Mech & a little fluids

[psyclones]

Hi,
This problem has been driving me mad!
Can anyone simplify the physics of this problem? Because I can't solve due to to many unknowns.
Question ref: Engineering Materials, Benham, Crawford & Armstrong.
Please refer to attachments.

[imatfaal]

I don't understand your quantification and direction of Buoyancy.

 

1. Buoyancy Force is opposite to gravity and not perpendicular to beam as you have drawn it

 

2. The Force that buoyancy exerts is equal to the gravitational force of a similar volume of water

 

[latex]F_{buoyancy} = \rho_{liquid}.V_{displaced}.g[/latex]

I am a little unsure of your classification of centre of gravity but you also need to consider centre of buoyancy; the single point at which buoyancy can be seen to act (CoB) does not necessarily coincide with the single point at which gravity acts (CoG) this is a major point of ship stability - in this case they obviously do not coincide.

I also do not see the need for your F_a and do not reasily understand which force this actually is. The beam is at equilibrium so I do not see the existence of F_a

I haven't had time to think too hard on the problem but at first glance I would think of using rotational mechanics rather than linear. You know that if the beam is at rest that there is no net torque (I use torque in the modern/American sense before Studiot weighs in with "correct" versions - which you can read about here ) AND because you get to choose where you do your calculations you can choose the pivot point A and thus ignore the forces there as they have no influence on the torque (forces which act through a pivot cannot affect the net torque). You can quantify the torque from gravity and from buoyancy - they will be equal and opposite at rest and you can find what angle this occurs

[elfmotat]

Indeed. The toughest part of this problem seems to be determining the submerged volume as a function of the angle, and the point through which the buoyant force acts.

Edited November 25, 2014 by elfmotat

[psyclones]

Thankyou both for your posts,

 

I agree, we could probably put aside the reaction forces at A, I only included them for completeness. I after some revision, I realised the direction and position of the buoyancy force was incorrect,

 

I think I found a way around this (refer to attached). Could we employ a sum of moments of area method to calculate the centroid distance? (Ignoring effects due to F_wa).

 

If this (sum of moments of area) isn't valid, then we have to use sum of moments (including force due to water), which will introduce another unknown, the CoB centroid.

 

Edited November 27, 2014 by psyclones

[imatfaal]

I would approach as follows. First off there are three forces to reckon with

 

Gravity

Buoyancy - which can be split into two - from large rectangular section under water and from triangular section underwater

 

The moments of those three forces are tough. Just the geometry is nasty. Pretty much sorted on this next diagram

 

 

 

The second diagram shows the triangle formed by the beam with the water. The inset triangles A and B show respectively the small triangle at the pivot and the triangle that we will calc seperately immediately under water.

I have to head off to lunch in a moment - but I am pretty sure that from the above I can quantify (in terms of theta) the forces involved. Then we can set opposing forces to be equal ( it is at rest) and solve for theta

 

using grams and centimetres

 

F_gravity = mass x acceleration due to gravity

 

mass = volume x density

mass = volume x specific density x density of water

mass = 400*15*15*0.6*1

 

F_grav = 54000g

 

R_grav = Perpendicular Distance from pivot to Point of Action of F_gravity

R_grav = P -> CofG

R_grav = 200

 

Torque_gravity = (Perpendicular Component of F_grav) * R_grav

Torque_gravity = 54000gCos(Theta)

 

(above not strictly true as the little triangle A is to the right of the pivot and thus should also be taken into account)