AIEM — ICSE Exam Prep

Chapter in a nutshell: A force can move a body in a straight line (translation) or turn it about a pivot (rotation). The turning effect is the moment of a force; a pair of equal opposite forces forms a couple; a balanced body obeys the principle of moments. We then meet the centre of gravity and uniform circular motion (constant speed, yet accelerated).

1. Force — recap

A force changes (or tends to change) a body's state of rest or motion, its direction, or its size/shape. Mathematically it is the rate of change of linear momentum: $$\vec{F}=\frac{d\vec{p}}{dt}=\frac{d(m\vec{v})}{dt}=m\vec{a}\ \ (\text{if } m \text{ constant})$$

Force is a vector. SI unit newton (N); gravitational unit kilogram-force (kgf), where $1\,\text{kgf}=g\,\text{N}=9.8\,\text{N}$.
On a rigid body a force can produce only motion; on a non-rigid body it can also change shape/size.

2. Translational vs Rotational Motion

Motion	When it occurs	Example
Translational (linear)	force on a body free to move	pushing a ball along the floor
Rotational	force on a body pivoted at a point	pushing a door, spinning a wheel about its axle

3. Moment (Turning Effect) of a Force — Torque

$$\textbf{Moment of force} = \text{Force} \times \text{perpendicular distance of its line of action from the pivot} = F \times d$$ Two factors decide the turning effect: (i) the magnitude of the force, (ii) the perpendicular distance of its line of action from the axis. Maximum turning is obtained when this perpendicular distance is greatest.

Units of moment of force

System	Unit	Relation
SI	newton metre (N m)	$1\,\text{N m}=10^{7}\,\text{dyne cm}$
CGS	dyne centimetre (dyne cm)	$1\,\text{kgf m}=9.8\,\text{N m}$
Gravitational	kgf m, gf cm	$1\,\text{gf cm}=980\,\text{dyne cm}$

(Note: although its unit is N m, a moment is never written in joule — torque is a vector, work is a scalar.)

Sign convention & direction: an anticlockwise moment is taken positive (direction along the axis, outward); a clockwise moment is negative (along the axis, inward). The direction of rotation depends on the point of application and direction of the force.

4. Common Examples of Moment of Force (and why)

Door: handle is fixed at the edge farthest from the hinges → larger d → a small force opens it. Near the hinge a much larger force is needed; at the hinge the door cannot be opened (d ≈ 0, torque = 0).
Hand flour-grinder: handle is near the rim so a small force gives a large moment about the central pivot.
Steering wheel: large diameter → larger d → easier to turn; sense of rotation changes by changing the point at which the force is applied.
Bicycle pedal: the toothed wheel is larger than the rear wheel so the pedal force acts at a large distance from the axle.
Spanner / wrench: a long handle gives a large moment to loosen a tight nut with a small force.

5. Couple

Two equal, opposite, parallel forces acting along different lines form a couple; it produces pure rotation (no translation, since the resultant force is zero).

Derivation (rod AB pivoted at O, forces F at each end, AB = couple arm = d):

moment $=F\times OA + F\times OB = F\times(OA+OB)=F\times d$. $$\boxed{\text{Moment of couple} = \text{either force} \times \text{perpendicular distance between them (couple arm)}}$$

Examples: turning a water tap, tightening a bottle cap, turning a key in a lock, winding a clock, turning a steering wheel with both hands.

6. Equilibrium of Bodies

A body is in equilibrium when several forces acting on it produce no change in its state of rest or of linear/rotational motion.

Static equilibrium — body at rest: a book on a table (weight balanced by reaction); a balanced beam balance.
Dynamic equilibrium — body in uniform motion: a raindrop falling at constant (terminal) velocity (weight = buoyancy + viscous drag); an aeroplane cruising at constant height (lift = weight); a stone whirled at uniform speed.

Two conditions for equilibrium:

The resultant of all forces acting on the body = 0 (no translation).
The resultant moment of all forces about any point = 0 (no rotation), i.e. anticlockwise moments = clockwise moments.

7. Principle of Moments

In equilibrium, sum of anticlockwise moments = sum of clockwise moments (about the same axis).

Verification (metre rule + two spring balances): suspend a metre rule horizontally from a thread at O. Hang weights $W_1$ (at distance $l_1$) and $W_2$ (at $l_2$) on either side via spring balances and adjust until the rule is horizontal. It is found that $$W_1 l_1 = W_2 l_2 \quad(\text{clockwise moment} = \text{anticlockwise moment})$$ A physical (beam) balance works on this principle.

8. Solving Numericals — method

Weight of a uniform rod/rule acts at its mid-point (50 cm mark).
Take moments about the pivot: $\Sigma(\text{anticlockwise}) = \Sigma(\text{clockwise})$.
Example: a uniform metre rule (pivoted at 0 cm) carries 40 kgf at the 40 cm mark, held up by a spring balance at the 100 cm end. If the rule is weightless: $40\times40 = F\times100 \Rightarrow F = 16\,\text{kgf}$. If the rule weighs 20 kgf (acts at 50 cm): $40\times40 + 20\times50 = F\times100 \Rightarrow F = 26\,\text{kgf}$.

9. Centre of Gravity (C.G.)

The centre of gravity is the point where the whole weight of the body appears to act (the resultant of the weights of all its particles), whatever the body's position.

Body	Position of C.G.
Uniform rod	mid-point
Circular disc / ring	centre
Square / rectangle / parallelogram	point of intersection of diagonals
Triangular lamina	centroid (intersection of medians)
Sphere / cube	geometric centre
Cone / cylinder	on the axis

- The C.G. may lie outside the material of the body (e.g. a ring, a horseshoe).

