

ΣF = 0 (net force = 0) AND ΣM = 0 (net moment about any point = 0)

📐 Worked Example 4 — Beam on Supports

A uniform beam AB of mass 20 kg and length 6 m rests on supports at A and at C, where AC=4m. A load of 30 kg is placed at B. Find the reactions at A and C.

Forces: Rₐ (up at A), Rc (up at C), Weight of beam=196N at midpoint (3m from A), Weight of load=294N at B (6m from A).

Take moments about A (eliminates Rₐ):
Rc×4 = 196×3 + 294×6
4Rc = 588+1764 = 2352 → Rc = 588 N

Vertical equilibrium:
Rₐ + Rc = 196+294 = 490
Rₐ = 490−588 = −98 N
Negative value means the support at A must push downward (the beam would tip without constraint at A).

3. Kinematics — Motion in a Straight Line M1

Kinematics is the study of motion without reference to its causes (forces). The key quantities are displacement s, velocity v, acceleration a, and time t.

SUVAT Equations — Constant Acceleration Only

v = u + at

s = ut + ½at²

s = vt − ½at²

v² = u² + 2as

s = ½(u+v)t

u=initial velocity, v=final velocity, a=acceleration, s=displacement, t=time.
IMPORTANT: These equations only apply when acceleration is CONSTANT. For variable acceleration, use calculus.

📐 Worked Example 5 — Vertical Projection

A ball is thrown vertically upward with speed 15 m/s from ground level. Taking g = 9.8 m/s², find: (a) the maximum height (b) the time to reach maximum height (c) the time to return to ground (d) the velocity when it reaches a height of 8 m (on the way up and down).

Take upward as positive. u=+15, a=−9.8.
(a) At max height v=0: v²=u²+2as → 0=225−19.6s → s=225/19.6=11.5 m

(b) v=u+at: 0=15−9.8t → t=15/9.8=1.53 s

(c) s=0 (returns to ground): 0=15t−4.9t²=t(15−4.9t)
t=0 (launch) or t=15/4.9=3.06 s

(d) s=8: v²=225−2×9.8×8=225−156.8=68.2
v=±√68.2=±8.26 m/s
Going up: v=+8.26 m/s. Coming down: v=−8.26 m/s

Kinematics Using Calculus (Variable Acceleration)

Variable Acceleration — Calculus Relationships:
v = ds/dt    a = dv/dt = d²s/dt²
s = ∫v dt + c    v = ∫a dt + c
v·(dv/ds) = a    (useful when a is a function of displacement s)

📐 Worked Example 6 — Variable Acceleration

A particle moves with velocity v = 3t² − 12t + 5 m/s (t ≥ 0). Find: (a) the acceleration at t=2 (b) when the particle is at rest (c) the displacement from t=0 to t=4 (d) the total distance in 0≤t≤4.

(a) a = dv/dt = 6t−12. At t=2: a=12−12=0 m/s²

(b) v=0: 3t²−12t+5=0 → t=(12±√(144−60))/6=(12±√84)/6
t=(12±9.165)/6 → t=0.472 s or t=3.528 s

(c) s=∫₀⁴(3t²−12t+5)dt=[t³−6t²+5t]₀⁴=(64−96+20)=−12 m (displacement)

(d) v changes sign at t=0.472 and t=3.528. Calculate distance in each section:
s(0→0.472): [t³−6t²+5t]₀^0.472=(0.105−1.336+2.360)=1.129 m
s(0.472→3.528): displacement = s(3.528)−s(0.472). s(3.528)=43.9−74.7+17.6=−13.2. s(3.528)−s(0.472)=−13.2−1.129=−14.33 (negative → moved 14.33m back)
s(3.528→4): s(4)−s(3.528)=−12−(−13.2)=1.2m
Total distance = 1.129+14.33+1.2 ≈ 16.7 m

Velocity-Time Graphs

v-t graph:
• Gradient = acceleration | Area under graph = displacement (signed) or distance (unsigned)
• Horizontal line: constant velocity | Straight line: constant acceleration
• Area above x-axis = positive displacement | Area below = negative displacement (moving backward)
• Total distance = sum of absolute areas

4. Newton's Laws of Motion M1

Newton's Three Laws

1st Law: A body remains at rest or moves with constant velocity unless acted on by a resultant force.

2nd Law: F = ma (resultant force = mass × acceleration)

3rd Law: Every action has an equal and opposite reaction (Newton pairs — act on different bodies).

