Question

Mechanics question help on forces?

A miners' cage of mass 420kg contains 3 miners of total mass 280kg. The cage is lowered from rest by a cable. For the first 10 seconds the cage accelerates uniformly and descends a distance of 75 m. The force in the cable during the first 10 seconds is
_N

Answer 1

The force on the cable = `700kg x (g -a)`# where a is acceleration of cage

Explanation:

the cage has travelled from rest ,initial velocity say u =0

it has travelled 75m in 10 sec.

`s = u.t + (1/2) .a .t^2` where a = acceleration

so `75 m = (1/2). a. 100 s^2`

calculate `a = (2 * 75)/100 m/s^2` `= 1,5 m/s^2`

now the cage is moving down with an acceleration

if the cable is having tension T then

Mass of cage with miners `(M +m) . a = (M+m) . g - T`

force on cable = (M+m) .( g-a) = 700 kg .(10 -1.5)`m/s^2`

`T =`700 . 8.5 N`= 5.95 ,`10^3 N#