Given
L->"Length of the rope"=25mL→Length of the rope=25m
M->"Mass of the man climbing"=72kgM→Mass of the man climbing=72kg
T->"Tension the rope can withstand"=2.9kNT→Tension the rope can withstand=2.9kN
Let the sagging be x m when the man is at the mid point of the rope and the rope then subtends angle thetaθ with the horizontal.
![modified]()
![my comp]()
The two vertical components of tension will balance the weight of the man.
So
2Tsintheta=Mxxg2Tsinθ=M×g
sintheta=(Mg)/(2T).....(1)
For theta being small sintheta~~tantheta=x/(L/2)=(2x)/L
Now imposing approximation the relation (1) becomes
(2x)/L=(Mxxg)/(2T)
=>x=(MxxgxxL)/(4T)
=>x=(72xx9.8xx25)/(4xx2900)~~1.52m
Without approximation the realation (1) can be written as
x/sqrt(x^2+(L/2)^2)=(Mxxg)/(2T)
=>(x^2+(12.5)^2)/x^2=((2xx2900)/(72xx9.8))^2=67.57
=>1+156.25/x^2=67.57
=>156.25/x^2=67.57-1=66.57
=>x^2=156.25/66.57
=>x=sqrt(156.25/66.57)~~1.53m
Another way of calculation
From (1)
theta=sin^-1((Mxxg)/(2T))=sin^-1((72*9.8)/(2*2900))
theta=6.99^@
So
(2x)/L=tantheta=tan6.99
x=(25xxtan6.99)/2=1.53m
