Question

. An elevator cab that weighs 27.8 kN moves upward. What is the tension in the cable if the cab’s speed is (a) increasing at a rate of 1.22 m/s2 and (b) decreasing at a rete of 1.22 m/s2?

Answer 1

Step-by-step explanation:

It is given that,
`F_(w)` = 27.8 kN =
`27.8 * 10^(3) N`

Now, formula for the mass of elevator is as follows.

`F_(w) = mg`

m =
`(F_(w))/(g)`

=
`(27.8 * 10^(3))/(9.8)`

=
`2.84 * 10^(3)` kg

(a) When acceleration is increasing at a rate of 1.22
`m/s^(2)`. Force according to the Newton's second law of motion is as follows.

F = T - mg = ma

`T - 27.8 * 10^(3) = 2.84 * 10^(3) * 1.22`

T =
`31.27 * 10^(3) N`

Hence, tension in the cable if the cab’s speed is increasing at a rate of 1.22
`m/s^(2)` is
`31.27 * 10^(3) N`.

(b) When acceleration is decreasing at a rate of 1.22
`m/s^(2)`. Force according to the Newton's second law of motion is as follows.

F = T - mg = ma

`T - 27.8 * 10^(3) = 2.84 * 10^(3) * (-1.22)`

T =
`24.34 * 10^(3) N`

Hence, tension in the cable if the cab’s speed is decreasing at a rate of 1.22
`m/s^(2)` is
`24.34 * 10^(3) N`.