Force of Wind from these variables

[Jetflyer0]

Homework Statement

I have a strip of foil with an area of 10 square cm and a mass of .561392 grams. The length of the foil is 5 cm and the width is 2 cm. The foil is hanging from a pivot point at the very top. The foil rotates 17 degrees backwards from where it started. The foil is in a sealed container where no outside factors are acting on it except for this wind. I cannot figure out how much force the wind that is pushing the foil back is exerting. Any help would be greatly appreciated. Thank you.

Homework Equations

F=ma

The Attempt at a Solution

I have tried to find ways to solve this, but none have worked. The wind speed is so slight, an extremely sensitive anemometer won't pick it up.

 

[srecko97]

Think about the first Newton's law instead of the second. Provide us with yor sketch with all the forces and angles.

 

[Jetflyer0]

Think about the first Newton's law instead of the second. Provide us with yor sketch with all the forces and angles.

Sorry it's crap, had to do on a computer using paint

 

[srecko97]

The foil is not moving if the wind is constant, right? So the vectorial sum of all the forces on the foil should be zero. (This is the 1st Newton's law). Is the sum really zero in your sketch? Is there no other force?

 

[CWatters]

Your sketch shows Fg acting in mid air. Where does it really act? Show it in that position with dimensions.

 

[Jetflyer0]

Your sketch shows Fg acting in mid air. Where does it really act? Show it in that position with dimensions.

The wind acts on the whole piece

 

[srecko97]

Where is the point of application?

 

[Jetflyer0]

I don't know, the wind is from a fan, and by the time the wind hit the foil, the wind was greatly dispersed

 

[srecko97]

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? And do not forget, one force is still missing in your sketch.

 

[srecko97]

please provide us with a new sketch.

 

[Jetflyer0]

Im assuming the point is in the center, and I am pretty sure I have all the forces

 

[srecko97]

Is the sum of Fwind and Fg zero?

 

[Jetflyer0]

At the time it stabilizes at 17 degrees

 

[Jetflyer0]

The force is not the force of gravity acting on it either, because the wind is blowing horizontally instead of vertically

 

[srecko97]

Which force prevents the foil from falling down (because of the Fg) and to being blown away (because of the Fwind)? Where is its point of application?

 

[Jetflyer0]

got it

 

[Jetflyer0]

 

[srecko97]

Such sketches are useless.

 

[Jetflyer0]

Such sketches are useless.

Then why did you want me to add the force?

 

[CWatters]

The wind acts on the whole piece

Fg is the force of gravity not the wind.

 

[srecko97]

Look, this is a useful sketch. You need to write F_ceiling as a sum of F_ceilingX in x direction and F_ceilingY in y direction. Use sine and cosine. Then you need to eqaulise forces in x and y.

 

[CWatters]

Then why did you want me to add the force?

I think he means your sketch is poor.

As I said before... Where does the force of gravity act? Hint: It does not act in mid air.

 

[CWatters]

I think I would sum the torques to zero.

 

[srecko97]

As I see he has problems even with elementary forces, I assume that he is not familiar with torques. And the result can be found from writing only forces equilibrium as well...
from F_ceiling * cos 17 = F_g you get F_ceiling ...
F_wind = F_ceiling * sin17

 

[CWatters]

As I see he has problems even with elementary forces, I assume that he is not familiar with torques. And the result can be found from writing only forces equilibrium as well...
from F_ceiling * cos 17 = F_g you get F_ceiling ...
F_wind = F_ceiling * sin17

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

 

[Jetflyer0]

I understand that you guys know more about how to solve this problem, but I am a high school student who is new to physics. I am also taking geometry this year, and we have not yet learned about sine or cos. If you could explain a way to solve this that would be easier to understand, I would be very grateful

 

[srecko97]

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

There would be no equilibrium of forces if the angle was not the same.

continuation of my way:
F_ceiling * cos 17 =F_g
F_ceiling = F_g / cos 17
F_wind = F_ceiling * sin 17 = F_g * sin 17 / cos 17 = F_g * tan 17

calculating with torques:
r1*F_wind* cos 17 = r1 * F_g * sin 17
F_wind* cos 17 = F_g * sin 17
F_wind = F_g * sin 17 / cos 17 = Fg * tan 17

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

I understand that you guys know more about how to solve this problem, but I am a high school student who is new to physics. I am also taking geometry this year, and we have not yet learned about sine or cos. If you could explain a way to solve this that would be easier to understand, I would be very grateful

If so, I think you have set yourself a too difficult problem.

 

[Nidum]

The force on the ceiling has little to do with solving this problem ?

All you actually have is a top hinged vane with wind pushing it to the left and gravity force pulling it downwards . @CWatters gave you the clue in post #22 .

You are not asked to work out how the wind generates the force so we can reasonably say that there is just an effective wind force acting at one point on the vane . Strictly we should investigate to find where that point is but for this problem just take it as being at the mid height point of the vane .

How to solve the problem without trig ? It can be done by using a graphic calculation method . Do you know how to use the parallelogram of forces ?

 

[srecko97]

Some helpful links:
trigonometry:
https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-trig-ratios-intro/v/basic-trigonometry
https://en.wikipedia.org/wiki/Trigonometry#Overview
vectors (forces are vectors):
http://www.physicsclassroom.com/class/vectors/Lesson-1/Vector-Addition

 

[srecko97]

The force on the ceiling has little to do with solving this problem ?

If there would be no F_ceiling (ceiling acts on the vane), which force would prevent the vane from falling down and being blown away? Maybe it would be better to name it F_reaction instead of F_ceiling, but name is not so important.

 

[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

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