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You stand in a lift and feel heavier as it starts going up. A pendulum in a car tilts backward when the car speeds up. Questions like these can be solved from the ground, but they are often much simpler from inside the lift or the car. To do that you need one extra force, the pseudo force. This post explains what it is, when you may use it, and works three standard problems.
What is a pseudo force?
Newton's laws hold in inertial frames, ones that are not accelerating. If you sit in an accelerating frame and pretend it is at rest, the laws stop working: a ball on the floor of a braking bus rolls forward with no horizontal force on it.
The fix is to add, to every body, a force of magnitude m a₀ pointing opposite to the frame's acceleration a₀. With that force included, Newton's second law works in the accelerating frame. OpenStax University Physics calls these inertial forces, "fictitious forces that merely seem to arise from motion, because the observer's frame of reference is accelerating or rotating." Its example: when a plane takes off, it feels as if you are pushed back into the seat, but in the ground frame the seat is pushing you forward.
NCERT's Work, Energy and Power chapter notes the same idea for energy: the work-energy theorem holds in all inertial frames and "can also be extended to non-inertial frames provided we include the pseudoforces."
Is this in the JEE syllabus?
The JEE (Advanced) 2026 syllabus lists "Inertial and uniformly accelerated frames of reference" under mechanics. The JEE (Main) 2026 syllabus lists the frame of reference and the laws of motion; lift problems appear in NCERT's own exercises. Check the bulletin for your exam year.
Worked example 1: a person in a lift
A 50 kg student stands on a weighing scale in a lift. Take g = 10 m/s². What does the scale read when the lift accelerates upward at 2 m/s²?
From inside the lift. The student is at rest in the lift frame. Forces: weight 500 N down, normal force N up, pseudo force 50 × 2 = 100 N down (opposite to the lift's upward acceleration). Balance: N = 500 + 100 = 600 N.
From the ground. N − 500 = 50 × 2, so N = 600 N. Same answer, as it must be.
If the lift accelerates downward at 2 m/s², the pseudo force points up and N = 400 N. In free fall (a₀ = g downward), N = 0: the scale reads zero.
Worked example 2: a pendulum in an accelerating car
A 0.2 kg bob hangs from the roof of a car that accelerates at 5 m/s² on a level road. At what angle does the string hang, and what is its tension?
In the car frame the bob is at rest. Forces: weight 2 N down, tension T along the string, pseudo force 0.2 × 5 = 1 N backward.
Horizontal: T sin θ = 1. Vertical: T cos θ = 2.
Divide: tan θ = 1/2, so θ ≈ 26.6° from the vertical, tilted backward.
Tension: T = √(1² + 2²) = √5 ≈ 2.24 N.
The general results are tan θ = a/g and T = m√(g² + a²). The bob behaves as if gravity were tilted and slightly stronger.
Worked example 3: a block resting on a moving wedge
A block sits on a smooth wedge of angle 37°. The wedge is pushed horizontally. What acceleration keeps the block from sliding? Take sin 37° = 0.6 and cos 37° = 0.8.
In the wedge frame the block is at rest. Forces: weight mg down, normal force N perpendicular to the incline, pseudo force ma backward, opposite to the push.
Along the incline, the pseudo force's component (ma cos θ, up the slope) must cancel the weight's component (mg sin θ, down the slope).
So a = g tan θ = 10 × 0.75 = 7.5 m/s².
Perpendicular to the incline: N = mg cos θ + ma sin θ = m(8 + 4.5) = 12.5m newtons. Equivalently, N = mg / cos θ.
Solving this from the ground means tracking the block's acceleration in two directions, which is where most errors come from.
What mistakes do students make with pseudo forces?
Adding the pseudo force in the ground frame, which counts the effect twice.
Getting the direction wrong. It points opposite to the frame's acceleration, not opposite to its velocity. A lift moving up but slowing down has its acceleration downward, so the pseudo force points up.
Mixing frames: writing forces in the lift frame but using the ground-frame acceleration.
Applying it to only one body. Every body you analyse in that frame gets its own m a₀.
Looking for a third-law partner. A pseudo force has none, because no body exerts it.
In rotating frames the pseudo force is the centrifugal force, mv²/r outward. Our post on circular motion and banking solves those problems from the ground frame instead, which is good practice for comparing the two views.
A practice routine for accelerating frames
For each problem, write "Frame: ground" or "Frame: lift (a₀ = ..., direction ...)" before the diagram.
Solve three lift problems both ways and confirm the answers agree.
Do two pendulum-in-vehicle and two wedge problems from the moving frame.
Mix in two problems where the frame moves at constant velocity and check that no pseudo force appears.
If the free-body diagram method is still shaky, fix that first. And if the idea of a frame itself is the gap, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revise frames of reference on their own.
Today, solve the lift example for a lift moving up and slowing at 2 m/s², from both frames.
