A block of mass 90 ,kg is suspended by three | vedclass

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(D) Given, mass of block, $m=90 \,kg$.Acceleration due to gravity, $g=10 \,ms^{-2}$.Let $T_A, T_B$ and $T_C$ be the tensions in strings $A, B$ and $C$

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$1200 \,N, 1500 \,N$ and $900 \,N$

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(D) Given, mass of block, $m=90 \,kg$.Acceleration due to gravity, $g=10 \,ms^{-2}$.Let $T_A, T_B$ and $T_C$ be the tensions in strings $A, B$ and $C$ respectively.The weight of the block acts downwards: $W = mg = 90 	imes 10 = 900 \,N$.Since the system is in equilibrium, the tension in string $C$ must balance the weight: $T_C = 900 \,N$.Now, consider the equilibrium of the junction point where the three strings meet. Resolving forces into horizontal and vertical components:Vertical equilibrium: $T_B \sin 37^{\circ} = T_C = 900 \,N$.Since $\sin 37^{\circ} = 0.6$, we have $T_B 	imes 0.6 = 900 \Rightarrow T_B = \frac{900}{0.6} = 1500 \,N$.Horizontal equilibrium: $T_A = T_B \cos 37^{\circ}$.Since $\cos 37^{\circ} = 0.8$, we have $T_A = 1500 	imes 0.8 = 1200 \,N$.Thus, the tensions are $T_A = 1200 \,N, T_B = 1500 \,N$ and $T_C = 900 \,N$.

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

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