Question

Question #bc133

Answer 1

Let the maximum uniform acceleration with which the man can ascend `5m` to reach the balcony with the help of the hanging rope (the other end of which is tied with a chandelier of mass `124kg` be `a m"/"s^2`
The accelerated motion of the man will create extra tension on the rope more than the weight of the man. This tension should not be greater than than weight of the chandelier of mass `124kg`.

So we can write `70xx(a+g)=124xxg`

where `g=` acceleration due to gravity `=9.8m"/"s^2`

So the above equation becomes

`70xx(a+9.8)=124xx9.8`

`=>(a+9.8)=124xx9.8/70`

`=>a=124xx9.8/70-9.8=7.56m"/"s^2`

So if time taken by the man to reach the balcony with this maximum acceleration staring from rest be `t` sec then by equation of kinematics we can write

`0xxt-1/2at^2=5`

`=t=sqrt(10/a)=sqrt(10/7.56)~~1.15s`