Question

A tank contains water upto a depth of $1.5m$. 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 bottom of the tank at the rear end.

• A. $24642.7N/m^2$
• B. $12003.0N/m^2$
• C. $34562.1N/m^2$
• D. $13256.8N/m^2$

Answer 1

Answer: (A)

$24642.7N/m^2$

Depth $\left(h\right)=1.5m$, Length $\left(L\right)=4m$, Breadth $\left(B\right)=2m$
acceleration $\left(a\right)=4m/s^2$
Inclination $\alpha =30$
$P_R=$ Pressure at bottom of tank at rear end.
Horizontal component of acceleration $a_x=a\cos \alpha$=3.464
Vertical component of acceleration $a_y=a\sin \alpha$=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 rear side $h_1=1.5+h_0=2.08m$
Pressure at rear end bottom = $P_R=\rho g{h}_1(1+\frac{a_y}{g})=1000\times 9.81\times 2.08(1+\frac{2}{9.81})=24642.7N/m^2$