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.
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...
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...
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.
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 =...
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
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.
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.