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Weight and weightlessness

Lessons 12 · 1 lessons · N. Sh. Turdiyev, K. A. Tursunmetov, A. G. Ganiyev, K. T. Suyarov, J. E. Usarov, A. K. Avliyoqulov. Physics Grade 10, 1st edition. Niso Poligraf Publishing House, Tashkent, 2017
12

How the weight of a body depends on the type of motion

Textbook: pp. 37–39
GoalExplain how the weight of a body depends on the acceleration of its support or suspension, and understand weightlessness and overload.
New words
weight of a body · jism og‘irligiweightlessness · vaznsizlikoverload · ortiqcha yuklamareaction force · reaksiya kuchi
Explanation

The weight of a body is the force with which it acts on a support or suspension; it equals the support’s reaction force N in size and is opposite in direction (third law). At rest or at constant velocity P = mg. If the support (a lift) accelerates downward with a, then mg – N = ma, so P = m(g – a); if it accelerates upward, P = m(g + a). If a = g and directed downward, then P = 0 – weightlessness: the body does not press on its support; free-falling bodies, orbiting satellites and the people inside them are in this state. The load factor n = P/(mg) = 1 + a/g (n > 1 for upward acceleration is overload): it is several times larger in a rocket launch and for pilots. On a curved path weight changes too: at the top of a convex bridge P = m(g – υ²/R), at the bottom of a dip P = m(g + υ²/R). If the acceleration is purely horizontal, the vertical weight does not change.

Worked examples
A 60 kg person in a lift: accelerating upward with a = 2 m/s², P = 60·(10 + 2) = 720 N; accelerating downward with a = 2 m/s², P = 60·(10 – 2) = 480 N; in a stationary lift P = 600 N.
A car (m = 1000 kg) crosses the top of a convex bridge at υ = 20 m/s, radius of curvature R = 40 m. P = m(g – υ²/R) = 1000·(10 – 400/40) = 0: at this point the car is weightless and the tyres do not press on the road.
Class activity

In class, from memory: recall moments on a swing or in a lift when you felt “lighter” or “heavier”, note in a table when each happened and explain them with P = m(g ± a). Do not try any risky jumps or experiments.

Practice
1
What is the weight of a 65 kg person in a stationary lift (N)? (g = 10 m/s²)
2
A 65 kg person is in a lift accelerating upward at a = 2 m/s². What is the weight (N)? (g = 10 m/s²)
3
A rocket accelerates upward at 30 m/s². What is the load factor n? (g = 10 m/s²)
4
Why is a person weightless in a lift that falls freely after its cable breaks?