In short

When you observe motion from an accelerating (non-inertial) frame, Newton's second law appears to fail — objects accelerate without any visible force acting on them. To fix this, you add a pseudo force \vec{F}_{\text{pseudo}} = -m\vec{a}_{\text{frame}} on every object, where \vec{a}_{\text{frame}} is the acceleration of the frame itself. This is not a real force — no physical agent produces it — but it makes F = ma work perfectly inside the accelerating frame. In a lift accelerating upward, your apparent weight increases; in free fall, it drops to zero; in an accelerating car, a hanging pendulum deflects backward.

You are standing in a lift on the 40th floor of a Mumbai highrise. The lift begins to descend — and for a brief moment, your stomach lurches upward, your knees feel light, and the bag in your hand feels like it lost half its weight. Nothing changed about gravity. The Earth did not suddenly pull you less. What changed is that the floor beneath your feet accelerated away from you, and your body — still obeying Newton's laws in the ground frame — has not caught up yet.

Now here is the question that makes this topic interesting: can you do physics inside the lift? Can you write F = ma for objects inside the lift, using the lift as your reference frame? The answer is yes — but only if you add a fictitious force that accounts for the fact that your frame itself is accelerating. That fictitious force is called a pseudo force, and it is one of the most useful problem-solving tools in mechanics.

Why Newton's laws break in accelerating frames

Stand on the platform at a Delhi Metro station and watch a train accelerate away. Inside the train, a passenger places a cricket ball on the smooth floor. From the platform (an inertial frame), the ball stays put — no horizontal force acts on it, so it does not accelerate. But from inside the train, the ball rolls backward even though nothing is pushing it. Newton's first law — an object at rest stays at rest unless a force acts on it — seems to fail.

It has not failed. The law is correct, but it was written for inertial frames — frames that are not accelerating. The train is accelerating forward, so the train is not an inertial frame. Newton's laws, in their standard form, do not apply inside it.

You have two choices:

  1. Work in an inertial frame (the platform). Newton's laws apply directly. The ball is stationary; the train floor slides forward under it. No mystery.
  2. Work in the accelerating frame (the train). To make Newton's laws work here, you must add a pseudo force on every object: \vec{F}_{\text{pseudo}} = -m\vec{a}_{\text{frame}}.

The second choice is often easier — especially when the problem involves objects that are stationary relative to the accelerating frame (a person standing in a lift, a pendulum hanging in a car). In the accelerating frame, these objects are in equilibrium, and equilibrium problems are simpler than acceleration problems.

The pseudo force — definition and direction

When a reference frame accelerates with acceleration \vec{a}_{\text{frame}}, every object of mass m inside that frame experiences an additional force:

\vec{F}_{\text{pseudo}} = -m\,\vec{a}_{\text{frame}}

Why the minus sign: the pseudo force always points opposite to the frame's acceleration. If the lift accelerates upward, the pseudo force on you points downward. If the car accelerates forward, the pseudo force on the pendulum bob points backward. This is why you feel "pushed back" when an autorickshaw accelerates — the pseudo force in the autorickshaw's frame acts backward on you.

Three things to remember about pseudo forces:

  1. They are not real. No physical agent — no rope, no spring, no gravitational field — produces them. They exist only because you chose an accelerating frame. An observer in an inertial frame sees no such force.

  2. They act on every object. Unlike gravity (which acts on mass) or tension (which acts through a string), the pseudo force applies to everything in the accelerating frame — every person, every ball, every molecule of air.

  3. They are proportional to mass. F_{\text{pseudo}} = ma_{\text{frame}}. This is the same dependence as gravitational force (F = mg), which is why pseudo forces feel like gravity. In a lift accelerating upward, you feel heavier — as if gravity increased. In free fall, you feel weightless — as if gravity vanished. The pseudo force is indistinguishable from a gravitational field, an insight that Einstein turned into general relativity.

Block in an accelerating lift — apparent weight

The classic application. You stand on a bathroom scale inside a lift. The scale does not measure your weight — it measures the normal force N the floor exerts on you. This normal force is what you feel as your weight, and it is called your apparent weight.

Free body diagrams for a person in a lift: stationary, accelerating up, accelerating down, and free fall Four FBDs showing the person as a dot with weight mg downward and normal force N upward. N is greater than mg when accelerating up, less than mg when accelerating down, and zero in free fall. Stationary mg N = mg a = 0 You feel normal Accel. up mg N > mg a ↑ You feel heavier Accel. down mg N < mg a ↓ You feel lighter Free fall mg N = 0 a = g ↓ Weightless
Four scenarios for a person in a lift. The normal force $N$ (what the scale reads) changes depending on the lift's acceleration. In free fall, $N = 0$ — you are weightless.

