Laws of Motion & Friction

Newton's laws look elementary, but Advanced weaponises them with pseudo-forces in accelerating frames, wedge-block systems, banked curves with friction, and coupled pulley problems. The core skill is drawing a correct free-body diagram (FBD) and choosing between the ground frame and a non-inertial frame.

Key results

• Pseudo-force: in a frame accelerating with a, add −ma on every body, then treat the frame as inertial.
• Friction: f ≤ μsN (static, self-adjusting up to the limit); f = μkN (kinetic). Static friction is whatever is needed to prevent sliding, not automatically μsN.
• Banked road: for radius r, safe speed range with friction is v² between rg(tanθ−μ)/(1+μtanθ) and rg(tanθ+μ)/(1−μtanθ).
• Block on wedge (smooth): acceleration of block relative to wedge and wedge itself are found by writing FBDs plus the constraint that the block stays on the incline.

Situation	Key relation
Minimum force to pull a block	F = μmg/√(1+μ²) at angle θ=arctan μ
Two blocks, contact force	analyse system, then one block
Lift accelerating up	apparent weight = m(g+a)
Belt/rough incline slipping	compare mg sinθ with μmg cosθ

Exam Tricks & Tips

• 🎯 Static friction is reactive: compute the force needed first; only if it exceeds μsN does the body slide.
• 🎯 Use the non-inertial frame of a wedge/lift to kill the wedge's acceleration and simplify the block's FBD — just add the pseudo-force.
• 🎯 Angle of repose = angle of friction: tanθ = μs; a block just slides when incline angle exceeds arctan μs.
• 🎯 Minimum pulling force is at angle arctan μ, not horizontal — a classic optimisation result worth memorising.
• 🎯 Check the direction of friction last: it opposes relative sliding (kinetic) or the tendency of relative sliding (static), which may be up the incline.
• ❌ Common mistake: using f = μN for static friction on a body that is not on the verge of slipping — that overestimates the force and gives a wrong acceleration.

Expected exam pattern

1–2 problems, usually a wedge-block or a multi-block pulley in the numerical/integer format, sometimes a banked-curve conceptual multiple-correct. Often the second part of a two-stage question.

Quick recap

Draw clean FBDs, decide inertial vs non-inertial, treat static friction as self-adjusting up to μsN, and remember angle of repose = arctan μs and the arctan μ minimum-force result.

Cover the answer, recall, then check. 12 cards on laws of motion and friction for JEE Advanced.

Q1. Is static friction always μsN?
A1. No. It self-adjusts to whatever is needed to prevent sliding, up to a maximum of μsN.

Q2. Apparent weight in a lift accelerating upward at a.
A2. m(g + a). Downward acceleration gives m(g − a); free fall gives zero.

Q3. Angle of repose in terms of μs.
A3. θ = arctan μs; the block just begins to slide at this incline angle.

Q4. What is a pseudo-force and when do you use it?
A4. A fictitious force −ma applied in a frame accelerating at a, so the frame can be treated as inertial.

Q5. Angle for minimum force to drag a block along the floor.
A5. θ = arctan μ; minimum force = μmg/√(1+μ²).

Q6. Direction of kinetic friction.
A6. Opposite to the relative velocity of the surfaces in contact.

Q7. On a rough incline, condition for a block to remain at rest.
A7. mg sinθ ≤ μs mg cosθ, i.e. tanθ ≤ μs.

Q8. Maximum safe speed on a banked road of angle θ, radius r, friction μ.
A8. v² = rg(tanθ + μ)/(1 − μ tanθ).

Q9. Newton's third-law pair for a book resting on a table — is it book's weight and normal force?
A9. No. Weight (Earth on book) pairs with book on Earth; normal (table on book) pairs with book on table.

Q10. Two blocks in contact pushed by force F — find contact force.
A10. Common acceleration a = F/(m1+m2); contact force on the second block = m2·a.

Q11. Why analyse a wedge problem in the wedge's frame?
A11. Adding the pseudo-force removes the wedge's motion, so the block's constraint (staying on the surface) is simpler.

Q12. Tension in a massless string over a frictionless pulley connecting m1, m2 (Atwood).
A12. T = 2m1m2 g/(m1+m2); acceleration a = (m1−m2)g/(m1+m2).

Newton's laws look deceptively simple, yet the JEE-Advanced version buries them in pulley trains, wedges that are themselves free to slide, and friction that must be checked rather than assumed. The skill being tested is a disciplined free-body diagram (FBD) plus the right constraint.

