Laws of Motion — NEET notes

Newton's First Law & Inertia

Every object resists changes to its state of motion — this resistance is called inertia. Newton's first law says an object stays at rest or moves at constant velocity unless acted on by a net external force. It also defines what an inertial frame is. NEET loves seatbelt/tablecloth/coin-on-card examples.

If F _ net = 0, then v = constant (including zero)

Statement of Newton's First Law

A body remains at rest or in uniform motion in a straight line unless a net external force acts on it. This is NOT just a special case of the second law — it defines what a valid reference frame is.

Newton's First Law: Every body continues in its state of rest or of uniform motion in a straight line, unless compelled to change that state by a net external force. The key word is 'net' — multiple forces can act, but if they cancel out ( F _ net = 0 ), the body behaves as if no force acts.

• The first law is also called the Law of Inertia
• It does NOT say 'a body at rest stays at rest' — it says 'at rest OR in uniform motion in a straight line'
• 'Uniform motion' means constant speed in a straight line — constant velocity
• The first law defines inertial frames: frames where this law holds true (no pseudo forces needed)

Types of Inertia

Inertia comes in three flavours: resistance to starting motion, resistance to stopping motion, and resistance to changing direction. Mass is the quantitative measure of inertia.

Inertia is the tendency of a body to resist any change in its state of motion. There are three types, each describing what the body 'resists':

• Inertia of rest: A body at rest resists being set into motion (dust flies off a carpet when beaten — carpet moves, dust stays)
• Inertia of motion: A moving body resists stopping (passengers lurch forward when a bus brakes suddenly)
• Inertia of direction: A moving body resists changing direction (sparks fly off a grinding wheel tangentially, not radially)

Mass is the measure of inertia. Greater mass greater inertia harder to change the state of motion. A 1000 kg car has 100 the inertia of a 10 kg box. This is why NCERT calls mass the 'quantitative definition of inertia.'

Quick recall

A coin is placed on a cardboard on top of a glass. The cardboard is flicked quickly. What happens to the coin and why?

The coin drops straight into the glass. The coin has inertia of rest — it resists the horizontal motion of the cardboard. When the cardboard is pulled out quickly, friction doesn't have enough time to set the coin in motion, so it stays in place and falls straight down due to gravity.

Inertial Frames of Reference

An inertial frame is one in which a free body (no net force) moves with constant velocity. Newton's laws are valid only in inertial frames. Accelerating frames are non-inertial and need pseudo forces.

A reference frame in which Newton's first law holds — i.e., a body with zero net force moves at constant velocity — is called an inertial frame. Any frame moving at constant velocity relative to an inertial frame is also inertial.

• A frame at rest or moving with constant velocity = inertial frame
• An accelerating frame (car braking, merry-go-round, elevator accelerating) = non-inertial frame
• In non-inertial frames, objects appear to accelerate without any real force — this is why we introduce pseudo forces
• Earth is approximately inertial for most problems (its rotation effects are tiny for NEET-level problems)

Newton's Second Law

The second law quantifies force: F _ net = m a . It also comes in the more general momentum form F = d p dt , which leads to the concept of impulse. NEET heavily tests unit conversions, momentum change, and impulse-momentum theorem problems.

F _ net = m a = d p dt | J = F t = p

F = ma

Net force equals mass times acceleration. This equation connects cause (force) to effect (acceleration) and is the backbone of nearly every mechanics problem in NEET.

Newton's Second Law: The net external force on a body equals the product of its mass and acceleration: F _ net = m a . Force and acceleration are in the same direction. Mass m is the proportionality constant — more mass means less acceleration for the same force.

F _ net = m a a = F _ net m

• F = ma is a vector equation: F_x = ma_x , F_y = ma_y (resolve into components)
• SI unit of force: newton (N) = kg m/s ^2
• CGS unit: dyne = g cm/s ^2 . Conversion: 1 N = 10^5 dyne
• The law is valid only in inertial frames — in non-inertial frames, you must add pseudo forces
• m in F = ma is the inertial mass — the resistance to acceleration

Momentum Form

Newton originally stated the second law in terms of momentum: force equals the rate of change of linear momentum. This form handles variable-mass systems (like rockets) that F = ma cannot.