Deriving apparent weight — the inertial frame approach

Take upward as positive. In the ground frame (inertial), apply Newton's second law to the person of mass m:

N - mg = ma

Why: two forces act on the person — the normal force N upward and weight mg downward. The net force equals mass times the person's acceleration a, which equals the lift's acceleration (since the person moves with the lift).

Solving for N:

N = m(g + a) \tag{1}

This is the apparent weight. Now examine three cases:

Case 1: Lift accelerates upward (a > 0).

N = m(g + a) > mg

The scale reads more than your true weight. You feel heavier. If a = 3 m/s² and m = 60 kg:

N = 60(9.8 + 3) = 60 \times 12.8 = 768 \text{ N}

Your true weight is 60 \times 9.8 = 588 N, but the scale reads 768 N — about 31% heavier.

Case 2: Lift accelerates downward (a < 0, so substitute a = -|a|).

N = m(g - |a|) < mg

The scale reads less than your true weight. You feel lighter. If the lift decelerates while going up (or accelerates downward) at 3 m/s²:

N = 60(9.8 - 3) = 60 \times 6.8 = 408 \text{ N}

You feel about 31% lighter.

Case 3: Free fall (a = -g, i.e., the lift falls with acceleration g downward).

N = m(g - g) = 0

The scale reads zero. You are weightless. The floor exerts no force on you. If you let go of your phone, it does not fall to the floor — it floats beside you, because both you and the phone are falling at exactly the same rate.

The same result from the accelerating frame

Now do the same analysis from inside the lift (the non-inertial frame). In this frame, the person is stationary — not accelerating. But to use F = ma = 0 (equilibrium), you must include the pseudo force.

The lift accelerates upward at a. The pseudo force on the person is F_{\text{pseudo}} = -ma (downward, opposing the frame's upward acceleration).

Forces on the person in the lift frame:

Equilibrium: N - mg - ma = 0

N = m(g + a)

The same result as equation (1). The two approaches always agree — that is the whole point of pseudo forces.

Weightlessness — what it really means

When the lift cable snaps and the lift falls freely, every object inside accelerates at g downward. In the lift's frame, the pseudo force is mg upward — exactly cancelling gravity. The effective gravity inside the lift is zero.

This is not just a thought experiment. ISRO's Chandrayaan spacecraft, coasting between the Earth and the Moon with its engines off, is in free fall — everything inside it is weightless. The astronauts, their tools, drops of water — all float. Gravity has not disappeared (the Earth is still pulling), but the spacecraft and everything in it are falling together, so there is no relative acceleration between objects inside. No relative acceleration means no normal forces, no tension in strings, no pressure on floors. Weightlessness.

The same principle explains the brief weightlessness you feel on a roller coaster at the top of a loop, or in the "Giant Frisbee" ride at an amusement park, or during the split second when a trampoline athlete reaches the peak of a jump. In each case, you and the reference frame (the cart, the seat, your own body) are in free fall for a moment.

Pendulum in an accelerating car

A small brass pendulum hangs from the rearview mirror of a car in Delhi. When the car accelerates forward, the bob swings backward — opposite to the car's acceleration — and settles at an angle \theta to the vertical. Why? And what determines the angle?

Pendulum bob deflecting in an accelerating car — two frames Left: ground frame view showing the bob's circular arc path with net force providing centripetal acceleration. Right: car frame showing the bob in equilibrium under tension, weight, and pseudo force. a θ FBD in car's frame T mg ma (pseudo) x y
Left: the car accelerates forward; the pendulum bob deflects backward at angle $\theta$. Right: FBD of the bob in the car's (non-inertial) frame. Three forces act — tension $T$, weight $mg$, and the pseudo force $ma$ backward. The bob is in equilibrium.

Analysis in the car's frame (non-inertial)

In the car's frame, the bob is stationary (it hangs at a fixed angle). Three forces act on it:

Since the bob is in equilibrium, resolve forces along two perpendicular directions.

Horizontal equilibrium:

T\sin\theta = ma \tag{2}

Why: the horizontal component of tension must balance the pseudo force. The pseudo force pulls the bob backward; the horizontal component of tension pulls it forward.

Vertical equilibrium:

T\cos\theta = mg \tag{3}

Why: the vertical component of tension must balance the weight. There is no vertical acceleration in the car's frame.

Finding the angle. Divide equation (2) by equation (3):

\frac{T\sin\theta}{T\cos\theta} = \frac{ma}{mg}
\tan\theta = \frac{a}{g} \tag{4}

Why: the tension T cancels, and you are left with a beautifully simple result. The deflection angle depends only on the ratio of the car's acceleration to gravitational acceleration. A larger acceleration means a larger angle.

