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

$30\sqrt 2$
$30$
$15$
$60$
Solution
$(i)$ Isolate the point $P$
$(iii)$ The forces acting on it are$:$
Unknown tension $\mathrm{T}_{2}$ in cord $2$
Unknown tension $\mathrm{T}_{1}$ in cord $1$
Known tension of $30 \mathrm{N}$ in the horizontal cord.
$(iii)$ $\mathrm{T}_{2}$ is resolved along $\mathrm{x}$ and $\mathrm{y}$ axes.
$(iv)$ Condition of equilibrium$:$
$\Sigma F_{x}=0 \Rightarrow 30-T_{2} \sin 45^{\circ}=0$ $…(1)$
and $\Sigma \mathrm{F}_{\mathrm{y}}=0 \Rightarrow \mathrm{T}_{2} \cos 45^{\circ}-\mathrm{T}_{1}=0$ $…(2)$
For body $B$ since the body $B$ is also in equilibrium,
Hence $\mathrm{T}_{1}=\mathrm{W}$ $…(3)$
$(v)$ After solving these equations,
we get $\mathrm{W}=30 \mathrm{N}$