Min Force Req !
Two hemispheres of radius R combined together by two minimum forces ' F ' to form a sphere (completely filled with water) as shown in figure. A small pin hole of radius a at top of the sphere as shown in figure. If the system is in equilibrium at tan θ = 2 x . Find the value of x .
(Gravity is present and no other support is given to hemispheres)
The answer is 3.
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1 solution
How do you arrive at the expression for F sin θ ?
@Spandan Senapati how did he directly found Fsin(theta)? I did the long way by integration ( though not that long)
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I too did by integration.although the title of the problem is a bit underrated.When he drew fbd of the hemisphere he neglected the Normal force on it due to the other hemisphere.I think thats on account of the minimum condition given in the question.Any better reasons?
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Your reasoning according to me seems to correct.
Btw since there is a small hole at top, atmospheric pressure also comes into play, so this result has neglected any atmospheric pressure.
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@Harsh Shrivastava – Na,atmospheric pressure if considered cancels out coz,it will act both from inside P o + h ρ g ,as well from outside.
Yep,that's clever.What,i think,is consider 2 symmetric points(area=dA) about the diameter,so one is x above it and other is x below it.So sum of pressures at these points is- ρ g ( R − x + R + x ) = 2 R ρ g =constant.So simply for force f = 2 R ρ g A , A is area of half circle...If you did by integration,the process you do can be minimised F = ∫ ρ g R c o s 2 θ ( 1 − s i n θ ) d θ .limits from ( − π / 2 to + π / 2 ),Now 2 F = ∫ ρ g R c o s 2 θ d θ ,this step is the same to our previous method of considering symmetric points.
The expression of Fsin(theta) comes easily by using centre of mass logic and it’s ( rho ) g Ycm. ( area) , it’s a general result you can avoid long integrations .