Pseudo Forces and Non-Inertial Frames
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Direct answer
Newton's second law holds only in inertial frames, so when you choose to work inside an accelerating frame — a lift, a braking car, a sliding wedge — you must add a pseudo force F = −ma on every body, with a the acceleration of the frame itself. This one bookkeeping force converts a dynamics problem into an equilibrium problem: the lift occupant feels N = m(g + a) while accelerating up and N = m(g − a) while accelerating down, the pendulum inside an accelerating car swings at tanθ = a/g and ticks with period 2π√(L/g_eff) where g_eff = √(g² + a²), and a freely falling frame has a = g downward, making everything inside apparently weightless.
What you must remember
- The rule: in a frame accelerating at a, add F_pseudo = −ma to each body's free-body diagram, directed opposite to the frame's acceleration, then use ΣF = 0 or ΣF = ma in that frame.
- Lift arithmetic: accelerating up N = m(g + a); accelerating down N = m(g − a); constant velocity N = mg.
- Effective gravity: in a vehicle accelerating horizontally at a, gravity is replaced by g_eff = √(g² + a²) tilted at tanθ = a/g from the vertical; a pendulum hangs along g_eff and has period 2π√(L/g_eff).
- Free fall and satellites: with a = g downward the pseudo force exactly cancels weight, g_eff = 0 — an orbiting satellite is a falling frame, hence apparent weightlessness without gravity being zero.
- Wedge problems: in the wedge's frame, add −ma on the block; the block then either rests or slides as a statics problem, which is usually far faster than ground-frame components.
- Rotating frames: a frame rotating at ω demands a centrifugal pseudo force mω²r outward on every body (and a Coriolis term for moving bodies) — the frame of a turntable rider.
- Direction discipline: the pseudo force opposes the frame's acceleration, not the body's acceleration; these coincide only for bodies at rest relative to the frame.
A lift, a car and a falling frame
A 60 kg person stands in a lift accelerating up at 2 m/s²: in the ground frame N − mg = ma gives N = 60 × 11.8 = 708 N; in the lift frame add pseudo force ma downward and demand balance — N = m(g + a) = 708 N. The lift-frame route scales trivially to a 1 m pendulum inside that lift: g_eff = 11.8 m/s² and T = 2π√(1/11.8) ≈ 1.83 s.
Next, a pendulum hanging from the roof of a train accelerating at 2 m/s². The bob lines up along effective gravity: tanθ = 2/9.8, θ ≈ 11.5° backwards from the vertical, and g_eff = √(9.8² + 2²) = 10 m/s². A 25 cm pendulum then has T = 2π√(0.25/10) ≈ 0.99 s — one second, a number examiners engineer deliberately around g = 10 and L = 0.25 m. Finally the falling frame: cut the lift cable and a = g, so g_eff = 0; the occupant floats, the pendulum stops oscillating (no restoring force), and every loose object drifts. The orbiting satellite is this same frame perpetually, which is why astronauts float even though gravity at their altitude is close to 90 per cent of its surface value.
Examiner's framing
The trap tested most often is direction: students draw the pseudo force opposite to the body's motion instead of opposite to the frame's acceleration; a lift moving up but decelerating still needs the pseudo force downward (frame acceleration is downward), giving N = m(g + a) with a now the deceleration magnitude. Second, constant velocity needs no pseudo force: "a car at 100 km/h" is an inertial frame, and questions phrase it precisely to check you add none. Third, weightlessness is misread as absence of gravity; in a satellite gravity is real and centripetal, the pseudo force merely cancels it locally. Advanced combines frames: a block on an accelerating wedge, solved elegantly in the wedge frame with −ma.
Frequently asked questions
When must a pseudo force be introduced?
Whenever you analyse motion from a non-inertial frame — accelerating linearly or rotating — you add F = −ma (frame acceleration) to every body so Newton's laws apply in that frame.
What is the apparent weight in a lift accelerating downward at 3 m/s²?
N = m(g − a) = m × 6.8 m/s², about 69 per cent of the true weight; if a = g the apparent weight falls to zero.
Why does a pendulum in an accelerating car tilt?
It hangs along effective gravity g_eff = √(g² + a²), tilted backwards by tanθ = a/g, because in the car's frame the pseudo force acts like an extra horizontal gravity component.
Are astronauts weightless because gravity is absent in orbit?
No; gravity supplies the centripetal force, and in the freely falling satellite frame the pseudo force −ma exactly cancels weight, producing apparent weightlessness.
Does a body at rest in a rotating frame need a pseudo force?
Yes: in that frame a centrifugal force mω²r outward balances the real centripetal contact forces, keeping the body in equilibrium as seen by the co-rotating observer.