Force of Wind from these variables

[CWatters]

I do not see any difference, CWatters. I do not understand your scepticism

My first instinct was to assume the angle of the reaction force at the ceiling is unknown, then prove its also 17 degrees. But how can you do that without writing a torque equation?...

Vertically you have
F_ceiling * cos(θ) - F_g = 0

Horizontally you have
F_ceiling * sin(θ) - F_wind = 0

Three unknowns and only two equations so the best I can get to is..

Tan(θ) = F_wind/F_g

 

[srecko97]

My first instinct was to assume the angle of the reaction force at the ceiling is unknown, then prove its also 17 degrees. But how can you do that without writing a torque equation?...

Vertically you have
F_ceiling * cos(θ) - F_g = 0

Horizontally you have
F_ceiling * sin(θ) - F_wind = 0

Three unknowns and only two equations so the best I can get to is..

Tan(θ) = F_wind/F_g

$\theta$ and $F_g$ are known ... So it is not hard to get $F_{wind}$

 

[Jetflyer0]

Tan(θ) = F_wind/F_g

I put this into an equation calculator with F_g equal to 0.0055016416 (mass of foil * 9.8 for force of gravity) and F_wind equal to x, and the calculator returned a value of 0 for the F_wind

 

[srecko97]

My first instinct was to assume the angle of the reaction force at the ceiling is unknown, then prove its also 17 degrees. But how can you do that without writing a torque equation?...

Vertically you have
F_ceiling * cos(θ) - F_g = 0

Horizontally you have
F_ceiling * sin(θ) - F_wind = 0

Three unknowns and only two equations so the best I can get to is..

Tan(θ) = F_wind/F_g

Well, everything I know about $F_{ceiling}$ is that it is the only force acting on a foil besides $F_{wind}$ and $F_g$. Its $x$ component $F_{ceiling X}$ must be equal (size) to $F_{wind}$ and $y$ component $F_{ceiling Y}$ must be equal (size) to $F_g$ in order to satisfy forces equlibrium. It is obvious then that the angle is the same. We really do not need to talk about torques here.

 

[srecko97]

I put this into an equation calculator with F_g equal to 0.0055016416 (mass of foil * 9.8 for force of gravity) and F_wind equal to x, and the calculator returned a value of 0 for the F_wind

$F_{wind}= m \cdot g \cdot \tan{\alpha} = 0.00056$ $kg$ $\cdot \tan{17 °}=0.000171$ $N$

 

[Jetflyer0]

Thanks, I haven't learned much about this yet, but this was helpful.

 

[Jetflyer0]

I didn't think this was possible without the acceleration of the foil

 

[David Lewis]

Acceleration doesn't come into play here. This is a static equilibrium free-body analysis. The weight of the foil should be written out to only 2 or 3 significant figures to match the precision of other variables. The point of application for the aerodynamic force will be a little above the centroid of the foil due to endplate effect and tip losses, but a fine point which I'm sure is not worth worrying about.

 

[haruspex]

I'll have to think about that. The angle of the reaction force at the ceiling isn't know

Handy rule: if three forces are in static equilibrium they must act through a common point. But...

My first instinct was to assume the angle of the reaction force at the ceiling is unknown, then prove its also 17 degrees. But how can you do that without writing a torque equation?...

Yes, it follows from the torque equilibrium. It cannot be proved otherwise.

 

[haruspex]

Wind does acts on every small piece of the foil, but the effect is the same if we assume it acts in one point? where is this point?

Strictly speaking, it is not necessarily in the centre. The pressure distribution across the surface will not be uniform.

 

[srecko97]

I do not understand why we need to talk about torque equilibrium. What is wrong with my deliberation:
I know the angle, I know the $F_g$, I know the direction of $F_{wind}$. There should be one more force to satisfy the forces equilibrium. I call this unknown force $F_{ceiling}$ ...(read the quote below)

Well, everything I know about $F_{ceiling}$ is that it is the only force acting on a foil besides $F_{wind}$ and $F_g$. Its $x$ component $F_{ceiling X}$ must be equal (size) to $F_{wind}$ and $y$ component $F_{ceiling Y}$ must be equal (size) to $F_g$ in order to satisfy forces equlibrium. It is obvious then that the angle is the same. We really do not need to talk about torques here.

 

[haruspex]

I do not understand why we need to talk about torque equilibrium. What is wrong with my deliberation:
I know the angle, I know the $F_g$, I know the direction of $F_{wind}$. There should be one more force to satisfy the forces equilibrium. I call this unknown force $F_{ceiling}$ ...(read the quote below)

View attachment 215177

Suppose instead that the hinge has some frictional torque. Now the angle of the reaction force cannot be in line with the plate. So your assumption that the angle is the same must effectively be using a torque balance equation.

 

[David Lewis]

I know the direction of F_wind.

You know the (free stream) direction of the wind. The presence of the foil is going to change it. The component of aerodynamic force parallel with the wind is drag, but you also have a perpendicular force component (lift).

 

[srecko97]

Ok, ok, ok, I know everything what you David Lewis and Haruspex have told me, but this task does not have enough given variables to consider all this. Do not forget, it is a high school task.