Draw interactions first, then test the contact regime: Newton’s laws, friction and circular dynamics.
JEE Main · JEE Advanced · NEET UG75 min study guide9 indexed sections
BY THE END, YOU CAN
Build consistent free-body diagrams and distinguish force pairs from balanced forces.
Solve connected-body and equilibrium problems with explicit constraints.
Determine static versus kinetic friction and its direction.
Apply radial and tangential dynamics with contact and tension limits.
BEFORE YOU START
Vector resolution and signed acceleration.
Basic kinematics; conservation of mechanical energy for the vertical-circle section.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Main 2015. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Matched topics: Friction; Newton's Laws and Equilibrium.
JEE Main · 2015 · 2015-04-04 · Offline Paper 1 · Code A · Physics · printed Q3
Paraphrased exam question
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
A horizontal force presses block A (weight 20 N) against B (weight 100 N), which touches a vertical wall. Friction coefficients are 0.1 between blocks and 0.15 at the wall. Under the intended stationary-block model, find the wall’s friction on B.
Reveal answer & worked solution
Answer: 120 N upward (independently derived; source option 3).
Treat A+B as one system. Their contact friction is internal and cancels, while total downward weight is 20+100=120 N.
Vertical equilibrium therefore requires the wall to exert 120 N upward. Static friction takes the needed value, not automatically μN.
This equilibrium needs sufficient push: 0.15F≥120 gives F≥800 N; then the A–B condition 0.1F≥20 is also satisfied.
Source: JEE (Main) 2015, 2015-04-04, Code A, Q3; original booklet scan archived by Resonance; independently derived solution; final-key status unverified. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. Newton’s laws begin with a system boundary
Choose the body or collection of bodies whose motion you want to explain. Draw only forces exerted on that system by its surroundings. A force is an interaction with an identifiable agent: Earth supplies weight, a surface supplies contact forces, and a taut rope supplies tension. Velocity and acceleration are descriptions of motion, not extra forces to add to the diagram.
Newton’s first law identifies inertial frames: in such a frame, a body with zero resultant external force maintains constant velocity. Rest is only one possibility. Newton’s second law relates the resultant to the rate of change of momentum; for constant mass it becomes ΣF = ma. Components of this vector equation apply simultaneously. Zero acceleration along one axis means zero net force along that axis, even when another component is nonzero.
Newton’s third law pairs forces from the same interaction acting on different bodies. The force of a table on a book and the force of the book on the table form a pair. The book’s weight and the table’s upward normal act on the same book, so they are not a third-law pair even when equal. Third-law partners cancel when summing forces over a system containing both bodies, but they never cancel within one body’s free-body diagram.
Mass measures inertia and is unchanged by taking an object into an accelerating lift. Weight mg is a gravitational force and can change with local g. Apparent weight is often the supporting normal force measured by a scale. Keep these meanings distinct instead of using “weight” for every downward or supporting force.
ΣFext = dp/dt; for constant mass, ΣFext = ma
Apply in an inertial frame to the stated system. Variable-mass problems require careful momentum-flux accounting.
ΣFₓ = maₓ; ΣFᵧ = maᵧ
A two-dimensional force balance is two equations, not an equality of force magnitudes.
02 / UNDERSTAND & APPLY
2. Equilibrium, normal reactions and ideal tensions
Translational equilibrium means zero net force, whether the body is at rest or moving at constant velocity. For an extended rigid body, full mechanical equilibrium also requires zero net torque. In particle models the rotational condition is not separately needed. Three coplanar forces can balance only if their vector sum closes; equal magnitudes alone do not guarantee balance.
A normal reaction is perpendicular to the local contact surface. Its magnitude is determined by the no-penetration constraint and the other forces. It is not always mg. On a fixed smooth incline it equals mg cosθ when acceleration perpendicular to the plane is zero. An upward pull can reduce it; downward pushing can increase it. An ordinary passive surface can push but cannot pull, so a calculated negative normal signals loss of contact or a wrong assumed regime.
An ideal massless string carries equal tension along an uninterrupted segment if it passes over ideal massless frictionless pulleys. A massive pulley may require unequal tensions to provide rotational torque. A string can pull but cannot push; if an equation demands negative tension, check whether the string goes slack. These physical inequalities are part of the solution.
A useful force sketch for a block on an incline contains mg vertically down, N perpendicular outward and any friction along the plane. Resolve mg into mg sinθ down the plane and mg cosθ into it. Do not rotate the weight arrow itself and then add its components as additional forces; components replace the vector in calculations.
ORIGINAL WORKED EXAMPLE
A scale in a lift
A 60 kg person stands on a scale in a lift accelerating upward at 2 m s⁻². Use g = 10 m s⁻². What does the scale measure?
