Elevators & Apparent Weight

Open a bathroom scale and you find a spring or a strain gauge. It deforms in proportion to the force pressing on it, and the electronics divide that force by 9.81 m s⁻² to print kilograms. By Newton's third law the force pressing the scale down has the same size as the normal force N the scale exerts up on you, so the instrument is a normal-force meter with a kilogram label. Your weight mg is a different force, exerted by the Earth, and inside a building it is constant: over a 100 m rise g falls by about 0.003%. The two numbers agree only when your vertical acceleration is zero, which covers standing on the pavement and also riding at constant velocity. Apparent weight is the name for N. The word apparent is a warning that the number describes a contact, and a contact force adjusts itself to whatever the motion requires.

Treat the rider as a particle with two vertical forces and take upward as positive. Then N − mg = m a_y, so N = m(g + a_y). An 80.0 kg passenger in a lift starting upward at 1.20 m s⁻² needs N = 80.0 × (9.81 + 1.20) = 881 N, against a weight of 785 N, and the display shows 881/9.81 = 89.8 kg. Starting downward at the same rate, a_y = −1.20 m s⁻² and N = 80.0 × 8.61 = 689 N, a display of 70.2 kg. Dividing by mg gives the load factor n = N/(mg) = 1 + a_y/g, here 1.12 and 0.88: the same 12% for every rider in the car, whatever their mass. The discipline is to let a_y carry its own sign. The usual wreck is to write N − mg = −ma because the acceleration is downward and then also substitute a = −1.20, which applies the minus sign twice and makes a lift that is dropping away read heavy.

A trip up has three phases: speeding up (a_y upward, heavier), cruising (a_y = 0, true weight) and slowing (a_y downward, lighter). A trip down runs the same list in reverse order: lighter at the start, true weight in the middle, heavier at the end, because a car that is moving down and slowing has an upward acceleration. On a velocity–time graph the reading follows the slope and ignores the height of the line. A lift descending at a steady 2.80 m s⁻¹ reads exactly mg, just as it does when parked. If it then stops in 1.60 s, a_y = (0 − (−2.80))/1.60 = +1.75 m s⁻² and n = 1 + 1.75/9.81 = 1.18: an 18% surge in the reading while the car is still travelling downward. Real lifts also limit the rate at which the acceleration changes, so the reading ramps between phases instead of jumping, and the ramps are what passengers notice.

Put a_y = −g into N = m(g + a_y) and the reading is zero. The floor has stopped pushing because it is accelerating downward exactly as the rider is; the weight mg is still there and is now the only force, which is what free fall means. Weightlessness in physics is this zero contact force. A drop tower supplies it for a few seconds, an aircraft flying a parabola for about twenty, and an orbit indefinitely, since an orbiting cabin is in continuous free fall: at 400 km altitude g is still 8.7 m s⁻², 89% of its ground value. Beyond −g the equation needs a guard. A floor can push but cannot pull, so N is never negative. If a car were driven downward at 12.0 m s⁻², the formula would return N = m(9.81 − 12.0), a negative number whose meaning is that contact has been lost: the rider falls at g while the floor pulls ahead at 12.0 − 9.81 = 2.2 m s⁻² relative to them, until the ceiling arrives.

Everything riding in the car shares one a_y, so the same line of algebra serves any system you care to box. For the cable, box the 950 kg car with its 450 kg of passengers: T − (M + m)g = (M + m)a_y, and at a_y = +1.00 m s⁻² the tension is 1400 × 10.81 = 1.51 × 10⁴ N against 1.37 × 10⁴ N at rest, a 10% rise that matches the load factor n = 1.10 felt inside. For one 60.0 kg rider's scale, box the rider: N = 60.0 × 10.81 = 649 N. In the first system the push between feet and floor is an internal pair, acting on two different members of the system, so it never enters the system's equation; in the second system it is the answer. A bag on a spring balance obeys the same rule with the tension in place of N, and it also reads zero in free fall. One caution: a real lift's counterweight hangs on the far side of the sheave, which makes the pair an Atwood machine from the previous topic. It changes what the motor must supply, not this equation for the car-side cable.

Because a_y = N/m − g, a logged reading tells you the motion. A 50.0 kg rider's scale shows a steady 560 N for the first 2.00 s of an ascent from rest, so a_y = 560/50.0 − 9.81 = 1.39 m s⁻², the speed reached is 1.39 × 2.00 = 2.78 m s⁻¹, and the car has risen ½ × 1.39 × 2.00² = 2.78 m. Near the top the log drops to 440 N: a_y = 8.80 − 9.81 = −1.01 m s⁻², and the 2.78 m s⁻¹ is shed in 2.78/1.01 = 2.75 s. When the reading varies, replace the products with integrals: Δv_y = (1/m) ∫ (N − mg) dt, the area between the record and the horizontal line N = mg divided by the mass, and a second integration gives the height. Over a complete trip from rest to rest that area must sum to zero, heavy patches exactly paying for light ones. What no scale record can supply is the starting velocity, since integration fixes only changes.