Forces

— Physics · AS Level (7408) · Newton's Laws, Equilibrium, Moments & Fluids —

★ Exam Tip: The A-level definitions here (force, momentum, principle of moments) must be quoted exactly as taught, not paraphrased — examiners check for the precise required structure, and a "principle" always needs its condition stated first, then the consequence.

1. Newton's Three Laws

Force (A-level def): the rate of change of momentum. Vector, measured in Newtons.

Common mistake: "push or pull" and "an agent that changes the state of motion" are O-level definitions — never write these at A-level, even though they aren't technically wrong; they will not be awarded marks.

F = dpdt = ΔpΔt | Impulse F×Δt = Δp

Momentum: the product of an object's mass and velocity. Vector, measured in kg m s⁻¹. p = mv

Gradient of a p–t graph = force. Area under an F–t graph = Δp.

① Newton's First Law

"An object will continue its state of rest or uniform motion until it is acted upon by an external force." AKA the Law of Inertia.

Inertia: the tendency of an object to continue its state of motion or uniform rest. Inertia ↑ as Mass ↑ — a heavier object more strongly resists having its motion changed.

② Newton's Second Law — two special cases of F=dp/dt

① Constant massF=dp/dt → F=d(mv)/dt → F=m(dv/dt) → F = ma

② Constant velocityF=dp/dt → F=d(mv)/dt → F = v(dm/dt) dm/dt = mass flow rate

Example (rocket):

constant velocity 1000 m/s, ejecting 500 kg of fuel each second: F = v(dm/dt) = 1000×(500/1) = 500,000 N

③ Newton's Third Law

"Every action has an equal and opposite reaction."

W-W pair (person↔Earth), R-R pair (person↔chair) — each pair is its OWN type of force

Action-reaction pairs — five properties:

They are equal (in magnitude)
They are opposite in direction
They act along the same line of action
They are of the same nature (type of force)
They act from different bodies onto each other

★ Exam Tip: Action-reaction pairs do NOT cause equilibrium — because they act on different bodies, not the same one. Equilibrium is about the forces on ONE object only.

2. Equilibrium — Two Conditions

Equilibrium: object is at rest or moving with uniform/constant velocity.

★ Exam Tip: examiners never say "net force is zero" directly — they describe the physical state ("at rest" / "constant velocity") and expect you to recognise this means ΣF=0. Remember: zero net force allows constant-velocity motion to continue, it doesn't mean the object must be still.

Condition 1ΣF = 0, i.e. ΣF_x = 0 and ΣF_y = 0

If you have a force in one direction, there must be an equal force in the opposite direction as well.

Condition 2 — Principle of MomentsΣmoment = 0, i.e. ΣCW = ΣACW

★ Exam Tip: a "principle" must be quoted with its CONDITION first, then the consequence: "For an object to be in equilibrium, the sum of clockwise moments must equal the sum of anticlockwise moments." Dropping the condition clause loses marks even if the equation itself is right.

Moment of a force: the turning effect of a force = Force × perpendicular distance from the pivot. Measured in N m, vector.

M = 15×(15×10⁻²) = 2.25 N m ACW

only 15sin60 (⊥ component) turns; 15cos60 passes near pivot, ignored if along the rod

Common mistake: forces whose line of action passes THROUGH the pivot contribute zero moment — they are not considered at all.

★ Major Exam Tip — choosing the pivot: 1) Place the pivot at the position of an unknown force (its own moment then vanishes). 2) Place the pivot where the MOST forces pass through, to cancel as many unknowns as possible at once. 3) Hinges/wall-attachments especially: you never know the direction of the force at a hinge — always put your pivot exactly there. This is the standard method whenever a question involves a hinge.

Worked example — beam, pivot PEquilibrium check

Beam: 20N down at 2m left of P; 50N at 30° to horizontal, 3m right of P (resolved to 50sin30 up, 50cos30 right).

Is it in equilibrium? Check ΣF_y=0: 50sin30 ?= 20 → 25≠20 — fails. (Also ΣCW≠ΣACW, no CW moment exists.) Not in equilibrium.

What is the net moment? (+ACW convention) +(20×2)+(50sin30×3) = +115 N m ACW

Torque of a Couple

A couple is: 1) forces equal in magnitude, 2) opposite in direction, 3) act on the same object, 4) have parallel lines of action, 5) separated by a distance.

T = F × d

Derivation:

pivot at the midpoint between the two forces (distance d/2 each side): ΣMoment = F(d/2) + F(d/2) = F(d/2+d/2) = Fd — true no matter where the pivot sits, for a pure couple.

