Physics, asked by PhysicsHelper, 1 year ago

Water is filled in a rectangular tank of size 3 m x 2 m 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 ẟx meter in this side, situated at a depth of x meter from the surface of the water. Find the force by the water on this strip. (c) Find the torque of the force calculated in part (b) about the bottom of 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.

Answers

Answered by tiwaavi
57

(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 × B

= 3 m × 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 = 2 × x


The force on this strip F =Area*pressure


=2 × x²ρg

=2 × x² × 1000 × 10

=20000x² N

___________________________

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


= 20000x × x × (1-x) Nm


= 20000 × x(1-x) × x Nm

________________________

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]        

=ρgH²


=1000 × 10 × 1²


=10000 N



______________________________

(e) Total Force = 1/2 × ρgH × H × 2

= 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 Nm


= 10000/3 Nm

_____________________________


Hope it helps.

Answered by lava88
17
(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 × B

= 3 m × 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 = 2 × x


The force on this strip F =Area*pressure


=2 × x²ρg

=2 × x² × 1000 × 10

=20000x² N

___________________________

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


= 20000x × x × (1-x) Nm


= 20000 × x(1-x) × x Nm

________________________

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]        

=ρgH²


=1000 × 10 × 1²


=10000 N



______________________________

(e) Total Force = 1/2 × ρgH × H × 2

= 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 Nm


= 10000/3 Nm

_____________________________


Hope it helps.

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