Question

a vessel containing a liquid has a constant acceleration of 19.6 metres per second square in horizontal direction. the free surface of water get sloped with horizontal at angle

Answer 1

The free surface of water get sloped with horizontal at angle

26.5 degrees

Answer 2

The free surface of water get sloped with horizontal at angle is $\bold{\sin^{-1}(\frac{2}{\sqrt{5}})}$

Given:

Acceleration of the liquid = a = 19.6 $\bold{m/s^2}$

To find:

The angle of the water = $\bold{\theta}$ = ?

Formula used:

$\bold{\sin\theta = \frac{Opposite \ side}{Hypotenuse}}$

Solution:

Refer the image given below.

$\bold{\sin\theta = \frac{Opposite side}{Hypotenuse}}$

$\bold{\sin\theta=\frac{a}{\sqrt{a^2+g^2}}}$

Where,

g = Acceleration due to gravity = $\bold{20 \ m/s^2}$

Now, let the acceleration be:

$\bold{a=19. 6 \ m/s^2 \approx 20 \ m/s^{2}}$

$\bold{\sin\theta=\frac{20}{\sqrt{20^2+10^2}}}$

$\bold{\sin\theta=\frac{20}{\sqrt{500}}}$

$\bold{\sin\theta= \frac{2}{\sqrt{5}}}$

$\bold{\therefore \theta=\sin^{-1}(\frac{2}{\sqrt{5}})}$