\theta = \tan^{-1}\!\left(\frac{a}{g}\right)

Finding the tension. Square and add equations (2) and (3):

T^2\sin^2\theta + T^2\cos^2\theta = m^2a^2 + m^2g^2
T^2(\sin^2\theta + \cos^2\theta) = m^2(a^2 + g^2)
T = m\sqrt{a^2 + g^2} \tag{5}

Why: \sin^2\theta + \cos^2\theta = 1 simplifies the left side. The tension is always greater than mg — the string must support the bob against both gravity and the pseudo force.

The concept of effective gravity

Look at equation (4) again. The bob behaves as if gravity has been tilted — instead of pointing straight down, the effective gravitational field points at an angle \theta to the vertical, with magnitude:

g_{\text{eff}} = \sqrt{g^2 + a^2}

This is not a coincidence. In the car's frame, the pseudo force ma acts horizontally and the real gravitational force mg acts vertically. Their vector sum is the effective gravitational force mg_{\text{eff}}, pointing at angle \theta = \tan^{-1}(a/g) from the vertical. The pendulum hangs along this effective gravity direction, just as it would hang vertically in an inertial frame.

This insight generalises: in any accelerating frame, you can combine gravity and the pseudo force into a single effective gravitational field, and all equilibrium problems reduce to the familiar form — just with g replaced by g_{\text{eff}}.

Effective gravity in a rotating frame

The Earth itself is a rotating frame. A point on the equator moves in a circle of radius R = 6{,}371 km at angular speed \omega = 7.27 \times 10^{-5} rad/s (one full rotation in 24 hours). This means there is a centrifugal pseudo force directed outward (away from the Earth's axis) at the equator:

F_{\text{centrifugal}} = m\omega^2 R

The effective gravity at the equator is reduced:

g_{\text{eff}} = g - \omega^2 R

Why: at the equator, the centrifugal force points directly opposite to gravity (radially outward). So the effective gravitational acceleration is simply g minus the centrifugal acceleration. At the poles, the centrifugal force is zero (you are on the rotation axis), so g_{\text{eff}} = g.

Plugging in the numbers:

\omega^2 R = (7.27 \times 10^{-5})^2 \times 6.371 \times 10^6 = 0.0337 \text{ m/s}^2
g_{\text{eff}} = 9.8 - 0.034 \approx 9.766 \text{ m/s}^2

The difference is small — about 0.3% — but it is measurable. A precise weighing scale calibrated at the equator would read slightly differently if moved to Delhi (latitude ~28°). This is one reason why g varies slightly across the Earth's surface (the other being the Earth's non-spherical shape).

At a general latitude \lambda, the distance from the rotation axis is R\cos\lambda, and the centrifugal acceleration is \omega^2 R\cos\lambda. But this centrifugal force is not purely radial (not along the line from you to the Earth's centre) except at the equator and poles. At intermediate latitudes, it has a small component along the surface that deflects a plumb line slightly away from the geometric centre of the Earth. The full treatment involves vector decomposition, which you can explore in the Going Deeper section.

Solving problems by switching frames

The real power of pseudo forces is as a problem-solving strategy. For many problems, working in the accelerating frame turns an acceleration problem into an equilibrium problem — and equilibrium problems are easier.

When to use the accelerating frame:

When to stay in the inertial frame:

The recipe for the accelerating frame:

  1. Identify the non-inertial frame and its acceleration \vec{a}_{\text{frame}}.
  2. Draw the FBD of the object in this frame.
  3. Include all real forces (gravity, normal, tension, friction).
  4. Add the pseudo force -m\vec{a}_{\text{frame}} on the object.
  5. Since the object is at rest in this frame, set the net force equal to zero (equilibrium).
  6. Solve.

Both frames must give the same answer for any physical observable — the trajectory of a particle, the reading of a scale, the tension in a string. If they don't, you have made an error.

Worked examples

Example 1: Person in a Mumbai highrise lift

A 60 kg person stands on a weighing scale inside a lift. Find the scale reading (apparent weight) when: (a) the lift accelerates upward at 3 m/s², (b) the lift accelerates downward at 3 m/s², (c) the lift cable snaps and the lift falls freely. Take g = 9.8 m/s².