Linear momentum: p = m v . Newton's second law in its general form: F _ net = d p dt . When mass is constant, this simplifies to F = m d v dt = m a . But when mass changes (rocket ejecting fuel), only the momentum form works.

p = m v | F _ net = d p dt

• SI unit of momentum: kg m/s = N s
• Momentum is a vector — direction matters
• If F _ net = 0 then p = constant — this is conservation of momentum (preview of Chapter 7)
• For constant mass: d p dt = m d v dt = m a (recovers F = ma )

Impulse

Impulse is force multiplied by the time interval over which it acts. It equals the change in momentum. This concept is critical for collisions, hitting balls, and catching problems in NEET.

Impulse J is defined as the product of force and the time interval during which it acts. For a constant force: J = F t . From Newton's second law: F = p t , so J = p .

J = F _ avg t = p = m v _f - m v _i

• Impulse has the same units as momentum: N s = kg m/s
• Impulse is a vector — direction of impulse = direction of p
• For variable force: J = _ t_1 ^ t_2 F dt = area under F-t graph
• Impulse-momentum theorem: The impulse of the net force equals the change in momentum

Quick recall

A 0.15 kg cricket ball moving at 20 m/s is hit by a bat and goes back at 30 m/s. If the contact time is 0.01 s, find the average force.

p = m(v_f - v_i) = 0.15(30 - (-20)) = 0.15 50 = 7.5 kg m/s. F_ avg = p t = 7.5 0.01 = 750 N . (Taking the bat's hit direction as positive.)

Newton's Third Law

Every action has an equal and opposite reaction — but the action-reaction pair acts on DIFFERENT bodies, which is why they don't cancel. This law explains walking, swimming, rocket propulsion, and recoil of a gun. NEET tests whether students can identify correct action-reaction pairs.

F _ AB = - F _ BA

Action-Reaction Pairs

When body A exerts a force on body B, body B simultaneously exerts an equal and opposite force on body A. These two forces always act on different bodies — that's why they never cancel each other.

Newton's Third Law: To every action, there is always an equal and opposite reaction. If body A exerts a force F _ AB on body B, then body B exerts a force F _ BA = - F _ AB on body A. These forces are simultaneous, equal in magnitude, opposite in direction, and act on different bodies.

F _ AB = - F _ BA (always, at every instant)

• Action and reaction are always equal — regardless of the masses of the objects
• They act on different bodies — this is why they DON'T cancel
• They are of the same type — if A pushes B with a contact force, B pushes A with a contact force (not gravity!)
• There is no time delay — action and reaction occur simultaneously
• You cannot have a single force in nature — forces always come in pairs

Applications

The third law is behind every form of locomotion. You push backward on the ground, the ground pushes you forward. Rockets push exhaust backward, exhaust pushes rocket forward.

Every mode of movement depends on Newton's third law. You can only accelerate yourself by pushing on something else — and letting the reaction force push you.

• Walking: Your foot pushes the ground backward (action). The ground pushes your foot forward (reaction) — this is what propels you
• Swimming: Your hands push water backward. Water pushes you forward
• Rocket propulsion: Rocket pushes exhaust gases out at high speed. Gases push the rocket forward. Works in vacuum — no air needed
• Gun recoil: Bullet goes forward (action on bullet by gun). Gun goes backward (reaction on gun by bullet). m_ bullet v_ bullet = m_ gun v_ recoil

Quick recall

A 50 kg astronaut in space throws a 5 kg tool at 10 m/s. What is the astronaut's recoil velocity?

By conservation of momentum (which comes from the third law): m_1 v_1 = m_2 v_2 5 10 = 50 v v = 1 m/s in the opposite direction.

Free Body Diagrams

An FBD is a picture of ONE body showing ALL forces acting on it — nothing else. It's the first step in solving any mechanics problem. Most NEET errors come from missing forces, adding wrong forces, or including forces that act on other bodies. Master FBDs and you master mechanics.

F = m a (apply to each FBD separately)

What is an FBD?

A free body diagram isolates one object and shows every external force acting on it. It strips away the environment and replaces it with force arrows. Every mechanics solution starts here.

