JEE Mains Physics · Laws of Motion

Lifts, Pseudo Forces and Circular Motion

In an accelerating frame, add a pseudo force −ma₀ to every body and solve as if the frame were at rest; in circular motion, the real forces must supply mv²/r toward the centre.

All Laws of Motion notes JEE Mains Physics strategy

Why this matters

Eighteen PYQs, fifteen of them multiple choice, and five from 2026. Seven are about apparent weight, in a lift or on a platform that sinks. Five work in an accelerating frame: a wedge pulled sideways, a bob in a car accelerating up a slope, a block held against the side of a moving cube. Six are circular motion: a pendulum in a turning car, a banked road, a coin on a turntable, a rotor drum.

Concept 1 of 3: Apparent weight in a lift

A weighing scale reads the normal reaction, the push of the floor on your feet. If the lift accelerates upward, the floor must push harder than your weight to speed you up with it; if it accelerates downward, it pushes less. Only the direction of the acceleration matters. A lift moving at constant speed, up or down, reads your true weight.

Definition

The reading is the normal reaction N.
Acceleration up: $N = m(g + a)$ . Acceleration down: $N = m(g - a)$ . Constant velocity: $N = mg$ . Free fall: $N = 0$ .
Speeding up while going down, or slowing down while going up, is a downward acceleration.
Every body on a platform accelerating down at a feels an effective gravity $g - a$ ; a stack presses on what is below it with its mass times $g - a$ .
A lift moving at constant velocity is an inertial frame: an incline inside it behaves exactly as on the ground.

Lift's motion	Acceleration	Scale reading	For 50 kg, a = 2 m/s², g = 10 m/s²
At rest	Zero	$mg$	500 N
Moving up or down at constant speed	Zero	$mg$	500 N
Starting upward, speeding up	Upward	$m(g + a)$	600 N
Moving down and slowing to a stop	Upward	$m(g + a)$	600 N Moving down, yet the reading goes UP: the acceleration points up.
Starting downward, speeding up	Downward	$m(g - a)$	400 N
Moving up and slowing to a stop	Downward	$m(g - a)$	400 N
Cable snaps (free fall)	Downward, equal to g	Zero	0 N

The reading depends only on the direction of the acceleration, never on the direction of motion.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 25 · Q30Moderate

Example 1 · Laws of Motion · Lifts, Pseudo Forces and Circular Motion

A person standing on a spring balance inside a stationary lift measures

60\text{ }kg

. The weight of that person if the lift descends with uniform downward acceleration of

1.8\text{ }m/s^{2}

will be ___ N.

\left\lbrack g= 10\text{ }m/s^{2} \right\rbrack

Velocity does not decide the reading

A lift moving down can read more than your weight, if it is slowing down. Ask which way the acceleration points, then add or subtract a.

A lift at constant velocity changes nothing

Uniform motion adds no pseudo force. An incline or a pendulum inside such a lift behaves exactly as on the ground, so time and acceleration are the ground values.

Drill 6 more on apparent weight in a lift

Concept 2 of 3: Pseudo force in an accelerating frame

Inside an accelerating car, a hanging bob swings back as if pushed. In the car's frame that push is the pseudo force, −ma₀, opposite to the car's acceleration. Add it to every body, and the problem becomes ordinary statics or an ordinary incline problem. Use it only in the accelerating frame; in the ground frame it does not exist.

Definition

In a frame accelerating at $\vec a_0$ , add $-m\vec a_0$ to every body.
Bob in a vehicle accelerating at a on a level road: the string makes $\tan\theta = a/g$ with the vertical, and $T = m\sqrt{g^{2} + a^{2}}$ .
Block on a smooth wedge accelerating horizontally at $a_0$ : along the slope $a_{\text{rel}} = g\sin\theta \mp a_0\cos\theta$ (minus when the pseudo force points up the slope). It stays at rest when $a_0 = g\tan\theta$ .
A free wedge on a smooth floor recoils as the block slides: solve the block and the wedge together; their momentum along the floor is conserved.
A block pressed on the front face of an accelerating body: $N = ma_0$ , and it does not slip if $\mu ma_0 \ge mg$ .

Pseudo force

\vec F_{\text{pseudo}} = -m\vec a_0, \qquad \tan\theta = \frac{a_0}{g}

Worked example

A van accelerates at

5\ \text{m/s}^{2}

on a level road. A 0.2 kg bob hangs from its roof. Find the angle of the string with the vertical and its tension. (g = 10 m/s²)

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 22 · Q30Moderate

Example 2 · Laws of Motion · Lifts, Pseudo Forces and Circular Motion

A car is moving on a plane inclined at

30^{\circ}

to the horizontal with an acceleration of

10{\text{ }ms}^{- 2}

parallel to the plane upward. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is (Take

g = 10{\text{ }ms}^{- 2}

)

The pseudo force points against the acceleration

A bob in a car speeding up forward swings BACK. Putting −ma₀ along the acceleration instead of against it reverses the answer, often to the other sign choice among the options.

