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Water is filled in a rectangular tank of size 3m x 2m x 1 m. Find the total force exerted by the water on the bottom surface of the tank.
Question :
Water is filled in a rectangular tank of size 3m x 2m x 1 m.
(a) Find the total force exerted by the water on the bottom surface of the tank.
(b) Consider a vertical side of area 2 m x 1 m. Take a horizontal strip of width ox metre in this side, situated at a depth of x metre from the surface of water. Find the force by the water on this strip.
(c) Find the torque of the force calculated in part (d) about the bottom edge of this side.
(d) Find the total force by the water on this side.
(e) Find the total torque by the water on the side about the bottom edge.
Neglect the atmospheric pressure and take g = 10 m/s2.
Answer :
Water is filled in a rectangular tank of size 3m x 2m x 1 m.
(a) Find the total force exerted by the water on the bottom surface of the tank.
(b) Consider a vertical side of area 2 m x 1 m. Take a horizontal strip of width ox metre in this side, situated at a depth of x metre from the surface of water. Find the force by the water on this strip.
(c) Find the torque of the force calculated in part (d) about the bottom edge of this side.
(d) Find the total force by the water on this side.
(e) Find the total torque by the water on the side about the bottom edge.
Neglect the atmospheric pressure and take g = 10 m/s2.
Answer :
( a) Given dimensions of the
tank ,
Length (L) = 3 m,
Breadth (B) = 2 m,
Height (H) = 1 m.
As 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
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