(a) The dimensions of the tank are Length L = 3 m,
Breadth B = 2 m, Height H = 1 m.
Since the tank is filled, the height of water in the tank = H = 1 m.
The pressure of water at the bottom = ρgH
=1000*10*1 =10000 N/m²
Area of the bottom of the tank = L x B
= 3 m x 2 m
= 6 m²
So the force on the bottom
= Pressure * Area =10000*6
= 60000 N
(b) The pressure at x m below the surface of the water = ρgx
Area of a horizontal strip ẟx wide and 2 m length =2*ẟx
The force on this strip F =Area*pressure
=2* ẟx*ρgx
=2*ẟx*1000*10*x
=20000.x.ẟx N
(c) The perpendicular distance of this force from the bottom of edge of this side = H-x =1-x m
Hence the torque of the force = Force*perpendicular distance
=20000.x.ẟx*(1-x) N-m
=20000.x(1-x)ẟx N-m
(d) The total force of the water on this side =∫Fda, {from x=0 to x=1 m, where F is the pressure at the depth x m and da=area of the strip}
=∫ρgx*(2*dx) {dx is the width of the strip}
=2ρg∫x.dx
=2ρg*[x²/2] {Limit from x=0 to x=H=1 m}
=ρgH²
=1000*10*1²
=10000 N
(e) The variation of the pressure will be triangular as shown in the figure.
We can also find out the force on the side of the tank by calculating the area of this triangle.
Total Force =½*ρgH*H*(2 m)
=ρgH² =1000*10*1² =10000 N
The height of this resultant force from the bottom of edge of the side will be one-third of H (height of the C.G. from the bottom) =H/3
Hence the torque =10000*H/3
=10000*1/3 N-m
=10000/3 N-m
