Question

Consider a rectangular tank of size (l×b) filled with a liquid of density ρ, to a height H, as shown in the figure. Find the ratio of the force acting on the base to that on the walls of the tank.

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Solution

The correct option is **B** 2lH

Whenever a liquid comes in contact with solid boundaries, it exerts a force on it. The force on the boundary may be obtained by integrating the pressure over the entire area of the boundary. The variation of liquid pressure, acting on the base (diagram (a) and on the wall (diagram (b), are shown below.

**(i) Force at the base **

Since the pressure is uniform at the base, force acting at the base is given by,

F=P× ( area of the base )

Since P=ρgH

F=ρgH(l×b)=ρg(lbH)

Since lbH=V ( Volume of the liquid ), so

F=ρgV = weight of the liquid inside the tank

**(ii) Force acting on the vertical wall**

Pressure acting on the vertical wall is not uniform but increases linearly with depth. Pressure at a depth h, from the free surface, is P=ρgh.

Force dF acting on a differential element of height dh is,

dF=P(b dh)=(ρgh)(b dh)=ρgbh dh

The total force is F=ρgb∫H0hdh=12ρgbH2

The total force acting per unit width of the vertical wall is,

Fb=12ρgH2

the ratio of forces is, =ρg(lbH)12ρgbH2=2lH

Hence, (B) is the correct answer.

Why this Question ?Thrust at the vertical surface :On a vertical surface in contact with the liquid, the pressure is not the same at all points. Points at greater depth experience the larger pressures and hence greater thrust.Fb=12(base)(height)=12(ρgH)(H)F [Pressure at CG of vertical surface×[Area of surface]

Whenever a liquid comes in contact with solid boundaries, it exerts a force on it. The force on the boundary may be obtained by integrating the pressure over the entire area of the boundary. The variation of liquid pressure, acting on the base (diagram (a) and on the wall (diagram (b), are shown below.

Since the pressure is uniform at the base, force acting at the base is given by,

F=P× ( area of the base )

Since P=ρgH

F=ρgH(l×b)=ρg(lbH)

Since lbH=V ( Volume of the liquid ), so

F=ρgV = weight of the liquid inside the tank

Pressure acting on the vertical wall is not uniform but increases linearly with depth. Pressure at a depth h, from the free surface, is P=ρgh.

Force dF acting on a differential element of height dh is,

dF=P(b dh)=(ρgh)(b dh)=ρgbh dh

The total force is F=ρgb∫H0hdh=12ρgbH2

The total force acting per unit width of the vertical wall is,

Fb=12ρgH2

the ratio of forces is, =ρg(lbH)12ρgbH2=2lH

Hence, (B) is the correct answer.

Why this Question ?Thrust at the vertical surface :On a vertical surface in contact with the liquid, the pressure is not the same at all points. Points at greater depth experience the larger pressures and hence greater thrust.Fb=12(base)(height)=12(ρgH)(H)F [Pressure at CG of vertical surface×[Area of surface]

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