A block of mass $90 \mathrm{~kg}$ is suspended by three strings $A, B$ and $C$ as shown in figure. Tensions…

A block of mass $90 \mathrm{~kg}$ is suspended by three strings $A, B$ and $C$ as shown in figure. Tensions in the strings $A, B$ and $C$ respectively are $\left(g=10 \mathrm{~ms}^{-2}, \sin 37^{\circ}=0.6, \cos 37^{\circ}=0.8\right)$

1. $400 \mathrm{~N}, 500 \mathrm{~N}$ and $300 \mathrm{~N}$
2. $500 \mathrm{~N}, 300 \mathrm{~N}$ and $900 \mathrm{~N}$
3. $300 \mathrm{~N}, 600 \mathrm{~N}$ and $900 \mathrm{~N}$
4. $1200 \mathrm{~N}, 1500 \mathrm{~N}$ and $900 \mathrm{~N}$

Solution

Given, mass of block, $m=90 \mathrm{~kg}$ Acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$ $T_A, T_B$ and $T_C$ are tensions in strings $A, B$ and $C$.
According to given diagram, $ \begin{aligned} \text { Weight } & =m g \\ & =90 \times 10 \\ & =900 \mathrm{~N} \\ \therefore \quad T_C & =900 \mathrm{~N} \end{aligned} $ By using Lami's theorem, $ \begin{array}{rlrl} & & \frac{T_A}{\sin \left(90^{\circ}+37^{\circ}\right)} & =\frac{T_B}{\sin 90^{\circ}}=\frac{T_C}{\sin \left(90^{\circ}+53^{\circ}\right)} \\ \Rightarrow & \frac{T_A}{\cos 37^{\circ}} & =T_B=\frac{900}{\cos 53^{\circ}} \\ \therefore & & T_B & =\frac{900}{\cos 53^{\circ}}=1500 \mathrm{~N} \\ & \text { and } & T_A & =\frac{900}{\cos 53^{\circ}} \cos 37^{\circ} \\ & & =1197=1200 \mathrm{~N} \end{array} $

Asked in: AP EAMCET 2021 (23 Aug Shift 1)

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