Max Wind Speed for Small Animals to Stay Upright

[Cypripedium]

Homework Statement

The posture of small animals may prevent them from being blown over by the wind. For example, with wind blowing from the side, a small insect stands with bent legs; the more bent the legs, the lower the body, and the smaller the angle . The wind exerts a force on the insect, which causes a torque about the point where the downwind feet touch. The torque due to the weight of the insect must be equal and opposite to keep the insect from being blown over. For example, the drag force on a blowfly due to a sideways wind is Fwind = cAv2, where v is the velocity of the wind, A is the cross-sectional area on which the wind is blowing, and c 1.3 N·s2 · m-4

http://img153.imageshack.us/img153/5154/fig087cq6.gif

If the blowfly has a cross-sectional side area of 0.10 cm2, a mass of 0.072 g, and crouches such that = 30.4°, what is the maximum wind speed in which the blowfly can stand? (Assume that the drag force acts at the center of gravity.)

Homework Equations

Since the body is at rest, I presume the only two relevant equations are Total Torque = 0 and Total Force = 0.

The Attempt at a Solution

First off, thanks for any future help provided. I do appreciate it. My main issue is that I feel like my free body diagram is incorrect. Here is what I drew:

http://img292.imageshack.us/img292/3930/windtorquesh7.jpg

However, I don't know if that is actually the correct way to draw this out. I don't really know how to calculate torque in this example because I am not given a length to use as r in the equation Torque = rFsin(theta). Thus, that makes me suspect that my free body diagram is incorrect. In addition, the free body diagram doesn't feel correct because the pivot point, the downwind legs, are at a lower level than the center of mass. I did great with the torque problems involving arms holding milk cartons, but nothing about this problem seems familiar. Could anyone shine some light on this? Thank you.

 

[PhanthomJay]

Rather than define torque as rFsin theta, it is often easier in these torque problems to define the torque (moment) of a force about a pivot point as the 'force times the perpendicular distance from the line of action of the force to the pivot point'. The wind force acts at the cg of the insect...if you call the horizontal distance between the downwind feet land the cg as 'L', then the perpendicular distance from the wind force to the pivot is L*(tan theta). Now find the moment of the insects weight about the pivot, and solve.

 

[Cypripedium]

Thanks for the guidance, Jay.

Ok, so following your first set of instructions (using Ltan(theta)), I come up with the following diagram. Please let me know if this is incorrect:

http://img148.imageshack.us/img148/8464/physicsbugck1.jpg

I realize I could get a value for the hypotenuse by following the Pythagorean theorem, but I don't think that would get me of the problem that I don't have a value for L. I know that a surface area is given, but I don't think I can pull anything from that. Unfortunately, I'm yet again stuck (and that's assuming that my little triangle up there is correct).

 

[PhanthomJay]

Thanks for the guidance, Jay.

Ok, so following your first set of instructions (using Ltan(theta)), I come up with the following diagram. Please let me know if this is incorrect:

http://img148.imageshack.us/img148/8464/physicsbugck1.jpg

I realize I could get a value for the hypotenuse by following the Pythagorean theorem, but I don't think that would get me of the problem that I don't have a value for L. I know that a surface area is given, but I don't think I can pull anything from that. Unfortunately, I'm yet again stuck (and that's assuming that my little triangle up there is correct).

Triangle looks good. So now what is the moment or torque of the wind force, F, about the pivot? Remember it is force times perpendicular distance. And what's the moment of the weight force about that same pivot?

 

[Cypripedium]

Ok, TorqueWind should be equal to (.000013)(v^2) * (.5867*L).

Since we aren't using rFsin(theta) for this one, I'm presuming TorqueWeight should be equal to (.0007056N) * (.5867*L). Since I'm dealing with two variables, I don't really think I can add them and set them to zero.

 

[PhanthomJay]

Ok, TorqueWind should be equal to (.000013)(v^2) * (.5867*L).

Since we aren't using rFsin(theta) for this one, I'm presuming TorqueWeight should be equal to (.0007056N) * (.5867*L). Since I'm dealing with two variables, I don't really think I can add them and set them to zero.

The wind torque (counterclockwise) is close, but you forgot to multiply it by the cross sectional area (watch units). Now for the weight torque, remember its line of action of the force times perpendicular distance. The perpendicular distance for the verticall acting weight is not .5867L. What is it? Then set the 2 torques equal , and the 'L's' will cancel out.

 

[Cypripedium]

The wind torque (counterclockwise) is close, but you forgot to multiply it by the cross sectional area (watch units).

Unless I'm missing something, I'm pretty sure I included the cross-sectional area. .1 square centimeter is equal to .00001 square meters. Multiply that times the constant 1.3. Multiply that by v^2 and get .000013v^2. Are you saying that I have to multiply the whole thing again by the cross-sectional area?

Now for the weight torque, remember its line of action of the force times perpendicular distance. The perpendicular distance for the verticall acting weight is not .5867L. What is it? Then set the 2 torques equal , and the 'L's' will cancel out.

Ah ok, the perpendicular distance is equal to the hypotenuse of the triangle, 1.159*L. Thus, the torque of the weight is = (.0007056N)*(1.159*L). I set this equal to the wind torque value I got and got v = 10.357 m/s, but that was wrong.

 

[PhanthomJay]

Unless I'm missing something, I'm pretty sure I included the cross-sectional area. .1 square centimeter is equal to .00001 square meters. Multiply that times the constant 1.3. Multiply that by v^2 and get .000013v^2. Are you saying that I have to multiply the whole thing again by the cross-sectional area?



Ah ok, the perpendicular distance is equal to the hypotenuse of the triangle, 1.159*L. Thus, the torque of the weight is = (.0007056N)*(1.159*L). I set this equal to the wind torque value I got and got v = 10.357 m/s, but that was wrong.

Yes, sorry, you have the wind torque correct, my error. But on the weight torque, the perpendicular distance is not the hypotenuse. A distance perpendicular to a vertical line is a horizontal one. What's the horizontal distance from the weight to the point?

 

[Cypripedium]

Ah HAH! The perpendicular distance is simply L (I had circular motion in mind, I suppose). Set them equal, L drops out, get 9.618 m/s! Your help has been much appreciated. Hopefully I'll be able to help out someone here with a similar problem sometime. Thank you.

 

[PhanthomJay]

Ah HAH! The perpendicular distance is simply L (I had circular motion in mind, I suppose). Set them equal, L drops out, get 9.618 m/s! Your help has been much appreciated. Hopefully I'll be able to help out someone here with a similar problem sometime. Thank you.

Great! The concept of 'Torque = Force times perpendicular distance' is at first perhaps confusing, but it's often a better approach than 'Torque = rFsintheta', because you can easily get messed up in determining the correct value of r and theta to use.