In the figure, the tension in the horizontal | vedclass

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(B) $1$. Isolate the point $P$ where the three cords meet.$2$. The forces acting on point $P$ are:- Tension $T_2$ in Cord $2$ (at an angle of $45^{\ci

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$30\,N$

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(B) $1$. Isolate the point $P$ where the three cords meet.$2$. The forces acting on point $P$ are:- Tension $T_2$ in Cord $2$ (at an angle of $45^{\circ}$ with the vertical).- Tension $T_1$ in Cord $1$ (acting vertically downwards, equal to the weight $W$ of body $B$).- Tension $T_h = 30\,N$ in the horizontal cord (acting horizontally).$3$. Resolve $T_2$ into components:- Horizontal component: $T_{2x} = T_2 \sin 45^{\circ}$- Vertical component: $T_{2y} = T_2 \cos 45^{\circ}$$4$. Apply equilibrium conditions ($\sum F_x = 0$ and $\sum F_y = 0$):- $\sum F_x = 0 \Rightarrow T_2 \sin 45^{\circ} = 30\,N$- $\sum F_y = 0 \Rightarrow T_2 \cos 45^{\circ} = T_1 = W$$5$. From the horizontal equilibrium equation:$T_2 \cdot \frac{1}{\sqrt{2}} = 30 \Rightarrow T_2 = 30\sqrt{2}\,N$$6$. Substitute $T_2$ into the vertical equilibrium equation:$W = T_2 \cos 45^{\circ} = (30\sqrt{2}) \cdot \frac{1}{\sqrt{2}} = 30\,N$Therefore, the weight of body $B$ is $30\,N$.

In the figure, the tension in the horizontal cord is $30\,N$. Find the weight of the body $B$.

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