Question

Two particles of masses $10kg$ and $35kg$ are connected with four strings at points B and D as shown in $Fig.6.22$.
Determine the tensions in various segments of the string.

Answer 1

The free-body diagram of the whole system is shown in $Fig.6.23$.

Analyzing the equilibrium of point D

$\sum F_x=0$ or $T_1\sin {45}^0-T_2\sin {37}^0=0$.....(i)

and $\sum F_y=0$ or $T_1\cos {45}^0+T_2\cos {37}^0=350$.....(ii)

From (i), we have $T_2=\frac{T_1\sin {45}^0}{\sin {37}^0}$

Now from (ii), $T_1\cos {45}^0+\frac{T_1\sin {45}^0}{\sin {37}^0}\times \cos {37}^0=350$

or $\frac{T_1}{\sqrt{2}}+\frac{T_1}{\sqrt{2}}\times \frac{4}{3}=350$

or $\frac{T_1}{\sqrt{2}}[1+\frac{4}{3}]=350$ $\Rightarrow$ $T_1=150\sqrt{2}N$

and $T_2=\frac{T_1\sin {45}^0}{\sin {37}^0}=\frac{150\sqrt{2}\times \left(\frac{1}{\sqrt{2}}\right)}{3/5}=250N$

Analyzing the equilibrium of point B

$\sum F_x=0$ or $T_2\sin {37}^0+T_3\sin {30}^0-T_4=0$.....(i)

and $\sum F_y=0$ or $T_3\cos {30}^0-T_2\cos {37}^0-100=0$......(ii)

From (ii), $T_3\cos {30}^0-250\times \frac{4}{5}-100=0$

$\Rightarrow$ $T_3=200\sqrt{3}N$

From (i), $T_4=150+100\sqrt{3}=50(3+2\sqrt{3})N$