Question
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:
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
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
Answers
bobpursley
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.
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.
myles
all solns to this please