Question

A block of mass $30kg$ is suspended by three strings as shown in $Fig.6.19$. Find the tension in each string.

Answer 1

Method I :
Considering equilibrium of each part of system
The whole system is in equilibrium; therefore, for each part $\sum \overrightarrow{F}=0$. From the free-body diagram of block C, $T_c=300N$.
Now consider the equilibrium of point O,
$\sum F_x=0$ or $T_B\cos {37}^0-T_A=0$
$\therefore$ $T_A=T_B\cos {37}^0=T_B\frac{4}{5}$....(i)
and $\sum F_y=0$ or $T_B\sin {37}^0-T_C=0$....(ii)
$\therefore$ $T_B=\frac{T_C}{\sin {37}^0}=\frac{300}{3/5}=500N$
From Eq. (i), we get
$T_A=\frac{4}{5}{T}_B=\frac{4}{5}\times 500=400N$
Method II :
Using Lami's theorem by Lami's theorem, we have
$\frac{T_A}{\sin (90+{37}^0)}=\frac{T_B}{\sin {90}^0}=\frac{T_C}{\sin (180-{37}^0)}$
But $TC=300N$
and $T_B=\frac{T_C}{\sin {37}^0}=\frac{300}{3/5}=500N$
$T_A=T_C\left(\frac{\sin (90-{37}^0)}{\sin (180-{37}^0)}\right)=300\left(\frac{\cos {37}^0}{\sin {37}^0}\right)=400N$