Mechanics question help on forces?

Question

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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Answer 1 · 2018-03-31T20:18:52

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

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

it has travelled 75m in 10 sec.

$s = u . t + (\frac{1}{2}) . a . t^{2}$ $s = u.t + (1/2) .a .t^2$ where a = acceleration

so $75 m = (\frac{1}{2}) . a . 100 s^{2}$ $75 m = (1/2). a. 100 s^2$

calculate $a = \frac{2 \cdot 75}{100} \frac{m}{s^{2}}$ $a = (2 * 75)/100 m/s^2$ $= 1, 5 \frac{m}{s^{2}}$ $= 1,5 m/s^2$

now the cage is moving down with an acceleration

if the cable is having tension T then

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

force on cable = (M+m) .( g-a) = 700 kg .(10 -1.5) $\frac{m}{s^{2}}$ $m/s^2$

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