Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tension and height of flagpole

  1. Jun 7, 2005 #1
    A uniform, horizontal flagpole of length 5.00 m and with a weight of 195 N
    is hinged to a vertical wall at one end. A stuntwoman weighing 600 N hangs
    from its other end. The flagpole is supported by a guy wire running from
    its outer end to a point on the wall directly above the pole.

    a) If the tension in this wire is not to exceed a force of 1070 N, what is
    the minimum height above the pole at which it may be fastened to the wall?

    b) If the flagpole remains horizontal, by how many newtons would the
    tension be increased if the wire were fastened a distance 0.520 m below
    this point?

    ** The moments about the point of hinging of the downward weights (that of the weight of the pole, and that of the stunt woman) must be balanced off by the upward moments at the end of the pole.

    Hence how do you calculate the theoretical perpendicular force exerted at the end of the pole to maintain equilibrum?

    Now, this theoretical upward force is the vertical component of the tension in the guy wire.

    Given that the tension may not exceed 1070, what is the minimum angle with the pole the wire can be?
    I calculated the downward forces = 195 N + 600 N = 795 N
  2. jcsd
  3. Jun 7, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    So.... now let's see an equation for the moments, and one for the horizontal force components, and one for the vertical force components.
  4. Jun 8, 2005 #3
    the point where the pole is attached to the wall as a hinge. The torque caused by the pole and the stuntwoman should be equal to the torque caused by the rope.
    So, only the vertical component of the tension will cause torque.

    Torque_pole = 1070N x sin(theta) x 5m
    --> 195N x 2.5m + 600N x 5m = 3442.5N -m
    3442.5N -m = 1070N x sin(theta) x 5m
    sin(theta) = 0.6435 and theta = 40.05

    Since the length of the flagpole is one side of the triangle formed by the wall, rope and pole we know that:
    cos(40.05) = d/5m
    d = 3.827m
    but that's not the minimum height. How do I solve this and part b?
  5. Jun 8, 2005 #4


    User Avatar
    Homework Helper

    Are you considering the reaction on the left of the pole?

    I can't seem to understand the geometry, if this a horizontal pole embedded to the wall or has a hinge (a support) on the left?
  6. Jun 8, 2005 #5
    ok, so if the minimum height above the pole at which it may be fastened to the wall is 4.30m finding it by solving tan(theta) = d/5
    how do I find part b?
  7. Jun 8, 2005 #6


    User Avatar
    Homework Helper

    Use a sum of torques with the changes introduced in your FBD.
  8. Jun 8, 2005 #7
    got it, thanks again
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook