Question

A tank contains water upto a depth of 1.5 m. The length and width of the tank are 4m and 2m respectively. The tank is moving up an inclined plane with a constant acceleration of $4m/s^2$. The inclination of the plane with the horizontal is ${30}^0$. Find the pressure at the botton of the tank at the front end.

• A. $10787.2N/m^2$
• B. $31457.9N/m^2$
• C. $25431.6N/m^2$
• D. $42561.3$

Answer 1

The correct option is A $10787.2N/m^2$

Depth $\left(h\right)=1.5m$, Length $\left(L\right)=4m$, Breadth $\left(B\right)=2m$

acceleration $\left(a\right)=$$4m/s^2$

Inclination $\alpha$$=30$

$P_F$= Pressure at bottom of tank at front end.

Horizontal component of acceleration $a_x=a\cos \alpha$$=3.464$

Vertical component of accleration $a_y=a\sin \alpha$=2

As the tank moves upward, the free surface of the water tank makes angle $\theta$ with horizontal which is given by

$\tan \theta =\frac{a_x}{a_y+g}=0.3$

$\theta =16.34$

Since the free surface makes an angle with the horizontal tank, so, the water has risen by some height at the rear end and fall by some height at front end.

Let $h_0$ be that raised height.

Then $\frac{h}{d}=\tan \theta$ where $d=\frac{L}{2}=2m$

Therefore , $h_0=2\tan \theta =0.58$

Total height on the front side $h_2=1.5-h_0=0.91m$

Pressure at front end bottom = $P_F=\rho g{h}_2(1+\frac{a_y}{g})=1000\times 9.81\times 0.91(1+\frac{2}{9.81})=10787.2N/m^2$