A tank is 180cm long, 90cm deep and 90cm wide at the top, tapering to 30cm wide at the base. If the tank is completely filled with water, calculate:

Question

a) the total weight of the water in the tank
b) The total force exerted by the water on the bottom
c) The total force exerted on one end

I managed to work out parts (a) and (b). However, I have no idea how to solve part (c). Can someone please explain? The answer for part (c) should be 1.99kN

Answer 1

I would integrate the force, in bands.

Consider the taper (we will do one half, then use symettry)..

let the x=0 axis be the center of bottom, so x will go from 0 to 15 cm at the bottom

find the equation of the slope of the side:
y=mx+b
90=(90-0)/(45-15) x + b
90=3x+b
at the bottom:
0=3*15+b or b=-45
y=3x-45
y is height, x is width

darea= width*dh= 2*xwidth*dh=2(h+45)/3*dh

Now weight of water above is (90-h)1g/cm^2*.0098N/g
so that times area is force on one end
integrate from h=0 to 90
force= INT .0098(90-h)*2*(h+45)/3 dh over limits.

force= .0098*2/3 INT (90-h)(h+45) dh
= ( ) int 90h+90*45-h^2-45h dh
= ( ) (45h^2 +90*45h-h^3/3-22.5h^2 over limits

Well, without a calc handy, I leave it to you, check my algebra and thinking, I did it in my head (not on paper), so the chance of an error is high.

Answer 2

all solns to this please