Free body diagrams for a 60 kg person in three lift scenarios Three FBDs side by side. Case (a): N = 768 N upward, mg = 588 N downward, net force upward. Case (b): N = 408 N upward, mg = 588 N downward, net force downward. Case (c): N = 0, mg = 588 N downward, free fall. (a) Accel. up, a = 3 m/s² LIFT 60 588 N 768 N a↑ Scale: 768 N (78.4 kgf) (b) Accel. down, a = 3 m/s² LIFT 60 588 N 408 N a↓ Scale: 408 N (41.6 kgf) (c) Free fall, a = g LIFT 60 588 N N = 0 a = g↓ Scale: 0 N Weightless!
FBDs for a 60 kg person in three lift scenarios. The scale reading equals the normal force $N$. In free fall, $N = 0$ — you float.

Given: m = 60 kg, g = 9.8 m/s².

True weight: W = mg = 60 \times 9.8 = 588 N.

(a) Lift accelerates upward at a = 3 m/s².

Step 1. Draw the FBD. Two forces act on the person: weight mg = 588 N downward, normal force N upward. The person accelerates upward with the lift.

Step 2. Apply Newton's second law (taking up as positive):

N - mg = ma
N = m(g + a) = 60(9.8 + 3) = 60 \times 12.8 = 768 \text{ N}

Why: the floor must push you up hard enough to both support your weight and accelerate you upward. The extra push is ma = 60 \times 3 = 180 N.

The scale reads 768 N, which is 768/9.8 \approx 78.4 kgf. You feel about 30% heavier.

(b) Lift accelerates downward at a = 3 m/s².

Step 3. Same FBD, but now the acceleration is downward. Taking up as positive, a = -3 m/s²:

N - mg = m(-3)
N = m(g - 3) = 60(9.8 - 3) = 60 \times 6.8 = 408 \text{ N}

Why: the floor does not need to push as hard — gravity is already accelerating you in the direction you need to go. The normal force only needs to make up the difference between g and the downward acceleration.

The scale reads 408 N, or about 41.6 kgf. You feel about 30% lighter.

(c) Lift in free fall (a = g = 9.8 m/s² downward).

Step 4. Now the lift accelerates downward at g:

N = m(g - g) = 0

Why: the lift floor falls at the same rate as you. It cannot push you. There is no contact force. You are weightless.

The scale reads 0 N. You float. If you release a coin from your hand, it does not fall to the floor — it hangs in mid-air beside you. This is the same condition as inside the Chandrayaan spacecraft in orbit: both the spacecraft and the astronaut are in free fall around the Earth.

Result:

Scenario Normal force N Scale reading Feeling
Accelerating up at 3 m/s² 768 N 78.4 kgf Heavier
Accelerating down at 3 m/s² 408 N 41.6 kgf Lighter
Free fall 0 N 0 kgf Weightless

What this shows: The normal force — not the gravitational force — determines what you feel. Gravity is always mg = 588 N. What changes is how hard the floor pushes back, and that depends entirely on the lift's acceleration.

Example 2: Pendulum in an accelerating car

A pendulum of mass 200 g hangs from the ceiling of a car. The car accelerates forward at a = 5 m/s² on a straight road. Find (a) the angle \theta the pendulum makes with the vertical, and (b) the tension in the string. Take g = 9.8 m/s².

FBD of pendulum bob in accelerating car frame The bob hangs at angle theta from the vertical. Three forces act: tension T along the string upward-right, weight mg straight down, and pseudo force ma to the left. The bob is in equilibrium in the car frame. m θ T mg ma (pseudo) T sin θ T cos θ car: a = 5 m/s² x y
FBD of the pendulum bob in the car's frame. The pseudo force $ma$ acts backward (opposing the car's forward acceleration). The bob hangs in equilibrium at angle $\theta$ where the three forces balance.

Given: m = 0.2 kg, a = 5 m/s², g = 9.8 m/s².

Step 1. Work in the car's (non-inertial) frame. The bob is stationary in this frame. Three forces act on it: tension T along the string, weight mg downward, and pseudo force ma backward.

Step 2. Horizontal equilibrium (taking the car's forward direction as positive):

T\sin\theta = ma
T\sin\theta = 0.2 \times 5 = 1.0 \text{ N} \tag{i}

Why: the horizontal component of tension must balance the pseudo force. The pseudo force pulls the bob backward; the string's horizontal component pulls it forward.

Step 3. Vertical equilibrium:

T\cos\theta = mg
T\cos\theta = 0.2 \times 9.8 = 1.96 \text{ N} \tag{ii}

Why: the vertical component of tension supports the bob's weight. No vertical pseudo force exists because the car accelerates only horizontally.

Step 4. Divide (i) by (ii) to find \theta:

\tan\theta = \frac{a}{g} = \frac{5}{9.8} = 0.5102
\theta = \tan^{-1}(0.5102) = 27.0°

Why: the angle depends only on the ratio a/g. A larger acceleration tilts the bob further. At a = g, the bob would hang at 45°.