3. Momentum-Time Graph MCQs

Trick: gradient of a p-t graph = force (F=dp/dt). Read what the force is doing (constant? increasing? in which direction relative to velocity?), then match the REQUIRED gradient behaviour to the graph options.

s17 qp11 q10MCQ · p-t straight line

p-t graph: straight line, 0 to 5000 kg m/s over 10s. What is the resultant force?

F = gradient = 5000/10 = 500 N.

Answer: A (500 N)

s20 qp12 q7MCQ · force slowly increasing

The resultant force acting on an object is slowly increased. Which graph could show p vs t?

F increasing ⇒ gradient of p-t must be INCREASING (concave up, steepening).

Answer: B

w21 qp11 q8MCQ · constant force, same direction as v

A constant resultant force acts on an object in the direction of the object's velocity (v, p, F all positive). Which graph could show p vs t?

F constant ⇒ p-t must be a STRAIGHT LINE with constant positive gradient (starting from a nonzero intercept, since the object already had momentum).

Answer: C

Golf ball — force-time humpMCQ · what can't be found

A golf ball is hit by a club; hump-shaped F-t graph shown. Which quantity, for the time of contact, CANNOT be found from the graph?

A ✓ average force — obtainableB ✓ change in momentum — area under curve C ✓ contact time — width of the graphD ✗ maximum acceleration — the ball's mass is unknown, so F=ma cannot give a

Answer: D

4. Connected Bodies & Free-Body Diagrams

Blocks X and Y in contactMCQ · connected system

Horizontal force F applied to block X, in contact with block Y; frictionless surface; X has greater mass than Y. Which statement is correct?

★ Trick: for a combined system pushed by one force, ALWAYS find the whole-system acceleration first (a = F/(m_x+m_y)), then apply F=ma to just the block you need, using that shared acceleration but only ITS mass. Never assume the single applied force acts undiminished on both blocks.

a = F/(m_x+m_y) (same for both, since connected) Force X exerts on Y: F_y = [F/(m_x+m_y)]×m_y Force Y exerts on X (reaction, Newton's 3rd law): F_x = [F/(m_x+m_y)]×m_x — equal in magnitude to F_y by N3L

A ✗ "a_X = F/mass of X" — wrong, ignores mass of YB ✗ "force X exerts on Y = F" — too large C ✓ "force X exerts on Y is less than F" — correct, since some of F also accelerates X itselfD ✗ "force X exerts on Y < force Y exerts on X" — violates Newton's 3rd law (must be equal)

Answer: C

Trick: for friction, first sketch the tendency of motion (even before anything moves) — static friction opposes that tendency, exactly as kinetic friction opposes actual sliding.

Free body diagrams: only mark the forces on the TARGET body.

smooth surface: R, W only

rough surface, moving right: F_r opposes v

rough incline, moving up: F_r acts down-slope

5. Types of Forces

Weight (W): always acts towards the centre of Earth. W = mg (g=9.81 m s⁻²). Marked vertically downwards, from the centre of gravity — "the point through which the entire weight of the object is assumed to act."

Normal contact force (R): a force generated between 2 rigid objects in contact. Always marked perpendicular to the surface of contact.

Friction: resistive force between 2 objects moving relative to each other (or with a tendency to). Always marked opposite to the direction of motion (or the tendency of motion).

Exam cue: "rough" in a question means CONSIDER friction; "smooth" means IGNORE it. This single word is the examiner's signal — read it every time.

Note: "Reaction/contact force" is the RESULTANT of R and friction combined at one point — keep it distinct from "normal contact force," which means the perpendicular part only.

Friction is a limiting force

F_r=F (static, rises 1:1) until the limiting value, then constant

F	F_r	ΣF
0N	0N	0N
+498N	−498N	0N
+500N (limiting)	−500N	0N
+501N	−500N (maxed out)	+1N

Common mistake: once F exceeds the limiting friction value, F_r stops growing (it's capped) — a net force appears and the object starts to accelerate.

Ladder on icy ground, against rough wallMCQ · sliding directions

A ladder rests on icy (frictionless) ground, leaning against a rough wall. At the instant of release it begins to slide. Which diagram shows the correct directions of P (wall friction+reaction), W, R (from the ground)?

Ground is smooth ⇒ only R (perpendicular, vertical) acts there, no friction. Wall is rough ⇒ friction acts there (up the wall, resisting the ladder-top's downward tendency) plus the wall's normal reaction. Sketch the TENDENCY of motion first (top slides down, base slides out), then draw friction opposing that tendency.

Answer: B

Why does a car move forward / why does a ball spin? The drive wheel pushes backward on the road via friction; by Newton's 3rd law, the road pushes the tyre (and car) FORWARD. Friction can propel motion, not just oppose it — the same idea explains why a spinning ball's contact-point friction converts sliding into rolling/spin.

6. Inclined Planes

Method: 1) Make conventions parallel and perpendicular to the plane. 2) Resolve everything (usually just W) into components parallel/perpendicular to the plane, then solve each direction separately.

W resolves to Wsinθ (parallel, down-slope) and Wcosθ (perpendicular, into the plane)

Perpendicular to plane: R = Wcosθ | Parallel to plane (smooth): ma = −Wsinθ ⟹ a = −gsinθ (down-slope)

★ Key physics fact: a block can NEVER stay stationary by itself on a SMOOTH incline — Wsinθ always acts down the slope with nothing to balance it. If a question says the surface is smooth AND the object is stationary, re-read it: another force (applied F, a string) must be present.

Rough incline, block stationaryEquilibrium on a slope

Block stationary on a ROUGH incline (no other applied force) — held only by friction against gravity.

ΣF_y=0: R=WcosθΣF_x=0: Wsinθ=F_r (static friction exactly balances the weight component — not necessarily at its limiting value)

Block sliding down a rough incline, applied force F up-slopeKinematics on a slope

At t=0: u=2 m/s down the slope. Convention: up-slope = +. Find v after t seconds.

ΣF_x = Wsinθ − F_r (up-slope positive)a_x = (Wsinθ−F_r)/m v=u+at ⟹ v = −2 + [(Wsinθ−F_r)/m]t (u is −2, since the initial motion is DOWN-slope under the up-slope-positive convention)

Full numeric example — 500N block on 30° inclineCalculation · net force & kinematics

Block on a rough 30° incline: applied F=500N up the slope, weight W=500N, limiting friction F_r=3N. Block moves up the slope.

Net force: ΣF_x = +500 − 500sin30 − 3 = +247N (up the slope) Mass: m=W/g=500/9.81; a=247/(500/9.81)=4.8464 m/s² If block ALSO had initial velocity 2 m/s DOWN the slope, distance in 5s: s=ut+½at² = (−2)(5)+½(4.8464)(5)² = −10+60.58 = 50.58 ⟹ s≈51m

Object M on a slope: P→Q→RStructured · constant deceleration on incline

M moves up the slope, stops at Q, returns to R. Constant deceleration 3.0 m/s² down-slope AT ALL TIMES (even on the way up). At t=0, M is at P with v=3.6 m/s up-slope. Total distance P→Q→R = 6.0m.

(a)(i) Time P→Q: v=u+at: 0=3.6+(−3)t ⟹ t=1.2s (a)(ii) Distance P→Q: v²=u²+2as: 0=(3.6)²+2(−3)s ⟹ s=2.16m (b) Speed at R (Q→R = 6.0−2.16=3.84m): v²=0+2(−3)(−3.84) ⟹ v=−4.8 ⟹ speed=4.8 m/s shown

(c) v-t graph P→Q→R: ONE straight line, constant gradient −3 throughout — from +3.6 at t=0, through 0 at t=1.2s, continuing down to −4.8 at the end.

Shopping trolley — constant force, then down a slopeStructured · deceleration then incline

42kg trolley pushed at 1.2 m/s, released, travels 1.9m before stopping (constant opposing force).

(a)(i) v²=u²+2as: 0=(1.2)²+2(a)(1.9) ⟹ deceleration=0.38 m/s² (a)(ii) F=ma=(42)(0.38) ⟹ 16N (b) Power at 1.2 m/s: P=Fv=16×1.2=19.2 W

(i) Component of weight down slope: mgsinθ=(42)(9.81)sin(2.8°)=20.1N (ii) ΣF_x=20.1−16=4.1N; a=4.1/42=0.098 m/s²; s=ut+½at²: 3.5=½(0.098)t² ⟹ t≈8.5s

(d) Why isn't the slope made steeper? "If made any steeper, there would not be enough time to stop it — so that anyone in the trolley's path can move out of the way."

Elastic between two pins, propelling a trolleyMCQ · resolving two tensions

Elastic stretched between fixed pins Y, Z and around pin X (on a trolley), tension 4N throughout. X is at the centre, propelled in direction P at right angles to YZ. What force accelerates the trolley when released?

Geometry gives a similar-triangle ratio cosθ=3/5. Both halves of the elastic contribute a component along P.

Force = 2 × T × cosθ = 2×4×(3/5) = 4.8N

Answer: C (4.8N)

7. Tension & Connected Particles

Tension: resistive force that develops in an object under deformation/stress. Always marked opposite to the direction of extension/compression (i.e. pulling back INTO the string/spring, never floating outside it). T = Ke (Hooke's law).

Rule: tension is always marked pointing back INSIDE the string/rope/spring at a cut point — never as an arrow drifting away from the material.

Trolley + engine, connected by a ropeWorked example · connected particles

2kg trolley connected by rope (tension T) to 3kg "engine," which has a 60N applied force. Smooth surface. Find T.

Method: 60N acts on the WHOLE system first (find shared a), then isolate ONE body with a free-body diagram to solve for T.

Whole system: 60=(2+3)a ⟹ a=+12 m/s² 2kg trolley alone: T=(2)(12) ⟹ T=24N Check via 3kg engine alone: 60−T=(3)(12) ⟹ T=24N ✓ (both methods agree)

Painting hung from two strings at 60°Worked example · equilibrium, two tensions

2kg painting hung from two strings, each at 60° to horizontal, attached to two nails. Show T₁=T₂, then find T.

ΣF_x=0: T₁cos60=T₂cos60 ⟹ T₁=T₂ (also shown via a closed vector triangle T₁+T₂+W) ΣF_y=0: T₁sin60+T₂sin60=W ⟹ 2Tsin60=(2)(9.81) ⟹ T=9.81/sin60 ⟹ T≈11N

Helicopter rescue — man + child on cableStructured · 3-stage lift, tension varies

Man (80kg) + child (50.5kg) on a cable: initial uniform acceleration → constant velocity → final uniform deceleration.

(a) Combined weight: W=(130.5)(9.81)=1280N (b)(i) Accelerating at 0.570 m/s²: (130.5)(0.570)=T−1280 ⟹ T≈1350N (b)(ii) Constant velocity 2.00 m/s: ΣF=0 ⟹ T=1280N (=W) (c) Tension=1240N during final decel: (130.5)a=1240−1280 ⟹ deceleration=0.31 m/s² (d)(i)1. Accel phase time (0→2.00 m/s): t=2/0.570≈3.5s (d)(i)2. Decel phase time (2.00→0): t=2/0.3065≈6.5s

(d)(ii) v-t graph: rises 0→2.0 over 3.5s, flat at 2.0 for the given 20s, falls 2.0→0 over 6.5s.

Motor drags a log up a 14° slopeStructured · tension on an incline

452kg log dragged up a slope by a cable (motor), incline 14°, constant friction 525N, accelerates at 0.130 m/s².

(a) Weight component down slope: mgsinθ=(452)(9.81)sin14° ⟹ 1070N (b)(i) Tension: ΣF=T−Wsinθ−F_r: (452)(0.130)=T−1070−525 ⟹ T≈1650N (b)(ii)1. Time for 10.0m from rest: s=ut+½at²: 10=½(0.130)t² ⟹ t=12.4s (b)(ii)2. Speed at 10.0m: v²=2(0.130)(10) ⟹ v≈1.61 m/s

8. Drag Force & Terminal Velocity

Drag force: resistive force on an object moving relative to the fluid around it. Always opposite to the direction of motion. As v↑, F_D↑. In air: "air resistance." In liquids: "viscous force."

Parachutist / object falling in air — full derivation

Convention: down = + throughout.

Stage	Analysis
① t=0	v=0 ⟹ no drag yet. ΣF=+W. Σa=+9.81 m/s²
② t>0	v↑ ⟹ F_D↑. ΣF=W−F_D. Σa<9.81, decreasing
③ t≫0	v↑↑ ⟹ F_D↑↑. Σa≪9.81
④ terminal velocity	F_D=W ⟹ ΣF=0 ⟹ Σa=0 — velocity stops increasing

a: decays from 9.81 to 0

s: concave up throughout (still speeding up)

v: rises then flattens at terminal velocity

Parachute opens mid-fall — SECOND (lower) terminal velocity

Once at terminal velocity①, the canopy suddenly opens: F_D jumps far above W.

Stage	Analysis
⑤ canopy opens	ΣF=W−F_D↑ (now hugely negative, i.e. upward) ⟹ Σa upward, large. v decreases ⟹ F_D↓ back down
⑥	Acceleration still upward but decreasing; v still decreasing
⑦ 2nd terminal velocity	v decreases until F_D=W again ⟹ ΣF=0, Σa=0 — new LOWER terminal velocity

v rises to terminal①, canopy opens, drops to a new lower terminal②

Structured Q — parachutist v-t graph (A→E)Full analysis + calculation

v-t graph: rises steeply A→B, flattens near terminal velocity B→C (≈50 m/s at t=15s), sharp drop C→D as canopy opens (≈20 m/s at t=17s), decays towards a second terminal D→E.

(b)(ii) Explain resultant force A→C: "Initially only weight acts. Velocity increases downwards, so drag force increases upwards. As ΣF=W−F_D, ΣF decreases. F_D increases until it equals W, so ΣF decreases to zero."

(b)(iii) Frictional/drag force: at t=15s (C) — "increases" (jumps once canopy opens); C→D — "constant"; D→E — "decreasing back to the value of weight"

(iv)1. Average acceleration C→D (mass 95kg): (v₂−v₁)/(t₂−t₁)=(50−20)/(15−17) ⟹ a=−15 m/s² (iv)2. Average drag force: ΣF=W−F_D: 95(−15)=95(9.81)−F_D ⟹ F_D=95(9.81+15) ⟹ F_D≈2360N

Trick: for "which graph" questions on falling-with-drag bodies, use the compulsory features: acceleration ALWAYS starts at g and decays towards 0 (never rises, never flat); displacement is ALWAYS concave up throughout (object never stops speeding up, even as it slows its rate); velocity ALWAYS rises and flattens (never straight-line, never overshoots).

s16 qp12 q7 — tennis ball, acceleration graphMCQ

Air resistance significant. Which graph shows a vs t as the ball falls?

Answer: A (starts at 9.81, decays towards 0)

s16 qp12 q16 — train resistive force ratioCalculation · graph reading

Total resistive force = wheel friction (constant) + wind resistance (grows with speed). At 200 km/h, total=40kN, wheel friction (flat part)=8kN. Find wind/wheel ratio.

Wind resistance = 40−8 = 32kN. Ratio = 32/8 = 4

Answer: A (4)

s17 qp11 q8 — match height/speed/force to graphs X,Y,ZMCQ

Object dropped, air resistance significant. Three graphs X, Y, Z given (unlabelled). Match to height, speed, resultant force.

X (decays to 0) = force (ΣF∝Σa, both decay to 0 at terminal v). Y (decreasing, curving) = height (steadily falling). Z (rises then flattens) = speed (rises to terminal v).

Answer: C (height=Y, speed=Z, force=X)

Four more graph-shape MCQs (a, s, x vs t for falling bodies)MCQ set

Question	Graph asked	Answer & reasoning
w17qp12q8 (stone, long fall)	a vs t	D — decays from 9.81 to 0
s18qp11q6 (tennis ball, free fall)	s vs t	A — concave up throughout (negligible drag)
w16qp11q10 (beach-ball, drag significant)	a vs t	C — decays from max to 0
w17qp13q6 (football, tall building)	x (distance) vs t	D — concave up, negligible drag
s18qp13q8 (sky-diver, drag→weight)	s vs t	B — concave up then straightens to constant gradient (terminal v)

s16 qp12 q10 — force diagram at point P on a v-t curveMCQ · match graph point to force diagram

Sphere falls through air; v-t graph shown with point P mid-rise (below terminal velocity). Which force diagram applies at P?

At P, v is still increasing ⇒ net downward force still exists ⇒ air resistance arrow must be visibly SHORTER than the weight arrow (not yet equal).

Answer: C

w16 qp12 q13 — resistive force graph, find key timesMCQ · reading a graph

F_R (drag) vs t graph: rises from 0, plateaus at t₂ (F_R=W there). At which times are speed zero, speed maximum, and acceleration zero?

Speed is zero only at the very start (t₁=0, release moment). F_R plateauing at t₂ means terminal velocity is reached there ⟹ both max speed AND zero acceleration occur at t₂.

Answer: B — zero speed at t₁, max speed at t₂, zero acceleration at t₂

s22 qp12 q9 — stone vs foam rubber ball, terminal velocities comparedMCQ · comparing two terminal velocities

Stone S and foam rubber ball R (same size/shape) released together, both reach terminal velocity. Compare their speed-time graphs.

Stone: m↑ ⟹ W↑ ⟹ needs larger F_D to balance ⟹ HIGHER terminal velocity Rubber ball: m↓ ⟹ W↓ ⟹ balances at smaller F_D ⟹ LOWER terminal velocity, reached sooner

Answer: B (stone reaches a higher terminal speed than the rubber ball)

9. Upthrust & Archimedes' Principle

Upthrust: an upward buoyant force experienced by an object immersed in a fluid. Marked upwards. Generated by a pressure difference (P is greater at the bottom face than the top, since pressure increases with depth).

U = F_B − F_T, generated by the pressure difference between top & bottom faces

U = F_B − F_T = (P_B×A) − (P_T×A) = A(P_B−P_T) = A(ρ_f g y − ρ_f g x) = ρ_f A g (y−x)

★ U = ρ_f A h g (h = submerged height) — constant force as long as the object is fully immersed

Alt formula: U = Area × (P_bottom − P_top) directly, when a pressure difference is given — saves re-deriving from ρgΔh each time.

Floating blockWorked example

Block floats (v=0, equilibrium), mass 30kg, area 100cm², fluid density 1000 kg/m³. Find submerged depth h₂.

ΣF_y=0: W=U: 30(9.81)=(1000)(100cm²)(9.81)h₂ ⟹ h₂=0.0003 m

Archimedes' Principle: the upthrust acting on an object equals the weight of the fluid it displaces.

★ Exam Tip: Archimedes' Principle has appeared more often in recent papers. It is not a separate topic — it's exactly the same upthrust result (U=ρ_fAhg), just derived a different way: W=U ⟹ mg=U ⟹ (V×ρ_f)×g=U ⟹ (A×h×ρ_f)×g=U. Don't be thrown by the name.

Floating body — force statementMCQ

A stationary body floats in water. Which statement about the forces acting on it is correct?

Only W and U act (equal, since floating = equilibrium); there's no motion, so NO drag/viscous force exists.

Answer: D (the viscous force is zero)

w18 qp13 q11 — sphere released underwaterMCQ · v-t graph shape

A rigid hollow sphere is released from rest deep in water; upthrust propels it towards the surface. Which graph shows v vs t?

Trick: "released from rest" instantly eliminates any option not starting at v=0. Physically this is the SAME shape family as free-fall-in-air — just mirrored: as v↑, drag↑, net force (U−W−F_D) decreases, acceleration decreases, curve flattens into a terminal (rising) velocity.

Answer: A (rising curve, flattening to a constant, dashed asymptote)

Cylinder suspended in water — upthrust + kinematicsStructured Q

Cylinder (weight 0.84N, diameter 0.031m) stationary in water, string tension T. Pressure difference top/bottom = 520 Pa.

(i) Height h: ΔP=ρgΔh: 520=(1×10³)(9.81)h ⟹ h=0.053m (ii) Upthrust (using U=A×ΔP directly): U=π(0.031/2)²×520 ⟹ U=0.39N (iii) Tension: T+0.39=0.84 ⟹ T=0.45N

(b) Cylinder now pulled up through the water; v-t graph given (triangle: rises to ~10 cm/s at t=2.5s, falls to 0 at t=4.0s).

(i) Acceleration at t=2.0s (gradient): (10×10⁻²−0)/(2.5−0) ⟹ a=0.04 m/s² (ii) Top face at 0.32m depth at t=0. Displacement = area under v-t graph = ½×4×(10×10⁻²) = 0.20m up ⟹ new depth = 0.32−0.20 = 0.12m

10. Lift & Thrust

Lift: an upward force generated on an object moving through a fluid (e.g. an aerofoil) — related to the Coandă effect (a fluid jet's tendency to stay attached to a curved surface).

Thrust: just a push force — e.g. a car's wheels pushing it forward, a rocket's exhaust pushing it upward.

11. Moments & Couples — MCQ Practice

★ Common misconception (O-level): "moment" and "torque" are NOT the same thing. Torque specifically means the moment of a COUPLE (two forces); moment more generally is the turning effect of a single force about a pivot.

★ Trick — is it really a couple? ALL five conditions must hold at once: equal magnitude, opposite direction, same object, parallel lines of action, separated by a distance. Missing even one (e.g. same direction instead of opposite) disqualifies it. For a genuine couple, both forces ALWAYS turn the SAME rotational way about any point between them — they never partially cancel, which is why Torque=F×d works using the FULL separation distance, regardless of exactly where the pivot sits.

Rigid body equilibrium — 4 force diagramsMCQ

Forces applied to a rigid body (½F/F combinations at various offsets λ, 2λ from a pivot). In which diagram is the body in equilibrium?

Check ΣF=0 first (fast), THEN Σmoment=0 (place pivot conveniently, e.g. where a force acts, to cancel it).

