The free-body diagram of the whole system is shown in Fig.6.23.
Analyzing the equilibrium of point D
∑Fx=0 or T1sin450−T2sin370=0.....(i)
and ∑Fy=0 or T1cos450+T2cos370=350.....(ii)
From (i), we have T2=T1sin450sin370
Now from (ii), T1cos450+T1sin450sin370×cos370=350
or T1√2+T1√2×43=350
or T1√2[1+43]=350 ⇒ T1=150√2N
and T2=T1sin450sin370=150√2×(1√2)3/5=250N
Analyzing the equilibrium of point B
∑Fx=0 or T2sin370+T3sin300−T4=0.....(i)
and ∑Fy=0 or T3cos300−T2cos370−100=0......(ii)
From (ii), T3cos300−250×45−100=0
⇒ T3=200√3N
From (i), T4=150+100√3=50(3+2√3)N
![1029145_983216_ans_82dd1fa7a96b48f5beb9ca676b928c55.png](https://search-static.byjusweb.com/question-images/toppr_invalid/questions/1029145_983216_ans_82dd1fa7a96b48f5beb9ca676b928c55.png)