find the net downward thrust on the curved surface of a bucket full of water of height h nd radii a nd b ???

i didnt quite understand ur Q. brother....but am still jus. trying......

the net force on the water column=0.........ie forces on water column are mg, normal force exerted by the bottom of the bucket, and the force exerted by the curved surface........and im asked to find out the force exerted on the curved surface downward......=force exerted by the curved surface on the liq. column in upward dirn......

normal force at bottom=force exerted by water column on the bottom(by 3rd law)=(P+Hdg)pi*b2......here P= atmospheric pressure......

as we said net upward force on the liq. column=0

=> mg=(P+Hdg)*pi*b²+ N_y............where we are interested in finding N_y........=mg-(P+Hdg)*pi*b²........

now m cud be found out by using V*d......and finding out V in terms of a and b is the donkey work which one have to do patiently.......hope my soln. is someway related to ur Q. if not nudge me.......

rate if useful.......

Its a huge solution, and so is the answer..

Answer is :

 2 * pi *(b-a) * h * (2a + b)/6 ....if you take density as 1...

I'll tell you how to do it..

Consider an element of d liquid at a depth x below surface.Let thickness of this be dx.

(Just like in moment of inertia derivations)

Find area of the element(dA) ....consider the trapezium as a rectangle..you get 2pi*r *dx/cos@...

@ is inclination of bucket wall...

Now find thrust on curved surface by this element usinf dF = dA * x*g(* d)...........where d is density..

Now integrate from 0 to h...to get DOWNWARD thrust..ie F = <integral > dF sin@.     ........sin@ since u need downward thrust..

Till here its straightforward..here you run into trouble.. 

u get a tan@ term...draw the figure and substitute values for that and r...

Solve and you get the answer..

Giving figure..

Red part is the considered element..

tan @ =(b - a ) /  h =(b - r) / x...        (REFER FIGURE)

Solve these two and you get r = b- (b - a)* x/h...substitute this value...

for tan@, substitute (b- a)/h...

.

.

I think that should clear it...

if there's still some doubt, temme,i'll show d integration..

do rate if it helped!

dekh yaara.......not that my answer is wrong.........but that its the shorter method of doin the same thing.......taking the pressure at an element of width dh ............finding the force at that pt and integrating the vertical component of that force from h=0 till h=H toh koyi bhi kar sakta hai........thats the technical soln. as posted very rightly by learner..........a much much simpler short cut will be to do it using the force balancing........net force on the water=0..............so the force due to the wt= normal force given by the ground + force given by the curved wall in the upward direction........normal force by bottom of beaker= Pressure exerted by the water* area of bottom............and wt of the liquid can be gotten by using M=V*d..............now for getting V ull have to do some additional work which i havnt done and left for u to do........REMEMBER i havnt posted the full soln. ........ive jus given a methodology to do the problem.....i cant attach paint files (due to technical probs. on my comp)other wise i wud hav posted a much detailed soln.........u have any debates to make abt this soln.....ur open to come on my nudgebook and i can prove my pt thru arguement............thnku........

Like I said, because of small @, the trapezium can be considered as a rectangle...

Now height of this rectangle is dx/cos@...refer to the figure..see the triangle at the side..

Hypot. is what we need here...since base is dx, hypot. becomes dx/cos@

And CSA of the cylinder is 2*pi* r * h...

h= dx/cos@..

so you get area as 2*pi*r*dx/cos@..

Pramod's method is perfect..though I think both work out to the same length..

For V, you can use this :

V =(pi * h/3)(a² +ab +b²)..

cheers!:-)