Question

A 770-kg boulder is raised from a quarry 106 m deep by a long uniform chain having a mass of 575 kg . This chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.80 times its weight without breaking.What is the maximum acceleration the boulder can have and still get out of the quarry?

Answer 1

Answer:

(a) $a_{y}= 1.931m/s^{2}$

(b) $t=10.478s$

Explanation:

For Part(a)

Apply ∑Fy=ma

$T-Mg=Ma_{y}$

Solve for a

$a_{y}=\frac{T-Mg}{M}=\frac{2.80mg-Mg}{M} \\a_{y}=\frac{(2.80*575kg-1345kg)}{1345kg}*(9.80m/s^{2} )\\a_{y}= 1.931m/s^{2}$

For Part(b)

Assume the maximum acceleration (maximum tension upwards is greater than the weight of the chain+boulder)

As the given data

$a_{y}=1.931m/s^{2}\\ y-y_{o}=106m\\v_{oy}=0$

Using equation of simple motion

$y-y_{o}=v_{oy}t+(1/2)a_{y}t^{2} \\t=\sqrt{\frac{2(y-y_{o})}{a_{y}} }\\ t=\sqrt{\frac{2(106m)}{1.931m/s^{2} } }\\t=10.478s$