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
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1 Answer
Mar 31, 2018

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#