A Free Body Diagram (FBD) is a diagram of a single body, isolated from everything else, showing all the external forces acting on it as arrows. 'Free' means you mentally free the body from its surroundings and replace every contact/interaction with a force arrow.

• Draw ONLY forces acting on the body — never include forces the body exerts on others
• Common forces to include: weight ( mg downward), normal force ( N perpendicular to surface), tension ( T along the string), friction ( f opposing relative motion), applied force
• The body is usually drawn as a simple dot or box — shape doesn't matter for FBDs
• Label every force with a symbol and, if known, its magnitude and direction

How to Draw an FBD

A systematic procedure to draw correct FBDs every time: isolate the body, identify contacts, draw weight, add normal/tension/friction, and verify the count.

Follow these steps every time and you'll never miss a force or add an extra one:

• Step 1: Identify the body and mentally isolate it from everything else
• Step 2: Draw weight mg vertically downward from the center of the body (gravity ALWAYS acts — don't forget it)
• Step 3: At every point of contact, draw the normal force perpendicular to the surface (always pushes, never pulls)
• Step 4: At every point of contact where surfaces can slide, draw friction parallel to the surface (opposing relative motion or tendency of motion)
• Step 5: Add any tension (along strings/ropes, away from the body), applied forces, spring forces, or other specified forces

Verification: Count the forces. For a body resting on a surface with no applied force, you should have exactly 2 forces (weight + normal) or 3 forces (weight + normal + friction if on an incline). If you have more or fewer, recheck.

Quick recall

Draw the FBD of a block of mass m on a rough inclined plane (angle ) about to slide down. How many forces act on it?

3 forces: (1) Weight mg vertically downward, (2) Normal force N perpendicular to the incline surface, (3) Static friction f_s along the incline pointing UP the slope (opposing the tendency to slide down). That's it — no component of gravity is a separate force.

Common FBD Setups

The most frequently tested FBD scenarios in NEET. Master these and you can handle any variation the exam throws at you.

NEET recycles the same setups with different numbers. Recognizing the setup instantly saves time.

• Block on horizontal surface: Forces = mg (down) + N (up) + F_ applied + friction. If no applied force and at rest: N = mg
• Block on inclined plane: mg (down), N (perpendicular to incline), friction (along incline). Along incline: mg f
• Hanging mass on a string: mg (down) + T (up along string). At rest: T = mg . Accelerating up: T = m(g+a)
• Two blocks in contact: Draw separate FBDs. The contact force on block 1 by block 2 is equal and opposite to the contact force on block 2 by block 1
• Block on top of another block: The bottom block feels: mg of itself + N from top block (downward) + N from ground (upward) + friction from top block + friction from ground

Friction

Friction is the force that opposes relative motion (or the tendency of motion) between two surfaces in contact. Static friction adjusts up to a maximum ( _s N ), kinetic friction is constant ( _k N ), and rolling friction is the smallest. NEET heavily tests friction on inclines, angle of repose, and the crucial fact that _s > _k .

f_s _s N | f_k = _k N | _s > _k

Static vs Kinetic Friction

Static friction prevents motion from starting (self-adjusting, up to a max). Kinetic friction opposes motion that has already begun (constant). It always takes more force to START sliding than to KEEP sliding.

Static friction f_s acts when the body is NOT sliding. It is self-adjusting — it matches the applied force exactly, up to a maximum: f_s _s N . Once the applied force exceeds _s N , the body starts moving and kinetic friction f_k = _k N takes over (constant value, independent of speed for NEET-level).

f_ s, = _s N | f_k = _k N | _s > _k

• Static friction is NOT always _s N — it equals the applied force when the body is at rest. _s N is only the MAXIMUM value
• Kinetic friction f_k = _k N is constant — doesn't depend on speed (NCERT approximation)
• _s > _k always — this is why a push feels easier once the object starts moving
• Both _s and _k are dimensionless and have no units
• Friction does NOT depend on the area of contact (NCERT-level approximation)

Direction of Friction

Friction opposes the relative motion (or tendency of relative motion) between the two surfaces in contact. This direction can be tricky — especially when one surface moves while another is on top.

