A block of mass `m=10kg` is suspended with the help of three strings as shown in fig. Find the tensions `T_(1)` and `T_(2)`.

Explanation

Method I: Considering equilibrium of each part of system
The whole system is in equilibrium, therefore, for each part `Sigma vec(F)=0`. From the free body diagram of block C, `t_(C )=300 N`.

Now consider the equilibrium of point O,
`Sigma F_(x)=0 or T_(B) cos 37^(@)-T_(A)=0`
`:. T_(A)=T_(B) cos 37^(@) =T_(B)*4/5` ...(i)
and `Sigma F_(y)=0 or T_(B) cos 37^(@)-T_(C)=0` ...(ii)
`:. T(B) = (T_(C ))/(sin 37^(@)) = (300)/(3//5) = 500N`
From Eq.(i), we get
`T_(A)=4/5 T_(B)=4/5 xx 500 = 400 N`.
Method II: Using Lami's theoram

By Lami's theoram, we have
`(T_(A))/(sin(90+37^(@)))=(T_(B))/(sin 90^(@)) =(T_(C))/(sin(180-37^(@)))`
But `TC = 300 N`
and `T_(B)=(T_(C))/(sin 37^(@))=(300)/(3//5)=500N`
`T_(A)=T_(C)((sin(90+37^(@)))/(sin(180+37^(@))))=300 ((cos 37^(@))/(sin37^(@)))=400N`.