Core concept: ΣF = ma applies to each body in an inertial frame; friction is a responsive force with a ceiling, not a fixed value.

The framework

Beginner — FBD discipline

Draw every real force on the body: weight, normal, tension, applied, friction. Never draw "ma" as a force — it is the result. Choose axes that make the acceleration lie along one axis (along the incline, not horizontal/vertical). For connected bodies, write ΣF = ma per body, then couple them with the constraint.

Intermediate — friction properly

Static friction adjusts to prevent sliding, up to a maximum f_s(max) = μ_s N. Kinetic friction is fixed at f_k = μ_k N and opposes relative sliding. The correct procedure: assume no sliding, compute the friction required, compare with μ_s N. If required > μ_s N, the body slides and you switch to μ_k N. Skipping this check is the number-one error.

For a block on an incline angle θ, it stays put if tanθ ≤ μ_s. The angle of repose is θ = arctan μ_s.

Advanced — pseudo-forces and free wedges

In a non-inertial frame accelerating with a, add a pseudo-force −ma on every body; then treat the frame as if inertial. This turns "block on an accelerating wedge" into a static problem in the wedge's frame. For a wedge free to slide, you have two unknown accelerations (block relative to wedge, and wedge itself) plus the constraint that the block stays on the wedge surface. Solve in the ground frame with a constraint linking the two, or in the wedge frame with pseudo-forces — either works, but be consistent.

Key subtlety: the normal force is not mg cosθ when the wedge accelerates. You must solve the coupled equations.

Worked example

A block of mass m rests on a frictionless wedge of mass M and angle θ, the wedge free to slide on a frictionless floor. Find the wedge's acceleration A. Let the block's acceleration relative to the wedge be a_r down the incline. Ground-frame constraint: horizontal momentum is conserved (no external horizontal force), so m(A + a_r cosθ)... solving the coupled Newton equations gives A = (m g sinθ cosθ)/(M + m sin²θ). Check limits: M → ∞ gives A → 0 (fixed wedge), as expected. This limit check is exactly what earns partial credit.

How JEE Advanced tests this

Two- or three-block pulley systems asking for tension and acceleration; "will it slide?" static-friction threshold questions; blocks on accelerating wedges/lifts; and problems where you must decide the direction of friction (e.g., a block on a wedge being pushed).

Exam tricks & shortcuts

• For an inclined plane, always resolve gravity into g sinθ (along) and g cosθ (perpendicular).
• Pulley constraint: total string length constant → sum of accelerations (with correct signs) relates them.
• Mnemonic: "Draw, resolve, constrain, solve" — in that order, every time.

Key formulas

• Newton's second law: F⃗ = dp⃗/dt = ma⃗ (constant m).
• Impulse: J = FΔt = Δp.
• Friction: f_s ≤ μ_sN; f_k = μ_kN; angle of repose tanθ = μ_s.
• Incline (rough): a = g(sinθ − μcosθ).
• Circular (banking): v = √(rg tanθ) (frictionless); v_max = √(μrg) (level road).
• Vertical circle top: v_min = √(gr).

Draw free-body diagrams and resolve along/perpendicular to motion; friction opposes relative sliding tendency.

Worked Example

Problem: A block A of mass 4 kg rests on a horizontal table (coefficient of kinetic friction μ = 0.25) and is connected by a light inextensible string over a frictionless pulley at the table edge to a hanging block B of mass 2 kg. Find the acceleration of the system and the tension in the string. (g = 10 m/s².)

Solution:
Friction on A opposes its motion:
f = μ m_A g = 0.25 × 4 × 10 = 10 N.

The hanging weight drives the system:
Driving force = m_B g = 2 × 10 = 20 N.

Since 20 N > 10 N, the system moves. Write Newton's second law for each block (a = common acceleration, T = tension):

• Block A (horizontal): T − f = m_A a → T − 10 = 4a
• Block B (vertical): m_B g − T = m_B a → 20 − T = 2a

Add the two equations:
20 − 10 = (4 + 2)a → 10 = 6a → a = 5/3 ≈ 1.67 m/s².

Tension from block B's equation:
T = 20 − 2a = 20 − 2(5/3) = 20 − 10/3 = 50/3 ≈ 16.7 N.

Answer: a ≈ 1.67 m/s²; T ≈ 16.7 N.