The direction of friction is determined by the relative motion (or tendency of relative motion) between the two surfaces. Not absolute motion. If block A sits on a moving belt, and the belt moves right, the tendency of A relative to the belt is to be left behind — so friction on A acts rightward (forward).

• Friction on a body opposes its motion relative to the surface it's on
• On a stationary surface: friction opposes the body's absolute motion (since the surface isn't moving)
• On a moving surface (belt, truck bed): friction can act in the SAME direction as the body's absolute motion — if it prevents sliding
• Static friction can act in ANY direction — whatever direction prevents relative sliding

Quick recall

A box sits on the back of a truck that accelerates forward. In which direction does friction act on the box?

Forward (same direction as the truck's acceleration). Without friction, the box would slide backward relative to the truck. Friction prevents this by pushing the box forward. Here, friction is the force that accelerates the box!

Angle of Repose & Friction on Inclined Planes

The angle of repose is the maximum angle of an inclined plane at which a block stays at rest. It directly gives the coefficient of static friction: _r = _s . Inclined plane friction problems are NEET favorites.

Place a block on a rough inclined plane and slowly increase the angle. At some critical angle _r , the block just begins to slide — this is the angle of repose. At this angle, the component of gravity along the incline exactly equals maximum static friction.

At angle of repose: mg _r = _s mg _r _r = _s

• For < _r : block stays at rest, f_s = mg (self-adjusting, less than max)
• For = _r : block is on the verge of sliding, f_s = _s N = _s mg _r
• For > _r : block slides down, friction = f_k = _k mg
• Acceleration when sliding: a = g( - _k )

Quick recall

A block stays at rest on a rough incline at 30° but slides at 35° . What can you say about _s ?

_s is between 30° and 35° , i.e., 0.577 < _s < 0.700 . The angle of repose is between 30° and 35° .

Rolling Friction

Rolling friction is much smaller than sliding friction. That's why wheels, ball bearings, and rollers are used everywhere. For NEET, know the hierarchy: rolling < kinetic < static.

When a body rolls on a surface (like a wheel on a road), the resistance is called rolling friction. It arises from the slight deformation of the surfaces at the contact region. Rolling friction is much smaller than sliding (kinetic) friction.

• Hierarchy: f_ rolling f_ kinetic < f_ static
• Rolling friction depends on the hardness of surfaces and radius of the rolling body
• Ball bearings replace sliding friction with rolling friction — reducing energy loss dramatically
• NCERT doesn't give a formula for rolling friction at this level — just know the concept and hierarchy

Circular Motion Dynamics

Any object moving in a circle needs a centripetal force directed toward the center — this is not a new force but is provided by existing forces (tension, gravity, friction, normal). NEET loves banking of roads, conical pendulum, and car-on-a-curve problems. Remember: centripetal force is the CAUSE, circular motion is the EFFECT.

F_c = mv^2 r = m ^2 r | _ bank = v^2 rg

Centripetal Force

Centripetal force is the net inward force required to maintain circular motion. It's always directed toward the center and is provided by real forces like tension, gravity, friction, or normal force. It's NOT a new type of force.

For a body of mass m moving at speed v in a circle of radius r , the net force toward the center must be F_c = mv^2 r . This is called centripetal force ('center-seeking'). It is NOT a separate force — it's the resultant of real forces that happen to point toward the center.

F_c = mv^2 r = m ^2 r = 4 ^2 mr T^2

• Centripetal force is always directed toward the center of the circular path
• It is NOT a new force — it's provided by tension (string), gravity (orbits), friction (car on road), normal force (wall of death), etc.
• If centripetal force vanishes, the body flies off tangentially (not radially outward!)
• Centripetal acceleration: a_c = v^2 r = ^2 r

Banking of Roads

On a banked road, a component of the normal force provides centripetal force. The ideal banking angle (no friction needed) is = v^2/rg . With friction, the safe speed range changes.

On a flat road, only friction provides centripetal force for turning — dangerous at high speeds. Banking tilts the road inward so that a component of the normal force ( N ) points toward the center, providing centripetal force even without friction.