Step 5. Find T by squaring and adding equations (i) and (ii):

T^2 = (ma)^2 + (mg)^2 = m^2(a^2 + g^2)
T = m\sqrt{a^2 + g^2} = 0.2\sqrt{25 + 96.04} = 0.2\sqrt{121.04} = 0.2 \times 11.0 = 2.20 \text{ N}

Why: since the two force components (ma horizontal and mg vertical) are perpendicular, their resultant has magnitude m\sqrt{a^2 + g^2} by Pythagoras. The tension equals this resultant because the bob is in equilibrium.

Result: The pendulum deflects at \theta \approx 27° from the vertical. The tension is T = 2.20 N, which is greater than the weight mg = 1.96 N.

What this shows: The pendulum acts as an accelerometer — you can read off the car's acceleration from the deflection angle. Indian Railways guards used to use a plumb bob hanging inside the guard's van to detect sudden changes in acceleration. The same principle is used in the accelerometer chip inside your smartphone.

Common confusions

If you came here to understand pseudo forces, derive apparent weight, and solve the pendulum problem, you have what you need. What follows is for readers who want the formal connection to general relativity's equivalence principle and the full treatment of effective gravity at arbitrary latitude.

The equivalence principle — Einstein's "happiest thought"

Einstein reportedly called the realisation that gravity and acceleration are locally indistinguishable "the happiest thought of my life." The argument is this:

Imagine you wake up inside a closed lift with no windows. You stand on a scale and it reads N = m \times 9.8 N/kg. Can you tell whether:

(a) The lift is stationary on the Earth's surface, and gravity is pulling you down at 9.8 m/s², or

(b) The lift is in deep space, far from any planet, accelerating upward at 9.8 m/s², and there is no gravity at all?

The answer is: you cannot. No experiment you perform inside the lift — dropping a ball, swinging a pendulum, measuring the period of oscillation, weighing objects on a scale — can distinguish between real gravity and a pseudo force from acceleration. The two are locally identical.

This is the equivalence principle, and it is the foundation of general relativity. The pseudo force -m\vec{a}_{\text{frame}} and the gravitational force m\vec{g} have the same dependence on mass — they are both proportional to m and independent of the object's shape, material, or charge. This is why free fall produces weightlessness: the pseudo force (from switching to the falling frame) exactly cancels gravity for every object, regardless of mass.

In Newtonian mechanics, this is a coincidence — gravitational mass just happens to equal inertial mass. In general relativity, it is not a coincidence; it is the starting axiom.

Effective gravity at general latitude

At a latitude \lambda on the Earth's surface, you are at distance R\cos\lambda from the Earth's rotation axis. The centrifugal acceleration is \omega^2 R\cos\lambda, directed outward from the axis (perpendicular to the axis, not radially away from the Earth's centre).

Decompose this centrifugal acceleration into two components:

  1. Radial component (outward from Earth's centre): \omega^2 R\cos^2\lambda
  2. Tangential component (along the surface, toward the equator): \omega^2 R\sin\lambda\cos\lambda = \frac{1}{2}\omega^2 R\sin(2\lambda)

The effective gravitational acceleration has magnitude:

g_{\text{eff}} = g - \omega^2 R\cos^2\lambda

Why: the tangential component is small and only deflects the plumb line direction slightly. The dominant correction is the radial component, which directly reduces g. At the equator (\lambda = 0), the reduction is maximum: \omega^2 R. At the poles (\lambda = 90°), \cos\lambda = 0 and there is no correction.

For a more precise result that accounts for the Earth's oblate shape, the measured values of g are:

Location Latitude g (m/s²)
Equator 0° 9.780
Mumbai 19° 9.791
Delhi 28° 9.800
North Pole 90° 9.832

The variation is about 0.5%, which matters for precision measurements but not for typical JEE problems (where g = 9.8 m/s² or g = 10 m/s² is used).

The Coriolis force — a brief note

When an object moves within a rotating frame (not just sits still), there is a second pseudo force: the Coriolis force, \vec{F}_{\text{Cor}} = -2m(\vec{\omega} \times \vec{v}'), where \vec{v}' is the velocity of the object relative to the rotating frame. This force is responsible for the clockwise rotation of cyclones in the Northern Hemisphere, the deviation of long-range artillery shells, and the swirl of water draining from a large tank. It is not significant for small-scale, short-duration problems (a ball thrown across a room, a pendulum in a car), which is why it does not appear in the main text. The full treatment belongs to the article on Coriolis effect.

Where this leads next