An elevator weighing 200 kilograms begins to descend at a uniform acceleration and travels 72 m in 12 seconds.

An elevator weighing 200 kilograms begins to descend at a uniform acceleration and travels 72 m in 12 seconds. Determine the tension of the elevator rope.

We find the tension force of the elevator rope from the ratio:
Fт – Fн = m * a, Fт – gravity (Fт = m * g, m – lift mass (m = 200 kg), g – free fall acceleration (g = 10 m / s²)), Fн – rope tension force lift (Н), a – lift acceleration (а = 2S / t², where S – distance traveled (S = 72 m), t – lift movement time (t = 12 s)).
Fн = -m * a + Fт = -m * 2S / t² + m * g = -200 * 2 * 72 / 12² – 200 * 10 = -200 + 2000 = 1800 N = 1.8 kN.
Answer: The pulling force of the elevator rope is 1.8 kN.