Without friction: = v^2 rg

With friction: v_ = rg _s + 1 - _s

• At the ideal speed ( v = rg ): no friction needed — the normal force alone provides centripetal force
• If v > ideal speed: friction acts inward (down the incline) to provide extra centripetal force
• If v < ideal speed: friction acts outward (up the incline) to prevent the car from sliding inward
• The v_ formula with friction is for the maximum safe speed

Quick recall

A road is banked at 30° for a curve of radius 100 m. Find the ideal speed (no friction, g = 10 m/s ^2 ).

v = rg = 100 10 30° = 1000 0.577 = 577 24 m/s 86 km/h.

Conical Pendulum

A mass on a string revolves in a horizontal circle with the string tracing a cone. Tension provides both the centripetal force (horizontal component) and balances weight (vertical component).

A conical pendulum has a bob of mass m attached to a string of length L that makes angle with the vertical. The bob moves in a horizontal circle of radius r = L . The tension in the string has two jobs: its vertical component ( T ) balances weight, and its horizontal component ( T ) provides centripetal force.

T = mg | T = mv^2 r

= v^2 rg | T = 2 = 2 L g

• Radius of circular path: r = L
• Tension: T = mg (always greater than mg )
• Time period: T_ period = 2 L g — depends on L and , NOT on mass
• As 90° : the string becomes horizontal, but this requires T and — impossible in practice

Connected Bodies & Pulleys

When bodies are connected by strings over pulleys, they share a constraint: the string length is constant, so their accelerations are linked. The Atwood machine ( a = (m_1 - m_2)g m_1 + m_2 ) is the most tested setup in NEET. The key skill is writing constraint equations and solving simultaneous equations from individual FBDs.

a = (m_1 - m_2)g m_1 + m_2 | T = 2m_1 m_2 g m_1 + m_2

Atwood Machine

The classic Atwood machine: two masses connected by a string over a frictionless pulley. The heavier mass goes down, the lighter goes up, both with the same acceleration and the same tension.

An Atwood machine has two masses m_1 and m_2 ( m_1 > m_2 ) connected by a light, inextensible string over a massless, frictionless pulley. Since the string is inextensible, both masses have the same acceleration magnitude a . The string has the same tension T throughout.

a = (m_1 - m_2) m_1 + m_2 g | T = 2m_1 m_2 m_1 + m_2 g

Derivation: FBD of m_1 (going down): m_1 g - T = m_1 a . FBD of m_2 (going up): T - m_2 g = m_2 a . Adding: (m_1 - m_2)g = (m_1 + m_2)a . Solving gives a and back-substituting gives T .

• If m_1 = m_2 : a = 0 and T = mg — the system is in equilibrium
• If m_2 = 0 (free fall): a = g and T = 0 — as expected
• Tension T is always between m_2 g and m_1 g (greater than the lighter weight, less than the heavier)
• The acceleration a is always less than g (since m_1 - m_2 m_1 + m_2 < 1 )

Quick recall

In an Atwood machine, m_1 = 5 kg and m_2 = 3 kg. Find acceleration and tension ( g = 10 m/s ^2 ).

a = (5-3) (5+3) 10 = 2 8 10 = 2.5 m/s ^2 . T = 2 5 3 5+3 10 = 30 8 10 = 37.5 N .

Pulley at Table Edge

A block on a smooth horizontal table is connected by a string over a pulley at the edge to a hanging mass. The hanging mass drives the system; the table block accelerates horizontally.

Setup: mass m_1 on a smooth table, connected to mass m_2 hanging off the edge via a string over a frictionless pulley. The constraint: both have the same acceleration a , and the string has the same tension T throughout.

a = m_2 g m_1 + m_2 | T = m_1 m_2 g m_1 + m_2

Derivation: FBD of m_1 (on table, smooth): T = m_1 a . FBD of m_2 (hanging): m_2 g - T = m_2 a . Adding: m_2 g = (m_1 + m_2)a .

• If the table has friction: m_2 g - m_1 g = (m_1 + m_2)a , so a = (m_2 - m_1)g m_1 + m_2
• For the system to move, the hanging mass must overcome friction: m_2 g > m_1 g m_2 > m_1
• Normal force on m_1 : N = m_1 g (the table supports m_1 's weight; the string pulls horizontally, not vertically)

Constraint Equations

When bodies are connected by inextensible strings, the total string length is constant. Differentiating this constraint twice gives the relationship between accelerations. Essential for complex pulley systems.