Use it in one frame only

Either work from the ground with the real acceleration, or from the accelerating frame with the pseudo force. Doing both counts ma₀ twice.

Drill 4 more on pseudo force in an accelerating frame

Concept 3 of 3: Forces in circular motion

Moving in a circle needs a net force toward the centre, mv²/r. It is not a new force: friction, the normal reaction or a tension supplies it. On a flat road friction does all of it; on a banked road the tilted normal reaction does part of it. In the rotating frame the same balance appears as a centrifugal force, mω²r outward.

Definition

Net inward force $= \dfrac{mv^{2}}{r} = m\omega^{2}r$ , supplied by real forces.
Flat road: $\mu mg \ge \dfrac{mv^{2}}{r}$ , so $v_{\max} = \sqrt{\mu rg}$ .
Banked road, no friction: $v^{2} = rg\tan\theta$ . With friction: $v_{\max}^{2} = rg\,\dfrac{\tan\theta + \mu}{1 - \mu\tan\theta}$ .
Pendulum in a turning car: $\tan\theta = \dfrac{v^{2}}{rg}$ with the vertical.
Coin on a turntable: slips beyond $r = \dfrac{\mu g}{\omega^{2}}$ . Rotor drum of radius R: needs $\mu \ge \dfrac{g}{\omega^{2}R}$ .
Rotating frame: add the centrifugal force $m\omega^{2}r$ outward. A ball in a smooth radial groove: $v^{2} = \omega^{2}(r_2^{2} - r_1^{2})$ .

Banked road with friction

v_{\max}^{2} = rg\,\frac{\tan\theta + \mu}{1 - \mu\tan\theta}

Worked example

A road of radius 100 m is banked at

37^{\circ}

(

\tan 37^{\circ} = 0.75

) and the coefficient of friction is 0.5. Find the maximum safe speed, and the speed at which no friction is needed. (g = 10 m/s²)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q10Moderate

Example 3 · Laws of Motion · Lifts, Pseudo Forces and Circular Motion

A car of mass '

m

' moves on a banked road having radius ' r ' and banking angle

\theta

. To avoid slipping from banked road, the maximum permissible speed of the car is

v_{0}

. The coefficient of friction

\mu

between the wheels of the car and the banked road is :-

μ = v²/(rg) only on a flat road

On a banked road the normal reaction supplies part of the inward force, so friction needs to supply less. Using the flat-road formula overstates μ.

Convert km/h before squaring

54 km/h is 15 m/s (multiply by 5/18). Squaring 54 instead gives an answer about 13 times too large.

Centrifugal force belongs to the rotating frame

From the ground, a body in a circle has an unbalanced inward force and no outward one. Add mω²r outward only when working in the rotating frame.

Drill 5 more on forces in circular motion

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Pseudo force in an accelerating frame
Pseudo force
$\vec F_{\text{pseudo}} = -m\vec a_0, \qquad \tan\theta = \frac{a_0}{g}$
Forces in circular motion
Banked road with friction
$v_{\max}^{2} = rg\,\frac{\tan\theta + \mu}{1 - \mu\tan\theta}$

Reference tables (1)

Apparent weight in a lift7 rows

Lift's motion	Acceleration	Scale reading	For 50 kg, a = 2 m/s², g = 10 m/s²
At rest	Zero	$mg$	500 N
Moving up or down at constant speed	Zero	$mg$	500 N
Starting upward, speeding up	Upward	$m(g + a)$	600 N
Moving down and slowing to a stop	Upward	$m(g + a)$	600 N Moving down, yet the reading goes UP: the acceleration points up.
Starting downward, speeding up	Downward	$m(g - a)$	400 N
Moving up and slowing to a stop	Downward	$m(g - a)$	400 N
Cable snaps (free fall)	Downward, equal to g	Zero	0 N

The reading depends only on the direction of the acceleration, never on the direction of motion.

Watch out for (7)

Velocity does not decide the reading
→ Apparent weight in a lift
A lift at constant velocity changes nothing
→ Apparent weight in a lift
The pseudo force points against the acceleration
→ Pseudo force in an accelerating frame
Use it in one frame only
→ Pseudo force in an accelerating frame
μ = v²/(rg) only on a flat road
→ Forces in circular motion
Convert km/h before squaring
→ Forces in circular motion
Centrifugal force belongs to the rotating frame
→ Forces in circular motion

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