# Swing of a hanging rope due to wind

1. Aug 10, 2011

### derek88

Hello Friends. I have a question that may have to do with wind lift and other effects.

I have two wooden poles set into the ground. Both poles are upright and has a height of H feet. The horizontal separation of these poles is S feet. Lets say I attach a rope in between these poles so that the ends of the rope are attached to the tops of the poles. This rope has a length of L feet and has a diameter of D feet.

If I have a sustained wind pressure of W lbs/ft^2, how high will the rope swing?

My work so far:

Wind load on the rope = (Diameter of rope, in feet)*(Wind Pressure on a cylindrical surface, in lbs/ft^2) [lbs/foot]

Total Force on rope = (Wind load on rope, lbs/foot)*(Length of rope, in feet) [lbs]

Any help/direction on this problem would be greatly appreciated! Also any helpful internet links. Thank you.

2. Aug 11, 2011

### gsal

briefly: force times distance

I am not sure if you need to work out some kind of integral or not, after all, the weight of the rope is nicely equally distributed and so is the force from the wind. So, maybe averages would be just find (i.e., rope as rigid body and force on center of mass)

But things are going to work out like this:

The weight of the rope is always point down
initially, the "parallel" distance from the weight to the pivot (top of posts) is zero and hence the torque is zero
the force of the wind is alway horizontal
the distance of the wind force to the pivot can be probably averaged and say that it is half way down the swing and so there is an initial force that overcomes the zero force from the weight
as the swing swings away from the vertical position, the weight of the swing starts producing its own torque and as the center of mass moves higher, it reduces the torque from the wind....at some angle, the two torques become the same...that is what you are looking for...so, right the equation equating the two torques as function of the angle of the swing from the vertical