Take upward positive. Forces on the person are upward normal N and downward weight 600 N.
Newton’s law gives N − 600 = 60 × 2.
Thus N = 720 N. The person’s gravitational weight is still 600 N; a kilogram-calibrated scale would indicate 72 kg under its usual calibration.
Answer Supporting force 720 N.
03 / UNDERSTAND & APPLY
3. Connect bodies without guessing every force at once
For connected objects, first write the geometric constraint. A taut inextensible string has fixed length. For a simple fixed pulley, one end rising by a distance requires the other to fall by the same distance, so their acceleration magnitudes match. A movable pulley can give a factor of two: the supporting string portions both change when the pulley moves. Equal tensions do not imply equal accelerations in every pulley arrangement.
Then decide whether to treat the bodies together or separately. A combined system equation eliminates internal tensions and contact forces, which is efficient for finding overall acceleration. A separate body equation then recovers the force transmitted between bodies. The sign of each acceleration component should follow the chosen constraint, not an intuition that all objects move in the same direction.
For a string joining masses across a fixed ideal pulley, each mass has weight and tension. Taking the anticipated motion as positive separately on each side is allowed if the coordinate choices are stated. If the resulting acceleration is negative, the motion accelerates opposite to the guess. This is information, not an automatic algebraic failure.
ORIGINAL WORKED EXAMPLE
An ideal Atwood system
Masses 3 kg and 2 kg hang from a light string over a fixed ideal pulley. Find acceleration and tension using g = 10 m s⁻².
For the 3 kg side downward positive: 30 − T = 3a. For the 2 kg side upward positive: T − 20 = 2a.
Add the equations to eliminate tension: 10 = 5a, so a = 2 m s⁻².
Substitute into the lighter side: T = 20 + 4 = 24 N. This lies between the two weights, consistent with their accelerations.
Answer Acceleration 2 m s⁻²; tension 24 N.
ORIGINAL WORKED EXAMPLE
Internal contact force
Two touching blocks of 2 kg and 3 kg lie on a smooth horizontal surface. A 20 N horizontal force pushes the 2 kg block toward the 3 kg block. Find their acceleration and contact force.
For both blocks together, external horizontal force is 20 N and total mass is 5 kg, giving a = 4 m s⁻².
For the 3 kg block alone, contact is its only horizontal force: C = 3 × 4 = 12 N.
The force on the 2 kg block from the 3 kg block is 12 N opposite the push. Its net force is 20 − 12 = 8 N = 2 × 4.
Answer Acceleration 4 m s⁻²; contact force 12 N.
04 / UNDERSTAND & APPLY
4. Static friction is a response, not a fixed value
Dry friction opposes relative sliding or the tendency toward relative sliding at the contact. It does not invariably oppose the centre-of-mass velocity. A moving belt can accelerate a stationary block forward by friction, and the road can exert forward static friction on a driven wheel. To determine direction, imagine the contact without friction and identify how the two surfaces would slip relative to one another.
While there is no slipping, static friction adjusts to whatever magnitude is required up to a limiting value μₛN. The equation fₛ = μₛN applies at impending slip, not to every resting body. A block under a small horizontal push can remain at rest with friction exactly equal to that push. If the required value exceeds the limit, the assumed no-slip state is impossible.
In the elementary kinetic model, sliding friction has magnitude μₖN and points opposite relative sliding. The coefficients are empirical approximations for specified surfaces and conditions; they are not universal constants independent of speed, wear or temperature. In many simple cases μₖ is less than μₛ, which explains why starting motion can require more force than maintaining it.
Use a regime test: assume rest or no slip, calculate required static friction, and check |f| ≤ μₛN. Only if that fails should you solve the sliding regime with the appropriate kinetic coefficient. If a sliding body slows to rest, retest the static condition rather than extending a kinetic-friction equation through zero speed indefinitely.
|fₛ| ≤ μₛN; |fₖ| = μₖN
Static friction obeys a bound; kinetic friction is the stated idealised sliding model. Both require contact.
tanθrepose = μₛ
For a block just about to slide down an otherwise unforced fixed incline.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Force directions are geometric; arrow lengths are not a numerical force triangle. Static equilibrium also requires the maximum-static-friction inequality. The shaded support is not part of the block’s free-body system.
ORIGINAL WORKED EXAMPLE
Test the static regime first
A 5 kg block rests on a horizontal floor with μₛ = 0.4 and μₖ = 0.3. A horizontal pull is first 15 N and then 25 N. Use g = 10.
Normal force is 50 N, so maximum static friction is 0.4 × 50 = 20 N.
At 15 N, static friction can match the pull: f = 15 N and acceleration is zero.
