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anonymous
 3 years ago
FLUID MECHANICS QUESTION:
A cylinder with liquid in it (not till the brim) is kept on a wedge(angle with horizontal is theta degrees) with friction coeff 'x'...is accelerating downwards. then the constant angle made by the surface of water with the incline will be????
anonymous
 3 years ago
FLUID MECHANICS QUESTION: A cylinder with liquid in it (not till the brim) is kept on a wedge(angle with horizontal is theta degrees) with friction coeff 'x'...is accelerating downwards. then the constant angle made by the surface of water with the incline will be????

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What you need first, is to find the acceleration of the system.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah i have tried tat ..bt the ans is not right could u please give the eqs for few steps Vincent

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What is your expression for the acceleration?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Using free body diagram we can find acceleration ,a=gsin(theta)xgcos(theta) Let the angle made by the surface with the incline be @. Now refer to the diagram in the file attached. We have considered the a water column of length L and area of cross section A. Let the two ends of the cross section be at a depth of h1 &h2 as shown. Pressure on the area at depth h1=p1=P+h1*d*gcos(theta) , where P is the atmospheric pressure Similarly pressure at depth h2=p2=P+h2*d*gcos(theta) Force at area A at depth h1=p1*A Force at area A at depth h2=p2*A Net force on the water column=F2F1=(h2h1)*d*gcos(theta) F2=F1=m*a=L*A*d*(gsin(theta)xgcos(theta)) Hence (h2h1)*d*gcos(theta) = L*A*d*(gsin(theta)xgcos(theta)) Rearranging and cancelling the like terms, (h2h1)/L=tan(theta)x Using trigonometry, tan@=tan(theta)x ****NOTE THAT tan(theta)>x [OTHERWISE CYLINDER WILL NOT SLIDE]. tan@ IS POSITIVE AND @ IS ACUTE ANGLE ******

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ Vincent a=gsintheta ugcostheta

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@ diwakar ..that seems to be right ...bt i got the answer from other method..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0At rest the angle between the surface of the water and the inclined plane would equal 180 deg.  angle of the wedge to the horizontal (theta). to that you would have to add the change in the angle of the water column due to acceleration. since that isn't given you would have to make numerous assumptions. Material that the wedge is made of, material that the cylider holding the water is made of, magnitude and direction of force causing the acceleration.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0tan inverse (sin(theta)  xcos(theta))/cos(theta)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0remember the fact that net force on a fluid is always perpendicular to it surface........ find all forces and then find the angle made by their resultant with surface of incline...... btw sorry the ans is 90  previous ans....
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