Question

A uniform ladder of length $10m$ and eight $100kg$ is kept with its lower end on a smooth horizontal floor and its upper end on a smooth vertical wall. A horizontal string tied to the bottom of the ladder keeps it in equilibrium. Find the tension along the string if a man weighing $60kg$ goes up by $3m$ in the ladder which is inclined at ${30}^o$ with the horizontal.

Answer 1

Given,

$AB=10m,w=100kg,BG=3m,BH=5m$

$angle={30}^0$

For the equilibrium of ladder,

$\sum f_x=0,R_x=T$

$\sum f_r=0,R_w=T$

$\sum m=0$

Now taking moment about B,

$RA\times AC-w\times BE=0$

$RA\times AC=w\times BE$

$AC=10sin{30}^0=10\times \frac{1}{2}=5$ Since $\frac{AC}{AB}=sin\theta$

$AC=8.66m$

Since $Sin\theta =\sqrt{1-co{s}^2\theta }$

so,

$T=w\times BC$

$=\frac{165\times 8.66}{\sqrt{4(10{)}^2-(8.66{)}^2}}$

$=\frac{1428.9}{36.055}=39.631N$