force on a submerged object?
a cylindrical drum with no top of height 3 meters and radiius 2 meters is half filled with water and capped. if the drum is turned to a horizontal position, what is the force on the cap?
b. if the drum were completely filled before being capped and turned what would the force be on the cap?
we have to use some formula ∫p.x.s(x)dx
from a to b where s(x) is the width of the object... I am confused as to how do I find the width of the object? i keep on thinking it is a 3D problem but in my book they make it seem like it is 2D.. the answer for a is 156 800/3, p is a constant = 9800 ..... and for b is 8*9800*pi
i actually dont understand anything. any help will do..(Headbang)(Crying)
Re: force on a submerged object?
Here's a diagram of the drum half full lying horizontally.
A point on the circle an angle a from the horizontal has the coordinate (2cos(a), 2sin(a)). The horizontal distance from the origin is 2cos(a) so the width of the circle is 4cos(a).
Note that in your function S(x) the variable x is the vertical distance.
You have the function for distance as a function of angle, now find it as a function of x
Also remember when doing the integration that the 0 point for x is the centre of the circle not the ground point.