Answer: A (only option with both ΣF=0 AND Σmoment=0)

Couple-only arrangement on a wheel rimMCQ

Two equal coplanar forces act on a wheel's rim. Which arrangement provides ONLY a couple?

Answer: D — parallel, opposite, offset (tangential top & bottom) — satisfies all 5 conditions with no net force

Rod XY on wall (X) + support (Z), unknown force direction at XMCQ · 3-force concurrency

Rod XY, weight W at Y, rests on a support at Z and a wall at X. What direction is the force on the rod at X?

★ Major Trick — three-force concurrency: extend the lines of action of the two KNOWN forces (R at Z, W at Y) until they cross at a point. For equilibrium, the THIRD (unknown) force's line of action MUST also pass through that same point — take the pivot there, and the two known forces' moments vanish automatically, forcing the third force's moment to vanish too, which only happens if its line passes through the same point. Combine this with a quick ΣF=0 check (e.g. "the horizontal force must point right, not left") to pin down the unique correct option.

Answer: D

Cylinder on a smooth slope, held by a threadMCQ · equilibrium of 3 forces

Weight W, contact force R (smooth slope), thread tension T attached to the cylinder surface, other end fixed. Which diagram shows equilibrium?

Since the slope is smooth, R always acts through the cylinder's centre. For equilibrium, W, R and T's lines of action must be concurrent (same point) AND sum to zero.

Answer: B

Wheel-and-axle: lever + disc + pulleyCalculation · reverse-engineering torque

Spindle: lever (1.20m) at one end, disc (radius 0.20m) at the other; string over disc lifts a 900N weight. Minimum force F at EACH end of the lever?

Method: work backwards from the known load. Tension in the string = 900N (same throughout an ideal string). This tension × disc radius gives the torque needed at the spindle. The two forces F (one at each lever end) form a COUPLE, whose torque = F × full lever length (1.20m). Set the two torques equal.

0.20×900 = F×1.20 ⟹ F=150N

Answer: B (150N)

Crane equilibrium — load moved further outMCQ · Wx=Ly moment balance

Crane: balancing load W (fixed x), load L (variable y), perfectly balanced with Wx=Ly. If the load moves further out (y increases) and the crane must not topple, which statement is correct?

Wx (ACW, about the original reaction point) stays constant; Ly (CW) increases as y increases. No new horizontal force is needed or consistent (options proposing one fail ΣF=0) — instead the ground reaction R's point of application must SHIFT to provide a compensating moment.

Answer: D (R moves to the right — satisfies both ΣF=0 and Σmoment=0)

Glossary — key exam keywords

Light	Has no weight
Elastic / Extensible	Hooke's law applies
Inelastic / Inextensible	Hooke's law does NOT apply
Uniform	Centre of gravity is at the geometric centre
Non-uniform	Position of centre of gravity is unknown
Rough	Consider friction
Smooth	Do not consider friction
Rigid	Fixed shape — normal contact & friction both apply

Rod AB on a wall, held by a string, hanging massStructured · moments about the hinge

Uniform rod AB (1.2m, 8.5N) fixed to a wall at A, held horizontal by a string to C; angle at B = 50°; mass M hung at B; string tension T=30N.

(i) Vertical component of T: 30sin50 ⟹ 23N (iii) Moments about A: ΣCW=ΣACW: 8.5(0.6)+W(1.2)=23(1.2) ⟹ W=[23(1.2)−8.5(0.6)]/1.2 ⟹ W≈19N (iv) Mass: 19=M(9.81) ⟹ M=1.9 kg

(c) Why must a force act on the rod at A? "There is a leftward force due to the tension, and the upward forces are not equal to the downward forces. As there are unbalanced forces, there must be forces at A."

Card hanging from a rodStructured · centre of gravity, equilibrium

Uniform card (weight W) suspended from a wooden rod, held to one side, then released to swing and settle.

(c)(i) Two forces (other than weight, air resistance) during swinging: 1) "Normal contact force from the rod onto the card." 2) "Friction force at the rod."

(c)(ii) Where does it come to rest, and why? "It comes to rest when the line of action [of W] passes through the rod, as it will be in equilibrium and there should be no net moment." (Centre of gravity ends up directly below the rod's contact point.)

Canopy held by two cables + hingeStructured · moments, 3-4-5 triangle

Canopy (1.30m wide) kept horizontal by two cables (tension 47.2N each) + a hinge. Each cable: 0.800m from wall on the canopy, 0.600m above the canopy on the wall.

