Hydrodynamics problem

  • Thread starter Waxterzz
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  • #1
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Hi fellas

1. Homework Statement

a wooden stick , length 4 meters , density = 600 kg/m³ and thickness = 225 cm² is attached on a hinge 1 meter above water

so the point is u got to find the angle the stick makes with the horizontal water

you will see the drawing in the jpg file

3. attempt
none , dont know where to begin
but i tried to set up an equitation of this system , i dont know how to involve the hydromechanical part in it
 

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Answers and Replies

  • #2
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your attachment is pending, so i ll give you a general hint.
the stick floats at an angle to the water surface, assume the angle as theta, you know the center of mass of the stick, find the center of bouyancy(its midway of the submerged part of the stick, you know the height of water(i am assuming this) you have assumed the angle, find the submerged length, simple trigo). take the moments about the point of contact of the stick with the container(reaction passes through that point, so it wont have any moment). solve
 
  • #3
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hi , thx for your reaction
but what do u mean with you know the height of water , u only know the lenght of the stick and the distance between the hingle and the water surface
 
  • #4
738
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distance between the hinge and water surface = height of water(sorry for silly terminology)

if theta is the angle to the vertical, cos (theta) = water height/length of submerged part of stick
 
  • #5
82
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actually its the angle to the horizontal :)

the sketch , still pending , will make things clear

everything about this problem is bothering me

is this basic idea right :

momentum of the non under water part = momentum underwater part
G pulls the non underwater part downwards and Archimedes pushes the underwater part upwards ?

so the weigth of the non underwater part multiplicate with the length of the non underwater part = archimedes force multiplicate the length of the non under water part

damn hydraulics
 
  • #6
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some part right but a large part wrong
you are confusing bw moment and momentum, they are different. here moment is considered
moment of the whole stick weight(weight acting at CoM, gravitational force) = moment of bouyant underwater part(bouyant force acting at midway of underwater part, archimedes force)
 
  • #7
82
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but u have to incalculate the moment of the tension force also , becos the stick is attached on an axis

and i guess ill need more equitations for this problem ...
 
  • #8
738
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geez i dont know what tension you are talking about. if you are talkin about the forces developed in the stick itself, dont worry boy, those forces pass through the hinge point, so no moment coz of them
 

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