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Equilibrium - Flagpole vs Rope Tension

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data

    A flagpole of height h has a flag attached that is being blown by the wind with force F at a height of 0.8h. In order to prevent the flagpole from blowing over a rope is run from an anchor in the ground to the top of the flagpole at an angle [tex]\vartheta[/tex]. What is the tension in this rope?

    2. Relevant equations

    [tex]\tau[/tex]=rFsin[tex]\vartheta[/tex]
    F[tex]_{}net[/tex] = F[tex]_{}flag[/tex]-F[tex]_{}rope[/tex]=0

    3. The attempt at a solution

    Since the objects are in equilibrium the net force of the system must be 0. Or rather the force of the wind on the flag must be equal, and in the opposite direction, of the force of the supporting rope attached at the top of the flagpole.

    Since the two forces act on different positions of the flagpole torque must be accounted for with the flagpole acting as the lever arm that is attached to the ground, but not supported by it.

    Since the weight of the flagpole = the normal force on the flagpole the Tension should only act in the x-axis. This lead me to believe that the tension in the rope equals the radius times the force in the x-direction, or T = 0.8Fsin(theta) which is not the case.

    This leads me to start thinking that any rope with any tension that is at an angle that is not perpendicular to an opposing force MUST have tension acting in both the x-axis and y-axis direction. In this questions the tension in the y-axis would be pulling the flagpole harder into the ground, and the normal force would increase so that the overall force of the flagpole in the y-axis direction would still be 0 overall. Now it seems like I am left with two unknowns of tension in the y-axis, and the overall tension along the hypotenuse. What am I missing?
     
  2. jcsd
  3. Nov 29, 2008 #2
    I figured out the solution. Obviously the two torques must cancel each other out.
    (0.8h*F)-(Ftension*1h*cos(theta))=0
    Ftension*cos(theta)=0.8F
    Ftension = 0.8F/cos(theta)

    NOT

    Ftension = 0.8Fcos(theta)
     
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