Min Force Req !

Two hemispheres of radius R R combined together by two minimum forces ' F F ' to form a sphere (completely filled with water) as shown in figure. A small pin hole of radius a a at top of the sphere as shown in figure. If the system is in equilibrium at tan θ = x 2 \tan \theta = \frac{x}{2} . Find the value of x x .

(Gravity is present and no other support is given to hemispheres)


The answer is 3.

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1 solution

Nishant Rai
May 28, 2015

How do you arrive at the expression for F sin θ F\sin \theta ?

Shaun Leong - 4 years, 11 months ago

@Spandan Senapati how did he directly found Fsin(theta)? I did the long way by integration ( though not that long)

Harsh Shrivastava - 3 years, 9 months ago

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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?

Spandan Senapati - 3 years, 9 months ago

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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.

Harsh Shrivastava - 3 years, 9 months ago

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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 P_{o}+h\rho g ,as well from outside.

Spandan Senapati - 3 years, 9 months ago

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 ) \rho g(R-x+R+x) = 2 R ρ g 2R\rho g =constant.So simply for force f = 2 R ρ g A f=2R\rho gA , 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 F=\int \rho gRcos^2\theta (1-sin\theta )d\theta .limits from ( π / 2 -π/2 to + π / 2 +π/2 ),Now 2 F = ρ g R c o s 2 θ d θ 2F=\int \rho gRcos^2\theta d\theta ,this step is the same to our previous method of considering symmetric points.

Spandan Senapati - 3 years, 9 months ago

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 .

Nitin Sachan - 1 year, 3 months ago

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