Question

An elevator cab and its load have a combined mass of 1200 kg. Find the tension in the supporting cable when the cab, originally moving downward at 10 m/s, is brought to rest with constant acceleration in a distance of 35 m.

Answer 1

Answer:

10044 N

Explanation:

The acceleration of the cab is calculated using the equation of motion:

$v^2 = u^2+2as$

v is the final velocity = 0 m/s in this question, since it is brought to rest

u is the initial velocity = 10 m/s

a is the acceleration

s is the distance = 35 m

$a = \dfrac{v^2-u^2}{2s} = \dfrac{(0 \text{ m/s})^2-(10 \text{ m/s})^2}{2\times (35\text{ m})} = -1.43\text{ m/s}^2$

Since it accelerates downwards, its resultant acceleration is

$a_R = g + a$

g is the acceleration of gravity.

$a_R = (9.8-1.43)\text{ m/s}^2 = 8.37\text{ m/s}^2$

The tension in the cable is

$T = ma_R = (1200\text{ kg})(8.37\text{ m/s}^2) = 10044 \text{ N}$