A lift of mass 1600 text{ kg} is supported by | vedclass

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Question

(A) The wire must support the weight of the lift and the additional force due to acceleration. The tension $T$ in the wire is given by $T = m(g + a)$.

Vedclass · Accepted Answer

$2.56$

Vedclass · Answer

(A) The wire must support the weight of the lift and the additional force due to acceleration. The tension $T$ in the wire is given by $T = m(g + a)$.Given: Stress $\sigma = 4 	imes 10^8 	ext{ N/m}^2$,mass $m = 1600 	ext{ kg}$,radius $r = 4 	ext{ mm} = 4 	imes 10^{-3} 	ext{ m}$,and $g = 10 	ext{ m/s}^2$.The cross-sectional area $A$ of the wire is $A = \pi r^2 = 3.14 	imes (4 	imes 10^{-3} 	ext{ m})^2 = 3.14 	imes 16 	imes 10^{-6} 	ext{ m}^2 = 50.24 	imes 10^{-6} 	ext{ m}^2$.The maximum tension $T$ the wire can withstand is $T = \sigma 	imes A = (4 	imes 10^8 	ext{ N/m}^2) 	imes (50.24 	imes 10^{-6} 	ext{ m}^2) = 20096 	ext{ N}$.Using the equation of motion $T = m(g + a)$,we have $20096 = 1600(10 + a)$.Dividing both sides by $1600$,we get $10 + a = \frac{20096}{1600} = 12.56$.Therefore,$a = 12.56 - 10 = 2.56 	ext{ m/s}^2$.

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