📐 Worked Example 7 — Connected Particles (Atwood's Machine)

Two particles of mass 3 kg and 5 kg are connected by a light inextensible string over a smooth fixed pulley. Find the acceleration of the system and the tension in the string.

Take downward direction for heavier particle as positive. Let acceleration = a and tension = T.

For 5 kg (moving down): 5g − T = 5a → 49−T=5a ...(1)
For 3 kg (moving up): T − 3g = 3a → T−29.4=3a ...(2)

Add (1)+(2): 49−29.4=8a → 19.6=8a → a=2.45 m/s²
From (2): T=29.4+3(2.45)=29.4+7.35=T=36.75 N

📐 Worked Example 8 — Connected Particles on an Incline

A 4 kg block on a rough incline at 25° (μ=0.2) is connected by a string over a smooth pulley to a 3 kg hanging mass. Find the acceleration and tension.

For block on incline:
R = 4g cos25° = 4×9.8×0.906 = 35.5 N
Friction (opposing motion up the slope): F=μR=0.2×35.5=7.10 N
Assume hanging mass falls (and block moves up incline).

Block (up the incline positive):
T − 4g sin25° − F = 4a
T − 16.58 − 7.10 = 4a → T−23.68=4a ...(1)

Hanging mass (down positive):
3g − T = 3a → 29.4−T=3a ...(2)
Add: 29.4−23.68=7a → 5.72=7a → a=0.817 m/s²
T=29.4−3(0.817)=29.4−2.45=T=26.9 N

Newton's Second Law with Vectors

📐 Worked Example 9 — Force as a Vector

A particle of mass 3 kg has position vector r=(2t²+t)î+(t³−4t)ĵ m at time t seconds. Find the force acting on it at t=2.

v=dr/dt=(4t+1)î+(3t²−4)ĵ
a=dv/dt=4î+6tĵ

At t=2: a=4î+12ĵ
F=ma=3(4î+12ĵ)=12î+36ĵ N
|F|=√(144+1296)=√1440=12√10≈37.9 N

5. Projectile Motion M1

Projectile: A body moving freely under gravity alone (no air resistance) after being given an initial velocity. The horizontal and vertical components of motion are independent.

Projectile Motion Equations

Horizontal: x = u cosα · t (constant velocity, no horizontal force)

Vertical: y = u sinα · t − ½g t² (constant downward acceleration g)

Vertical velocity: vᵧ = u sinα − gt

Range on horizontal ground: R = u² sin2α / g

Maximum height: H = u² sin²α / (2g)

Time of flight: T = 2u sinα / g

Maximum range: when α = 45°, R_max = u²/g

📐 Worked Example 10 — Projectile

A ball is projected from ground level with speed 20 m/s at 60° to the horizontal. Find: (a) time of flight (b) range (c) maximum height (d) speed and direction at t=1 s.

u=20, α=60°. uₓ=20cos60°=10 m/s, uᵧ=20sin60°=10√3 m/s.

(a) T=2×10√3/9.8=20√3/9.8=3.53 s

(b) R=uₓ×T=10×3.53=35.3 m
Or: R=400×sin120°/9.8=400×(√3/2)/9.8=200√3/9.8=35.3 m ✓

(c) H=(10√3)²/(2×9.8)=300/19.6=15.3 m

(d) At t=1: vₓ=10, vᵧ=10√3−9.8=17.32−9.8=7.52
Speed=√(100+56.5)=√156.5=12.5 m/s
Direction: θ=arctan(7.52/10)=arctan(0.752)=36.9° above horizontal

📐 Worked Example 11 — Projectile from a Height

A particle is projected horizontally at 12 m/s from a cliff 45 m high. Find the time to reach the sea and the horizontal distance traveled.

Vertical: y=45 (downward). uᵧ=0, a=9.8.
45=½×9.8×t² → t²=45/4.9=9.184 → t=3.03 s

Horizontal: x=12×3.03=36.4 m

6. Driving Force, Resistance, and Inclinations M1

For a vehicle of mass m moving along a road:
Net force = Driving force − Resistance force − Component of weight along slope
Net force = ma (Newton's Second Law)

📐 Worked Example 12 — Vehicle on an Incline

A car of mass 1200 kg moves up a slope inclined at sin⁻¹(1/20) to the horizontal. The driving force is 2400 N and resistance is 800 N. Find the acceleration.

