🎯 Key Points
- F=ma (Second Law); action-reaction pairs act on DIFFERENT bodies, never cancel each other on the same body (Third Law)
- Static friction (f_s ≤ μ_s N) adjusts to prevent sliding; kinetic friction (f_k = μ_k N) is constant once sliding, and μ_k < μ_s
- Centripetal force F=mv²/r always points toward the centre — it's not a new force, just the net result of existing forces (tension, gravity, normal, friction)
- Banking angle: tan θ = v²/rg — designed so friction isn't needed at the design speed
- Impulse J = FΔt = Δp; momentum is conserved when net external force is zero
Newton's Laws
- First Law: An object remains at rest or in uniform motion unless acted on by an external force (inertia)
- Second Law: F = ma (net force = mass × acceleration)
- Third Law: Every action has an equal and opposite reaction
Friction
Free-body diagram of a block on an incline: weight (mg) resolves into mg sinθ along the slope and mg cosθ into the slope, balanced by the normal force N and friction f.
- Static friction: f_s ≤ μ_s·N (maximum just before sliding)
- Kinetic friction: f_k = μ_k·N (constant during sliding, μ_k < μ_s)
- Angle of friction: tan(λ) = μ
Circular Motion
- Centripetal acceleration: a = v²/r = ω²r
- Centripetal force: F = mv²/r (directed toward centre)
- Banking angle: tan(θ) = v²/rg
- Critical speed at top of loop: v = √(rg)
Impulse and Momentum
- Impulse J = F·Δt = Δp (change in momentum)
- Conservation of momentum: if ΣF_ext = 0, then Σp = constant
Free Body Diagrams (FBD) — The Method
- Isolate ONE body and draw only the forces acting ON it: weight mg (always downward), normal reaction N (perpendicular to the contact surface), tension T (along the string, away from the body), applied force, and friction f (along the surface, opposing relative motion/tendency).
- Choose convenient axes (for inclines, take axes ALONG and PERPENDICULAR to the slope), then write ΣF = ma separately along each axis.
- Never include forces the body exerts on others — an action-reaction pair acts on two DIFFERENT bodies, so the two never appear in the same FBD.
- Solve the resulting simultaneous equations for the unknowns (acceleration, tension, normal force, etc.).
Connected Bodies: Pulley, Lift, and Chain
- Bodies on a table joined over a pulley: for a hanging mass m₁ pulling a mass m₂ on a smooth horizontal surface, a = m₁g/(m₁ + m₂) and tension T = m₁m₂g/(m₁ + m₂).
- Atwood machine (two masses over a pulley): a = (m₁ − m₂)g/(m₁ + m₂); T = 2m₁m₂g/(m₁ + m₂).
- Apparent weight in a lift: N = m(g + a) when accelerating up, N = m(g − a) when accelerating down; N = mg at rest or constant velocity; N = 0 (weightlessness) in free fall (a = g).
- Inextensible string: all connected bodies share the same magnitude of acceleration, and tension is uniform in a massless string over a frictionless pulley.
Friction on an Inclined Plane
- On an incline of angle θ, the weight resolves into mg sinθ (down the slope) and mg cosθ (into the surface), so N = mg cosθ.
- Angle of repose (α): the maximum incline angle at which a block just stays at rest → tan α = μ_s. If θ ≤ angle of repose, the block does not slide.
- Block sliding DOWN a rough incline: acceleration a = g(sinθ − μ_k cosθ).
- Block pushed UP a rough incline (friction acts down-slope): retardation a = g(sinθ + μ_k cosθ).
Banking of Roads (With Friction)
- For a frictionless banked road, the safe speed is fixed: v = √(rg tanθ), i.e. tanθ = v²/(rg).
- With friction (coefficient μ), a RANGE of safe speeds is allowed: v_max = √[rg(tanθ + μ)/(1 − μ tanθ)] and v_min = √[rg(tanθ − μ)/(1 + μ tanθ)].
- On a flat (unbanked) road, the turn relies on friction alone: v_max = √(μ rg).
Pseudo Forces in Non-Inertial Frames
- Newton's laws hold directly only in inertial frames (non-accelerating). In an accelerating (non-inertial) frame, they can still be applied by adding a pseudo force = −m·a_frame on every body, opposite to the frame's acceleration.
- Example: inside a car accelerating forward with acceleration a, a passenger feels a backward pseudo force ma; a hanging pendulum settles at tanφ = a/g from the vertical.
- Pseudo force explains the "apparent weight" in a lift and the outward "centrifugal force" felt in the rotating frame of a turning vehicle — it is a bookkeeping force, not a real interaction.
🚀 JEE Advanced Edge
Pseudo force in non-inertial frames: When analysing motion from inside an accelerating frame (like a lift or accelerating car), add a pseudo force = −ma_frame (opposite to the frame's acceleration) to make Newton's laws apply in that frame. This is the fastest way to solve "apparent weight in a lift" or "pendulum hanging in an accelerating car" problems without switching to the ground frame.
Connected bodies / Atwood machine with pulleys: For two masses connected over a pulley, write F=ma for EACH mass separately (using the SAME tension T and SAME magnitude of acceleration a, since the string is inextensible), then solve the two equations simultaneously: a = (m₁−m₂)g/(m₁+m₂) for a simple Atwood machine.
Worked problem: A block of mass 2 kg on a rough incline (θ=30°, μ=0.5) — will it slide? Approach: Compare tan θ to μ: tan(30°)≈0.577 > μ=0.5? No, actually tan(30°)=0.577 > 0.5, so since tan θ > μ, gravity's component down the slope EXCEEDS the maximum available static friction, and the block DOES slide. (If tan θ ≤ μ, it would stay put.)