Question

Safety engineers estimate that an elevator can hold a maximum mass of 1000 kg (including the mass of the elevator). Tensile strength (force) tests show that the cable supporting the elevator can tolerate a maximum upward force of 25,000 N. What is the maximum acceleration that the elevator's motor can produce without breaking the cable?​

Answer 1

a = 15.2 m/s2

Step-by-step explanation:

• Assuming that the elevator behaves as an isolated system, there are two forces acting on it at any time: the tension T in the cable (acting upward), and gravity, acting downward with a constant acceleration g = 9.8 m/s2 over the total mass (in this case 1000 kg).
• So, according 2nd Newton's Law, F = m*a, we can write the following expression:
• F = T - m*g (1)

⇒
`T- m*g = m*a (2)`

• The maximum force, will be applied when T reaches to its maximum possible value, 25,000 N.
• Replacing the values of Tmax, m and g in (2) we can solve for a, as follows:

`a = (T-m*g)/(m) = (25,000N- 1000kg*9.8m/s2)/(1000kg) = 15.2 m/s2 (3)`

• So, the maximum acceleration possible without breaking the cable is 15.2 m/s2.