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A vertical lift of total mass 750 kg acquires an upward velocity of 3m/s over a distance of 4m moving with constant acceleration starting from rest. Calculate the tension in the cable.
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Given:

$m=750 kg$

$u=0 m/s$

$v=3 m/s $

$S=4m$

$a=constant$

$v^2=u^2+2aS $

${\left(3\right)}^2=0+2\times{}a\times{}\left(4\right) $

$\therefore{}a=1.125\ m/s^2\ \uparrow{} $

Applying D’Alemberts Principle -

$\displaystyle\sum F_y=0 $

$T-ma-mg=0 $

$\therefore{}T=m(a+g) $

$=750\times{}(1.125+9.81) $

$T=8201.25\ N $

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