Triangle: hyp=√(0.6²+0.8²)=1.0 ⟹ sinθ=0.6, θ=36.87° Moments about the hinge: m(9.81)(0.65) = 2×47.2×(0.6)×0.8 ⟹ m≈7.1 kg

12. Force Exerted by a Jet of Fluid

jet: length l, area A, density ρ, constant velocity v

F = dp/dt = d(mv)/dt = v(dm/dt) [v constant, pulled outside] ρ=m/V ⟹ m=ρAl (volume = area × length) dm/dt = v×d(ρAl)/dt = ρAv×(dl/dt) [ρ, A constant] — and dl/dt = v itself

★ F = ρAv² | ★ P = F/A = ρv²

★ CRITICAL Exam Tip: F=ρAv² and P=ρv² are ONLY valid when the fluid's speed is reduced to ZERO after impact (100% momentum transfer, nothing bounces back). If any speed remains afterward, you must compute the actual Δp = m(v_final−v_initial) using real velocities instead — always check for wording like "speed is reduced to zero" before reaching for this formula.

Mass flow rate (dm/dt): simply "how much mass escapes per unit time" — e.g. 2kg/s.

Wind on a wallMCQ

Wind 33 m/s, air density 1.2 kg/m³, wall area 12m² (right angles to wind), speed reduced to zero. Force on the wall?

F=ρAv²=(1.2)(12)(33)² ⟹ 16,000N

Answer: D

Alpha-particle beam on lead sheetMCQ · discrete particles

Each α-particle: mass 6.6×10⁻²⁷kg, speed 1.5×10⁷ m/s. 5.0×10⁴ particles/s hit 1.0cm² of lead, all absorbed (v→0). Estimate the average pressure.

Note: for discrete particles, compute dm/dt from (mass per particle)×(particles per second) rather than jumping straight to ρv² — the underlying physics is the same, but this is the given-data route.

dm/dt = (6.6×10⁻²⁷)(5×10⁴) F = v×(dm/dt) = (1.5×10⁷)(6.6×10⁻²⁷)(5×10⁴) = 4.95×10⁻¹⁵N P = F/A = 4.95×10⁻¹⁵/(1×10⁻⁴) ⟹ 5.0×10⁻¹¹ Pa

Answer: C

Firefighter's hose — projectile + momentum combinedStructured Q

Water leaves the nozzle at 13.1 m/s at 15.0° to horizontal.

(a) Velocity vs speed: "Velocity needs a direction to be specified whereas speed does not."

(b)(i) Vertical component: 13.1sin15 ⟹ 3.39 m/s (b)(ii) Vertical distance to max height: v²=u²+2as: 0=(3.39)²+2(−9.81)s ⟹ s≈0.586m

(d)(i) Momentum in 1s (water density 1000, nozzle diameter 4.00cm): Δp=ρAv²×t = 1000×[π(2×10⁻²)²]×(13.1)²×1 ⟹ Δp≈216 kg m/s

(d)(ii) Why isn't this equal to the actual force on the nozzle? "1) Weight not being considered. 2) There is friction between the nozzle and the water."

✓ Quick Recap — Full Chapter Cheat-Sheet

F = dp/dt (A-level def, exact wording). Special cases: F=ma (const. mass); F=v(dm/dt) (const. velocity).
Newton's 1st law = Law of Inertia. Newton's 3rd law pairs: equal, opposite, same line of action, same nature, on DIFFERENT bodies — never cause equilibrium.
Equilibrium: ΣF=0 AND Σmoment=0. A "principle" needs its condition stated first.
Moment = F × ⊥distance from pivot. Torque (couple) = F × full separation distance, position-independent.
★ Pivot placement: put it at an unknown force / hinge / most-forces-through point to eliminate unknowns from the moment equation.
★ Three-force equilibrium: unknown force's line of action passes through the intersection of the other two known forces' lines.
Weight (→centre of gravity, always down), Normal contact force (⊥ to surface), Friction (opposes relative motion/tendency, limiting force), Tension (into the string), Drag (opposes motion through a fluid), Upthrust (U=ρ_fAhg, up), Lift, Thrust.
Rough = consider friction; Smooth = ignore it.
Terminal velocity: a decays 9.81→0, v rises and flattens, s stays concave-up. A parachute opening mid-fall creates a SECOND (lower) terminal velocity.
Archimedes' Principle = upthrust, derived via displaced-fluid weight instead of pressure difference — not a separate topic.
Force from a fluid jet: F=ρAv², P=ρv² — ONLY when speed is reduced to zero on impact.