Accelerated along inclined plane

Consider a tank open at top, containing a liquid and moving upwards along inclined plane with a uniform acceleration as shown in fig-1(a).

We know that when the tank starts moving, the liquid surface falls down on the front side and rises up on the back side of the tank as shown in fig-1(b).

Let,

• $\phi$ = Inclination of the plane with the horizontal
• $\theta$ = Angle, which the liquid surface makes with the horizontal, and
• $a$ = Acceleration of the tank

Now consider any particle A on the liquid surface. We know that the forces acting on the liquid particle are :

1. Weight of the particle (W) acting vertically down.

2. Accelerating force (F) on an angle $\phi$ with the horizontal.

3. Pressure (P) exerted by the liquid particles normal to the free surface.

Now resolving the accelerating force (F) horizontally and vertically, we get

$F_H = F \cos\phi$ and $F_V = F \sin\phi$

We know that the weight of the particle,

$W = mg$

Similarly, the accelerating force,

$F = ma$

Now resolving all the forces horizontally at A,

$P\sin\theta = F_H = F\cos\phi = ma\cos\phi$

And resolving all the forces vertically at A,

$P\cos\theta = F_V = F\sin\phi + W = ma\sin\phi + mg$

Dividing equation (1) by (2)

$\frac{P\sin\theta}{P\cos\theta} = \frac{ma\cos\phi}{ma\sin\phi + mg} = \frac{a\cos\phi}{a\sin\phi + g}$ $\tan\theta = \frac{a_H}{a_V + g}$

where,

• a\_H = horizontal component of the acceleration
• a\_V = vertical component of the acceleration