Effort supported by a rope-bridge ?

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Answer 1 · 2016-08-30T05:38:18

Given
$L->"Length of the rope"=25m$
$M->"Mass of the man climbing"=72kg$
$T->"Tension the rope can withstand"=2.9kN$

Let the sagging be x m when the man is at the mid point of the rope and the rope then subtends angle $theta$ with the horizontal.

The two vertical components of tension will balance the weight of the man.

So

$2Tsintheta=Mxxg$

$sintheta=(Mg)/(2T).....(1)$

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
$(2x)/L=tantheta=tan6.99$

$x=(25xxtan6.99)/2=1.53m$