At 25 N, the static bound is exceeded. During sliding, kinetic friction is 15 N, so net force is 10 N and acceleration is 2 m s⁻².
Answer 15 N pull: rest with 15 N friction. 25 N pull: sliding acceleration 2 m s⁻².
05 / UNDERSTAND & APPLY
5. Friction, slopes and angled pulling interact through the normal force
For a block on a fixed incline, gravity’s down-slope component is mg sinθ. If it remains at rest without other forces, friction must act up the slope with that magnitude. The normal is mg cosθ, so equilibrium requires tanθ ≤ μₛ. The mass cancels because both the driving force and limiting friction scale with mass.
For a block sliding down the incline, kinetic friction is up the slope and a = g(sinθ − μₖ cosθ). If it is sliding up, both gravity and kinetic friction point down the slope, so the acceleration component along an uphill-positive axis is −g(sinθ + μₖ cosθ). A formula without a specified motion direction can conceal the wrong friction sign.
On a horizontal surface, a pull F at angle α above horizontal gives N = mg − F sinα while contact and zero vertical acceleration persist. The horizontal driving component is F cosα. Thus increasing the pull angle trades a reduced horizontal component against reduced friction. A downward push produces the opposite change in N. Always verify N ≥ 0.
ORIGINAL WORKED EXAMPLE
An angled pull at impending motion
A 10 kg crate is pulled at 37° above horizontal with μₛ = 0.5. Use g = 10, sin37° = 0.6 and cos37° = 0.8. Find the pull for impending sliding.
At the threshold, N = 100 − 0.6F and horizontal balance requires 0.8F = 0.5N.
Substitute: 0.8F = 50 − 0.3F, so 1.1F = 50 and F ≈ 45.5 N.
The normal is about 72.7 N, positive, so the contact assumption is consistent.
Answer Approximately 45.5 N.
06 / UNDERSTAND & APPLY
6. Centripetal describes the net radial requirement
A body moving on a circle of radius r at speed v has inward radial acceleration v²/r. Consequently the inward resultant force must be mv²/r. “Centripetal force” is not a new interaction to add beside tension, gravity and friction. It names the inward resultant of those actual forces. Adding a separate centripetal arrow double-counts the physics.
Choose inward radial positive and sum the radial components. Separately apply the tangential equation to determine changes of speed. A horizontal turn at constant speed needs an inward force even though tangential acceleration is zero. A vertical circle generally has changing speed because gravity has a tangential component away from the top and bottom.
On an unbanked level road, static friction supplies the lateral force for a car turning without skidding. Since N = mg, the friction bound yields v ≤ √(μₛgr). This ideal estimate assumes the full available static friction is used laterally; simultaneous braking or acceleration also consumes the contact’s friction capacity.
For a frictionless banked turn at angle θ, the normal has vertical component N cosθ = mg and inward horizontal component N sinθ = mv²/r. Dividing gives tanθ = v²/(rg). The design speed is independent of vehicle mass. At other speeds friction may point up or down the bank depending on the tendency to slip.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Front-view force schematic. The inward direction is left on the page. There is no additional centripetal-force arrow because centripetal force names the net radial force.
ORIGINAL WORKED EXAMPLE
A banked road without friction
A frictionless circular road of radius 50 m is banked so tanθ = 0.2. Find its design speed using g = 10 m s⁻².
Resolve the normal: N cosθ = mg and N sinθ = mv²/r.
Divide to eliminate both N and m: tanθ = v²/(rg).
Thus v² = 50 × 10 × 0.2 = 100, giving v = 10 m s⁻¹.
Answer 10 m s⁻¹.
07 / UNDERSTAND & APPLY
7. Check tension, contact and the observer
For a particle on a light string in a vertical circle, at the top both tension and weight point toward the centre, so Ttop + mg = mvtop²/r. A taut string needs Ttop ≥ 0; the limiting top speed is √(gr). At the bottom, tension points inward upward while weight points outward, so Tbottom − mg = mvbottom²/r. Speeds at top and bottom are related by energy when only gravity does work.
The often-used minimum bottom speed √(5gr) follows for a mass on a flexible string completing a full vertical circle without slack: energy from bottom to top requires vbottom² = vtop² + 4gr, and the limiting top condition supplies vtop² = gr. A rigid rod can push as well as pull, so it does not share that string constraint. Do not transfer a threshold between different physical supports.
You may also analyse a problem in a frame translating with acceleration A, but then an inertial or pseudo force −mA must be included to use a Newton-like balance in that frame. This force has no interaction partner; it arises from the accelerating coordinates. In an inertial ground frame, omit it and retain the actual acceleration. Mixing the two frame descriptions counts acceleration twice.