In any pulley system with an inextensible string: the total length of the string is constant. If you write this length in terms of the positions of the masses and differentiate twice with respect to time, you get the constraint equation linking their accelerations.

• Simple case (Atwood/table edge): Both masses have the same acceleration magnitude: a_1 = a_2 = a
• Movable pulley: If a pulley itself moves, the string wraps twice — so the mass attached to the pulley moves at half the acceleration: if block moves with a , the pulley (and attached mass) moves with a/2
• General method: Write string length L = sum of segments. Since L = constant, d^2L dt^2 = 0 gives the constraint
• Virtual work method (advanced): Sum of T v for all segments of a string = 0 (for massless string)

Quick recall

A string goes from block A on a table, over a fixed pulley, down to a movable pulley, and back up to the ceiling. A mass B hangs from the movable pulley. How are a_A and a_B related?

The string wraps around the movable pulley, so two segments support B. If A moves by x , the movable pulley (and B) moves by x/2 . Therefore a_B = a_A 2 .

Pseudo Forces

In a non-inertial (accelerating) frame, Newton's laws don't work unless you add a fictitious 'pseudo force' = -m a _ frame on every body. The classic NEET application is the apparent weight in an elevator: heavier when going up, lighter when going down, weightless in free fall.

F _ pseudo = -m a _ frame | W_ apparent = m(g a)

Pseudo Force

When you solve problems from inside an accelerating frame (non-inertial), objects seem to accelerate without any real force. To fix this, add a pseudo force = -m a _ frame on every body. Then Newton's laws work as usual.

In a non-inertial frame (accelerating car, rotating merry-go-round, accelerating elevator), Newton's second law F = m a doesn't hold unless we add a correction: the pseudo force (also called fictitious or inertial force). For a frame accelerating at a _0 , the pseudo force on every mass m is: F _ pseudo = -m a _0 .

F _ pseudo = -m a _ frame

• Pseudo force acts on EVERY body in the non-inertial frame — not just the one you're analyzing
• It is proportional to mass (like gravity) — all objects experience the same pseudo acceleration
• Direction: opposite to the acceleration of the frame
• Pseudo force has NO reaction pair (Newton's third law doesn't apply to it)
• If you solve the problem from an inertial frame, you do NOT add pseudo force

Elevator Problems

When an elevator accelerates, a person's apparent weight (what the scale reads) changes. Going up: feels heavier. Going down: feels lighter. Free fall: feels weightless. This is the most tested pseudo force application in NEET.

A person of mass m stands on a scale in an elevator. The scale reads the normal force N , which is the apparent weight. From the ground (inertial frame), apply F = ma to the person: N - mg = ma (taking up as positive). So N = m(g + a) .

N = m(g + a) (where a > 0 means upward acceleration)

• Elevator accelerating upward ( a > 0 ): N = m(g+a) > mg — feels heavier
• Elevator accelerating downward ( a < 0 , i.e., a is downward): N = m(g-|a|) < mg — feels lighter
• Elevator in free fall ( a = g downward): N = m(g-g) = 0 — weightlessness!
• Constant velocity (up or down): a = 0 , so N = mg — no change in apparent weight
• If a > g downward: N becomes negative — person lifts off the floor (needs to be strapped in)

Quick recall

A 60 kg person stands on a scale in an elevator accelerating upward at 2 m/s ^2 . What does the scale read? ( g = 10 m/s ^2 )

N = m(g + a) = 60(10 + 2) = 60 12 = 720 N (apparent weight = 72 kgf). The real weight is 600 N.

Rotating Frames

In a rotating frame, two pseudo forces appear: centrifugal (outward, m ^2 r ) and Coriolis ( 2m v ). NEET only tests centrifugal force in simple problems — Coriolis is beyond syllabus but good to know conceptually.

In a frame rotating with angular velocity , a body at distance r from the rotation axis experiences a pseudo force pointing radially outward: F_ centrifugal = m ^2 r . This is why you feel 'pushed outward' on a merry-go-round.

