Question

in the fig, the tension in the horizontal cord is 30N Find the weight of the body B​

Answer 1

$\displaystyle\large\boxed{\sf{(B)\quad\!30\ N}}$

Let the tension in the cord 1 = $\displaystyle\sf{T_1.}$

Let the tension in the cord 2 = $\displaystyle\sf{T_2.}$

Since cord 2 makes an angle 45° with the vertical, the vertical component of the tension in it will be $\displaystyle\sf{T_2\cos45^{\circ},}$ upwards opposite to $\displaystyle\sf{T_1,}$ and the horizontal component is $\displaystyle\sf{T_2\sin45^{\circ},}$ leftwards opposite to 30 N.

The weight of the body B is balanced by the tension in the cord 1, i.e.,

$\displaystyle\longrightarrow\sf{W_B=T_1\quad\quad\dots(1)}$

Since the point of intersection is in equilibrium, the net horizontal force acting on the point is zero. Thus,

$\displaystyle\longrightarrow\sf{T_2\cos45^o=T_1}$

From (1),

$\displaystyle\longrightarrow\sf{W_B=T_2\cos45^o}$

$\displaystyle\longrightarrow\sf{W_B=\dfrac{T_2}{\sqrt2}\quad\quad\dots(2)}$

And the net vertical force acting on it is also zero. Thus,

$\displaystyle\longrightarrow\sf{T_2\sin45^o=30\ N}$

$\displaystyle\longrightarrow\sf{\dfrac{T_2}{\sqrt2}=30\ N}$

Therefore (2) becomes,

$\displaystyle\longrightarrow\sf{\underline{\underline{W_B=30\ N}}}$