ORIGINAL WORKED EXAMPLE
A vertical circle with an explicit constraint
A 0.5 kg mass on a 1 m light string just completes a vertical circle. Use g = 10. Find minimum bottom speed and bottom tension.
At the limiting top point, tension is zero, so vtop² = gr = 10.
Energy conservation from bottom to top gives vbottom² = vtop² + 4gr = 50.
At the bottom T − mg = mvbottom²/r, so T = 5 + 0.5 × 50 = 30 N.
Answer Minimum bottom speed √50 ≈ 7.07 m s⁻¹; bottom tension 30 N.
08 / UNDERSTAND & APPLY
8. Original practice: identify the physical regime
Use g = 10 m s⁻² throughout. These are original problems with idealisations stated or implied by the lesson models.
ORIGINAL PRACTICE · 1
A 4 kg body moves at constant velocity on a smooth horizontal plane. What is the net force?
Reveal answer & reasoning
Zero.
Constant vector velocity means zero acceleration. Motion itself does not require a resultant force.
ORIGINAL PRACTICE · 2
A 2 kg block is released from rest on a fixed smooth 60° incline. Find normal force and down-slope acceleration.
Reveal answer & reasoning
10 N and 5√3 m s⁻².
N = mg cos60° = 10 N. The unbalanced along-plane force is mg sin60°, yielding a = g sin60°.
ORIGINAL PRACTICE · 3
A 50 kg person accelerates downward in a lift at 3 m s⁻². Find scale force.
Reveal answer & reasoning
350 N.
With upward positive, N − 500 = 50(−3).
ORIGINAL PRACTICE · 4
A 2 kg block with μₛ = 0.5 has a 6 N horizontal force applied while resting on a level floor. Find friction.
Reveal answer & reasoning
6 N opposite the force.
Maximum static friction is 10 N. The actual value only needs to balance 6 N.
ORIGINAL PRACTICE · 5
A block just begins sliding on an incline at 45°. Find μₛ.
Reveal answer & reasoning
1.
At impending motion mg sinθ = μₛmg cosθ, so μₛ = tan45°.
ORIGINAL PRACTICE · 6
A 1 kg block slides down a 37° incline with μₖ = 0.25. Use sin37° = 0.6 and cos37° = 0.8. Find acceleration.
Reveal answer & reasoning
4 m s⁻² down the incline.
Gravity contributes 6 N down; friction is 0.25 × 8 = 2 N up. Net force is 4 N.
ORIGINAL PRACTICE · 7
A car turns on a level road of radius 40 m with μₛ = 0.4. Find ideal maximum no-skid speed.
Reveal answer & reasoning
√160 ≈ 12.6 m s⁻¹.
The lateral requirement mv²/r cannot exceed μₛmg. Mass cancels.
ORIGINAL PRACTICE · 8
At the top of a vertical circle, a string-supported mass has speed √(2gr). Find its tension.
Reveal answer & reasoning
mg.
Inward positive gives T + mg = m(2gr)/r = 2mg, so T = mg.
ORIGINAL PRACTICE · 9
Two blocks of 1 and 4 kg are pushed on a smooth floor by a 15 N force applied to the 4 kg block. Find contact force on the 1 kg block.
Reveal answer & reasoning
3 N.
Combined acceleration is 15/5 = 3 m s⁻²; the contact force accelerates the 1 kg block.
ORIGINAL PRACTICE · 10
Why can equal and opposite third-law forces fail to keep either body at rest?
Reveal answer & reasoning
They act on different bodies.
Each body accelerates according to the resultant of forces on that body. The interaction pair cancels only in the combined-system force sum.
09 / UNDERSTAND & APPLY
9. Finish by testing the assumptions
A correct algebraic solution can describe an impossible contact regime. Check normal forces and tensions for nonnegative values, static friction against its limiting magnitude, and the direction of kinetic friction against relative sliding. If a condition fails, change the physical model and solve again rather than accepting an unphysical negative contact force.
For circular motion, identify the actual radial force providers and keep the tangential equation separate. For connected bodies, write a string-length constraint before assuming accelerations. For equilibrium, distinguish a particle balance from full rigid-body force-and-torque balance. The force diagram is the reasoning; the equations are its compact record.
Every real force must have an agent outside the chosen system.
Normal force is determined, not automatically mg.
Static friction is bounded and self-adjusting.
Centripetal force is a resultant, not an additional force.
State the frame before adding or omitting pseudo forces.
Third-law partners act on different bodies.
KEEP THE BIG IDEAS
Quick revision
A system boundary decides which forces are external.
Friction and contact require regime checks, not automatic substitution.
Radial acceleration is produced by the sum of actual radial force components.
Constraints and frame choice must remain consistent from diagram to answer.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.