F_ centrifugal = m ^2 r (radially outward in rotating frame)

• Centrifugal force is a pseudo force — exists only in the rotating frame, NOT in an inertial frame
• In the inertial frame, the same situation is explained by centripetal force (real, inward)
• Centrifugal force = m ^2 r outward; Centripetal force = m ^2 r inward — same magnitude, opposite directions, different frames
• Coriolis force ( 2m v ) acts on moving objects in rotating frames — responsible for cyclones, trade winds. NOT in NEET syllabus

Equilibrium

A body is in translational equilibrium when the net force on it is zero ( F _ net = 0 ). For three concurrent forces, Lami's theorem provides a powerful shortcut. NEET tests equilibrium on inclines, hanging signs, and ring/knot problems where multiple strings meet.

F = 0 | F_1 = F_2 = F_3 (Lami's theorem)

Conditions for Equilibrium

Translational equilibrium means zero net force, which means zero acceleration (not necessarily zero velocity). The body can be at rest OR moving at constant velocity.

A body is in translational equilibrium when the vector sum of all forces on it is zero: F = 0 . In component form: F_x = 0 and F_y = 0 . The body then has zero acceleration — it is either at rest or moving at constant velocity.

• Equilibrium does NOT mean 'at rest' — a car cruising at constant velocity on a highway is in equilibrium
• Static equilibrium: at rest + zero net force
• Dynamic equilibrium: constant velocity + zero net force
• For equilibrium in 2D, you need TWO equations: F_x = 0 and F_y = 0
• A body at rest is NOT in equilibrium if a net force acts on it (it's about to accelerate)

Lami's Theorem

When exactly three concurrent coplanar forces keep a body in equilibrium, each force is proportional to the sine of the angle between the other two. This avoids resolving into components entirely.

Lami's Theorem: If three concurrent coplanar forces F_1 , F_2 , F_3 are in equilibrium, then each force divided by the sine of the angle opposite to it is constant.

F_1 = F_2 = F_3

Here is the angle between F_2 and F_3 (opposite to F_1 ), is the angle between F_1 and F_3 (opposite to F_2 ), and is the angle between F_1 and F_2 (opposite to F_3 ). Note: + + = 360° .

• Works ONLY for exactly three concurrent coplanar forces in equilibrium
• The angles are between the force vectors as they meet at the point — NOT the angles in a triangle
• + + = 360° (the angles go all the way around the point)
• This is essentially the sine rule applied to the force triangle

Quick recall

A 10 kg lamp hangs from two strings making 30° and 60° with the ceiling. Find the tensions ( g = 10 m/s ^2 ).

Weight = 100 N (downward). Angle opposite to T_1 (string at 30° ) is 180° - 30° = 150° ... Actually, use components: T_1 60° + T_2 30° = 0 (horizontal) and T_1 60° + T_2 30° = 100 (vertical). Wait — simpler: the string at 30° from ceiling makes 60° with vertical. Resolve: T_1 30° + T_2 60° = 100 , T_1 30° = T_2 60° . From the second: T_1 = T_2/ 3 . Substituting: T_2 = 86.6 N , T_1 = 50 N .

Equilibrium on an Inclined Plane

A block on a rough incline is in equilibrium when friction exactly balances the downhill component of gravity. This ties together FBDs, friction, and equilibrium in one neat package.

A block of mass m on a rough incline (angle , coefficients _s , _k ) stays at rest when static friction balances the gravitational pull along the incline. Equilibrium along the incline: f_s = mg . Perpendicular to incline: N = mg .

• Block stays at rest if _r (angle of repose), where _r = _s
• At rest: friction f_s = mg (self-adjusting, less than _s N )
• At the verge: f_s = _s N = _s mg = mg
• If an external force F is applied up the incline to hold the block: F + f_s = mg (friction may act down the incline if F is large enough)

Quick recall

A 2 kg block is on a rough incline at 30° , _s = 0.6 . Is the block in equilibrium? ( g = 10 m/s ^2 )

Force down the incline = mg 30° = 2 10 0.5 = 10 N. Max friction = _s mg 30° = 0.6 2 10 0.866 = 10.4 N. Since f_ s, = 10.4 N > 10 N, friction can balance gravity. Yes, the block is in equilibrium. Actual friction = 10 N (not 10.4).