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Maximum possible force exerted on a sail

  1. Jan 31, 2014 #1
    This isn't actually homework. It's just been a very long time since I've been in mechanics/calculus class and I was trying to figure this out for my own reference / sanity checking. I've made up this problem to try and understand what amount of force is exerted on a sailboat under the most ideal of conditions. A sail is limited by the mass and speed of the air that moves past it. What is the absolute maximum possible force that could be exerted by the wind on a sail if all momentum of the wind were transferred to the sail? The point of this problem is to throw away all other considerations about aerodynamics and come up with a value which would serve as a sanity checking value for other calculations of force on a sail. Basically "if you've calculated a force greater than this, you've done something wrong".

    1. The problem statement, all variables and given/known data
    Let's say I have a stationary sailboat with a sail that has area A. The sail is perpendicular to the wind which has a velocity of V. The density of the air is q.

    My sail is a magical sail which has the property that it absorbs all the momentum from all air that touches it. Let's not worry about where the air with no momentum goes. That momentum is absorbed by the sail which exerts a force on my sailboat. What is the net force exerted on the sailboat while the sailboat is not in motion?

    Given these values, how many pounds of force are exerted on the sailboat at the instant that the sails are opened into the wind and the sailboat is not moving?

    A = 2m2
    V = 4.5 m/s
    q = 1.29 g/liter = 1.29 kg/m3

    2. Relevant equations
    p = m*v
    F = dp/dt
    k = 1/2*m*v2
    P = ΔW/Δt = F*v

    3. The attempt at a solution
    My first thought was to figure out the "change in momentum per second" which I think is equal to the force.

    Mass of air per second = q*A*V

    Momentum per second = (mass per second)*v = q*A*V * V

    Force = Momentum per second = dp/dt = q*A*V2

    F = qAV2 = (1.29kg/m3)*(2m2)*(4.5m/s)2 = 52.2 Newtons = 52.2N * (1lb/4.44822162N) = 11.7lbs

    So that would say that in a 10ish mile per hour wind with a 2 square meter sail, the MAXIMUM possible force that could be exerted by the sail is only 12 pounds? In a 22 mph wind (10m/s) it would be 60 pounds of force? These values seem far too small to be correct, and I'm sure I'm doing something completely wrong here.

    It's occurred to me to look at this from the point of view of kinetic energy per second and Watts, but I have no idea how I would then turn that back into a force acting on the sailboat.
     
  2. jcsd
  3. Jan 31, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Slowing down air in front of the sail can help to slow down air elsewhere, too, but for a rough approximation your approach should be fine.
     
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