Two particle of masses 10 kg and 35 kg are connected with four strings at points B and D as shown in fig.
Determine the tension in various segments of the string.

Explanation

The free-body diagram of the whole system is shown in fig.

Analyzing the equilibrium of point D
`Sigma F_(x)=0` or `T_(1) sin 45^(@)=T_(2) sin 37^(@)=0` …(i)
`Sigma F_(y)=0` or `T_(1) cos 45^(@)=T_(2) cos 37^(@)=350` …(ii)
From (i), we have `T_(2)=(T_(1)sin 45^(@))/(sin 37^(@)) `
Now from (ii) , `T_(1) cos 45^(@)+(T_(1)sin 45^(@))/(sin 37^(@)) xx cos 37^(@)=350`
or, `(T_(1))/(sqrt(2))+(T_1)/(sqrt(2)) xx 4/3 = 350`
or , `(T_(1))/(sqrt(2))[1+4/3]=350 impliesT_(1)=150sqrt(2)N`
and `T_(2)=(T_(1)sin 45^(@))/(sin 37^(@))=(150sqrt(2)xx((1)/(sqrt(2))))/(3//5) = 250 N`
Analyzing the equilibrium of point B
`Sigma F_(x)=0` or `T_(2) sin 37^(@)=T_(3) sin 30^(@)-T_(4)=0` …(i)
and `Sigma F_(y)=0` or `T_(3) cos30^(@)-T_(2) cos 37^(@)-100=0` …(ii)
from (ii) `T_(3) cos 30^(@)-250 xx 4/5 -100 =0`
`implies T_(3)=200sqrt(3)N`
from(i) `T_(4)=150+100sqrt(3)=50(3+2sqrt(3))N`.