C.G. of an irregular lamina is found by the plumb-line method: suspend the lamina freely from different points; the C.G. lies at the intersection of the vertical plumb lines drawn from each point of suspension.

10. Uniform Circular Motion (UCM)

Motion of a body along a circular path at constant speed.

The direction of velocity changes continuously (always along the tangent) → velocity changes → the motion is accelerated, even though the speed is constant.
Centripetal force is the net force directed towards the centre that keeps the body on the circle:

$$F=\frac{mv^2}{r}$$ Supplied by: tension (stone on a string), gravitation (planet/satellite, electron round nucleus), friction (car on a curve).

Centrifugal force is the apparent (pseudo) outward force experienced in the rotating frame; it is not a real force (no real reaction partner).

Centripetal vs Centrifugal

	Centripetal force	Centrifugal force
Direction	towards the centre	away from the centre
Nature	real force	apparent / pseudo force
Frame	acts in an inertial frame	felt only in the rotating frame

11. Quick Recall / Exam Tips

Moment $=F\times$ perpendicular distance; unit N m (never joule). $1\,\text{kgf m}=9.8\,\text{N m}$, $1\,\text{N m}=10^{7}\,\text{dyne cm}$.
Anticlockwise = +ve, clockwise = −ve.
Couple needs two equal-opposite forces; moment of couple $=F\times$ couple arm.
Equilibrium needs both ΣF = 0 and Στ = 0.
Principle of moments: ACW moments = CW moments; uniform rod's weight acts at the mid-point.
UCM: speed constant but velocity changes → accelerated; name the real centripetal force; centrifugal force is fictitious.

12. Worked Numerical Examples (ICSE pattern)

Q1. A force of 10 N acts at a perpendicular distance of 30 cm from a pivot. Find the moment. Solution: Moment $=F\times d = 10 \times 0.30 = \mathbf{3\ N\,m}$.

Q2. The moment of a 5 N force about a point P is 2 N m. Find the perpendicular distance. Solution: $d = \dfrac{\text{Moment}}{F} = \dfrac{2}{5} = \mathbf{0.4\ m}$.

Q3. A mechanic opens a nut with a force of 150 N using a 40 cm handle. What handle length lets him use only 50 N? Solution: Required moment $=150\times0.40 = 60\ \text{N m}$. Then $50\times L = 60 \Rightarrow L = \mathbf{1.2\ m}$.

Q4. An iron door 3 m broad is opened by a 100 N force applied at the middle. Find (a) the torque, (b) the least force (and where) to open it. Solution: (a) $r = \tfrac12\times3 = 1.5\ \text{m}$, torque $=100\times1.5 = \mathbf{150\ N\,m}$. (b) Least force acts at the free end (r = 3 m): $F' = \dfrac{150}{3} = \mathbf{50\ N}$.

Q5. On a see-saw, children of 30 kg and 50 kg sit at 2 m and 2.5 m on one side of the centre. Where must a 74 kg man sit to balance it? Solution: Anticlockwise moment $=30\times2 + 50\times2.5 = 60+125 = 185\ \text{kgf m}$. For balance $74\times x = 185 \Rightarrow x = \mathbf{2.5\ m}$ on the other side.

Q6. A uniform metre rule (weight 20 kgf) is pivoted at the 0 cm mark; a 40 kgf weight hangs at the 40 cm mark and a spring balance supports the 100 cm end keeping it horizontal. Find the spring-balance reading. Solution: The rule's weight (20 kgf) acts at the 50 cm mark. Taking moments about the pivot: $40\times40 + 20\times50 = F\times100 \Rightarrow F = \dfrac{1600+1000}{100} = \mathbf{26\ kgf}$.

13. Key Terms — Quick Glossary

Term	One-line definition
Force	a push/pull; rate of change of momentum ($F=ma$).
Moment of force (torque)	force × perpendicular distance from the axis.
Couple	two equal, opposite, parallel forces on different lines → pure rotation.
Couple arm	perpendicular distance between the two forces of a couple.
Equilibrium	no change in state of rest or of motion (ΣF = 0 and Στ = 0).
Principle of moments	in equilibrium, ACW moments = CW moments.
Centre of gravity	point where the whole weight of the body acts.
Centripetal force	net inward force keeping a body in circular motion ($mv^2/r$).
Centrifugal force	apparent outward (pseudo) force in a rotating frame.

14. Likely Exam Questions (with crisp answers)

Define moment of force; state its SI unit. → Force × perpendicular distance from the axis; unit N m.
Is moment of force a scalar or vector? → Vector.
State two factors affecting the turning effect. → Magnitude of force; perpendicular distance of its line of action from the axis.
Why is a door's handle at its outer edge? → To maximise the perpendicular distance so a small force gives enough moment.
Why is it easier to turn a steering wheel of larger diameter? → Larger distance → larger moment for the same force.
Define a couple; give two examples. → Two equal, opposite, parallel forces on different lines; e.g. turning a tap, winding a clock key.
State the two conditions for equilibrium. → Resultant force = 0; resultant moment = 0.
Differentiate static and dynamic equilibrium. → Static = body at rest; dynamic = body in uniform motion. (Both satisfy ΣF = 0, Στ = 0.)
Where does the C.G. of a triangular lamina lie? → At its centroid (intersection of medians).
Why is uniform circular motion an accelerated motion? → The direction of velocity changes continuously, so velocity changes even though speed is constant.
Name the force that keeps a planet in orbit. → Gravitational force (acting as the centripetal force).
Is centrifugal force real? → No; it is a pseudo (apparent) force felt only in the rotating frame.

Force, Turning Effects of Force, and Uniform Circular Motion