Component of weight along slope (opposing motion up) = mg sinθ = 1200×9.8×(1/20) = 588 N

Net force = 2400−800−588 = 1012 N
a = F/m = 1012/1200 = 0.843 m/s²

📝 Exam Practice Questions

Q1 [4 marks] — A block of mass 6 kg rests on a rough inclined plane at 35° to the horizontal. A force of P N acts up the slope. The coefficient of friction is 0.3. Find the range of values of P for which the block remains in equilibrium.

R=6g cos35°=6×9.8×0.819=48.2N. Friction limit: F_max=μR=0.3×48.2=14.5N
Weight component along slope: W sinθ=6×9.8×0.574=33.7N

Minimum P (friction acts up slope, preventing slide down):
P+F_max=W sinθ → P=33.7−14.5=19.2N

Maximum P (friction acts down slope, preventing slide up):
P=W sinθ+F_max=33.7+14.5=48.2N
Range: 19.2 ≤ P ≤ 48.2 N

Q2 [4 marks] — A uniform plank AB of mass 15 kg and length 8 m is supported at A and at D, where AD=6m. A person of mass 60 kg stands at B. Find the reactions at A and D, and determine whether the plank is about to tip.

Forces: Rₐ at A, Rᴅ at D, 147N at midpoint (4m from A), 588N at B (8m from A).
Moments about A: Rᴅ×6=147×4+588×8=588+4704=5292 → Rᴅ=882N
Vertical equil: Rₐ+882=147+588=735 → Rₐ=−147N
Rₐ is negative → support at A would need to push down.
The plank would tip about D unless the support at A exerts a downward force of 147N (e.g. is bolted).

Exam Tip: A negative reaction force means the support must exert a downward force (or the object tips). Always comment on the physical meaning of a negative answer in moments questions.

Q3 [4 marks] — A stone is thrown vertically upward with speed 14 m/s from the top of a tower 20 m high. Find: (a) the maximum height above the ground (b) the time to reach the ground.

Take upward positive, origin at top of tower. u=14, a=−9.8.
(a) v²=196−19.6s=0 → s=10m above tower → max height=30m above ground
(b) s=−20 (ground is 20m below): −20=14t−4.9t²
4.9t²−14t−20=0 → t=(14+√(196+392))/9.8=(14+√588)/9.8=(14+24.25)/9.8=3.90 s

Q4 [5 marks] — Two particles A (4 kg) and B (6 kg) are connected by a light inextensible string passing over a smooth fixed pulley at the edge of a smooth horizontal table. A rests on the table; B hangs vertically. Find the acceleration of the system and the tension. Find the speed of B when it has fallen 2 m from rest.

A on table (horizontal): T=4a ...(1) (no friction, no weight component)
B hanging (down positive): 6g−T=6a → 58.8−T=6a ...(2)
Add: 58.8=10a → a=5.88 m/s²
T=4×5.88=T=23.5N
v²=u²+2as=0+2×5.88×2=23.52 → v=4.85 m/s

Q5 [4 marks] — A ball is projected from a point O at ground level with speed 25 m/s at angle α to the horizontal. It just clears a wall of height 10 m at a horizontal distance of 30 m. Find α and the range on the ground.

At x=30: t=30/(25cosα). Substitute into y=10:
10=25sinα·(30/25cosα)−½×9.8×(30/25cosα)²
10=30tanα−4.9×900/(625cos²α)
10=30tanα−7.056sec²α=30tanα−7.056(1+tan²α)
7.056tan²α−30tanα+17.056=0
tanα=(30±√(900−4×7.056×17.056))/(2×7.056)=(30±√(900−481.4))/14.112
=(30±√418.6)/14.112=(30±20.46)/14.112
tanα=3.575 → α=74.4° or tanα=0.677 → α=34.1°
Range for α=34.1°: R=625sin(68.2°)/9.8=625×0.929/9.8=59.3m

Q6 [4 marks] — A particle has velocity v=(6t−t²) m/s for 0≤t≤6. Find: (a) maximum velocity (b) total distance in 0≤t≤6 (c) displacement at t=6.

(a) dv/dt=6−2t=0 → t=3. v_max=18−9=9 m/s
(c) s=∫₀⁶(6t−t²)dt=[3t²−t³/3]₀⁶=108−72=36m
(b) v≥0 for 0≤t≤6 (v=t(6−t)≥0). So distance=displacement=36m

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