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Calculating the force acting on a pulley in a hoist system

  1. Aug 21, 2015 #1
    In the diagram below there is a hoisting system set up to lift a 1000 lb. unit up a radio tower. This unit is used for painting towers and is lifted using a capstan hoist. As the unit is lifted, it will be tagged with a 6ft. (adjustable) tag line with a pulley that travels up the end of the load line that goes to the capstan hoist. The purpose of this is to hold the unit away from the radio tower as it raises. I am trying to get an idea of how much force the traveling pulley on the tag line will see as it travels up the load line. As the unit is raised higher, the tag line will pull the load line closer to the tower and the ground. This pull will create a sag in the load line and I am having difficulty with calculating the force on the pulley. Do you have any ideas of how to calculate the force on the tag line/tag line pulley?
    Tower hoisting system_1.jpg
  2. jcsd
  3. Aug 21, 2015 #2


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    Gold Member

    Here is how it will look like

    To find out the force on the tag line you have to solve a set of equations. The best to do is in Cartesian coordinates.
    Lets pick X axis going from tower to the capstan, and Y axis vertical.
    The capstan is at the position (100,0)
    The bridle is attached to the tower at the position (0, 980)
    We can ignore the presence of the short rope between the top pulley and the tower and fix the top pulley position as (0, 980)
    Let us use the following variable
    The position of the unit is (x1,y1)
    The position of the pulley attached to the tag line is (x2,y2)

    Let's call the tension of the pull line T and the tension of the tag line as F
    We have a total of 6 unknowns

    The forces acting on the unit are the following:
    100 lbs weight directed down
    Tension T directed along the direction (0 - x1, 980 - y1)
    Tension of the tag line F, directed along direction (x2 - x1, y2 - y1)

    Similarly, the forces on the piece of wire touching the tag line pulley are
    Tension T, acting along direction (0 - x2, 980 - y2)
    Tension T acting along direction (100 - x2, 0 - y2)
    Tension of the tag line, F acting along direction (x1 - x2, y1 - y2)

    At equilibrium, forces have to add up to zero as vectors that gives us the following equations:
    At (x1,y1) x components have to add up to zero
    T *(-x1) / sqr [x1^2 + (980-y1)^2] + F * (x2 - x1) / sqrt [ (x2-x1)^2 + (y2 - y1)^2 ] = 0
    For y component of the forces we have
    T *(980-y1) / sqr [x1^2 + (980-y1)^2] + F * (y2 - y1) / sqrt [ (x2-x1)^2 + (y2 - y1)^2 ] - 100 lbs = 0

    At the other point we have
    T * ( -x2) / sqrt[ x2^2 + (980 - y2)^2] + T * (100 - x2) / sqrt [ (100 - x2)^2 + y2^2 ] + F * (x1 - x2) / sqrt [ (x2-x1)^2 + (y2 - y1)^2 ] = 0
    T * (980 -y2) / sqrt[ x2^2 + (980 - y2)^2] + T * (-y2) / sqrt [ (100 - x2)^2 + y2^2 ] + F * (y1 - y2) / sqrt [ (x2-x1)^2 + (y2 - y1)^2 ] = 0

    We also have the length of the tag line of 6 ft, that gives us another equation
    (x1 - x2) ^2 + (y1-y2)^ = 6^2

    We have 6 unknown and 5 equations, that gives us 1 degree of freedom: the unit can be at any height as long as it above ground and below 980 ft.

    You can also include the effects of the short line between upper pulley and the tower. That will give you 2 more unknowns and 2 more equations.

    I hope it helps.
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