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Safety engineers estimate that an elevator can hold 20 persons of 75kg average mass. The elevator itself has a mass of 500kg. Tensile strength test how the cable supporting the elevator can tolerate a maximum force of 2.96X10^4N. What is the greatest acceleration that the elevators motor can produce without breaking the cable? (show work)

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onyebuchinnaji

The greatest acceleration that the elevators motor can produce without breaking the cable is 5 m/s².

The given parameters;

  • average mass of the 20 persons = 75 kg
  • mass of the elevator, m = 500 kg
  • tensile strength of the elevator, T = 2.96 x 10⁴ N

The total mass the elevator can support;

M = 500 kg + 20(75 kg)

M = 2000 kg

The greatest acceleration that the elevators motor can produce without breaking the cable occurs when the elevator is moving upwards;

T = M(a + g)

[tex]2.96 \times 10^4 = 2000 (a + 9.8)\\\\2.96 \times 10^4 = 2000a + 19600\\\\29600 - 19600 = 2000a\\\\10,000 = 2000a\\\\a = \frac{10,000}{2000} \\\\a = 5 \ m/s^2[/tex]

Thus, the greatest acceleration that the elevators motor can produce without breaking the cable is 5 m/s².

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