Question

A mass of $10$ kg is suspended vertically by a rope of length $5$ m from the roof. A force of $30$ N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $θ$ = $tan^{–1} (x × 10^{–1})$. The value of $x$ is _______.(Given, $g = 10\; m/s^2$)

Answer 1

The value of $x$ is $\underline{3}$

$T$ $cosθ = mg$

$T$ $cosθ$ = $100$ $N$ …(i)

$T$ $sinθ$ = $30$ …(ii)

$⇒\frac{Tsin\;θ}{Tcos\;θ}=\frac{30}{100}$

$⇒tan\;θ=\frac{3}{10}$

$∴ x = 3$