A cylindrical oil storage tank 12 feet in diameter and 17feet long is lying on its side. Suppose the tank is half full of oil weighing 85 lb per cubic foot. What's the total force on one endof the tank?
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I'm currently on the same question after spending over 2 hours on it... What did you get?
12,240 lbs is the answer.
ok, first off, since you only need to consider one face of the cylinder, draw a circle for visualization centered at the point (0,6) on the y axis.
Next, draw a horizontal rectangle (trapezoid, sort of) that extends from the left of the tank, to the right. Call that length "w" and make the height of it "dh" Now, you know the radius is 6, and since such is the case, that means the center of the circle to the outside is 6 also. from such, draw a line, from the center of the circle, to the end of "w" and draw a line vertically downwards to "w"
Now you may notice you have a right triangle, with height (6-h) and base (w/2). Now use the pythagorean thereom to find a function in which w= something with h. So, first you have (6-h)^2+(w/2)^2=6^2. Do some algebra and you get w=2*(-h^2+12h)^(1/2). Ok, from here, you need to look at that slice drawn earlier that has length w and height dh. Assume that it is a rectangle and the formula for area is A=w*dh. now subsititute what w is equal to and you have A=2*(-h^2+12h)^(1/2)*dh
Now, you know F=P*A, so now we need to find P(pressure).
P=weight density * depth
Depth is the distance from the horizontal slab to the surface.
Since the tank is half full, the distance from the horizontal slab to the surface is 6-h.
P=85*(6-h)
so we have both P and A
so F=P*A
dF=85*(6-h)*2*(-h^2+12h)^(1/2)*dh, but now, you need to take the integral from 0 to 6
integrate dF from 0 to 6 and you get 12240 lbs.
Alternatively a simpler solution is to draw the circle centered at the origin, so that the surface of the water is at the x axis. The oil is half full below the x axis.
Then
dF=85*(-y)*2*(36 - y^2)^(1/2)*dh
integrate dF from -6 to 0 and you get 12240 lbs.
Next, draw a horizontal rectangle (trapezoid, sort of) that extends from the left of the tank, to the right. Call that length "w" and make the height of it "dh" Now, you know the radius is 6, and since such is the case, that means the center of the circle to the outside is 6 also. from such, draw a line, from the center of the circle, to the end of "w" and draw a line vertically downwards to "w"
Now you may notice you have a right triangle, with height (6-h) and base (w/2). Now use the pythagorean thereom to find a function in which w= something with h. So, first you have (6-h)^2+(w/2)^2=6^2. Do some algebra and you get w=2*(-h^2+12h)^(1/2). Ok, from here, you need to look at that slice drawn earlier that has length w and height dh. Assume that it is a rectangle and the formula for area is A=w*dh. now subsititute what w is equal to and you have A=2*(-h^2+12h)^(1/2)*dh
Now, you know F=P*A, so now we need to find P(pressure).
P=weight density * depth
Depth is the distance from the horizontal slab to the surface.
Since the tank is half full, the distance from the horizontal slab to the surface is 6-h.
P=85*(6-h)
so we have both P and A
so F=P*A
dF=85*(6-h)*2*(-h^2+12h)^(1/2)*dh, but now, you need to take the integral from 0 to 6
integrate dF from 0 to 6 and you get 12240 lbs.
Alternatively a simpler solution is to draw the circle centered at the origin, so that the surface of the water is at the x axis. The oil is half full below the x axis.
Then
dF=85*(-y)*2*(36 - y^2)^(1/2)*dh
integrate dF from -6 to 0 and you get 12240 lbs.
'the oil is half full below the x axis'
meant to say the cylinder is half full
meant to say the cylinder is half full
1 answer