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Hanging Sign Problem and Pulley Problem

  1. Mar 17, 2005 #1
    Okay, we've been studying this for a while but I still always seem to forget how to do these and they aren't anywhere in my Physics book.

    Problem 1: Sign hanging at rest from two cables. One is angled at zero degrees while the other is at 135 degrees. The sign is 10kg. Find the tension of both cables.

    Problem 2: Sign hanging at rest from two cables. One is angled at 170 degrees while the other is at 35 degrees. The sign is 500kg. Find the tension of both cables.

    Problem 3: There's a frictionless pulley with two weights on either end of a string. One weight is 20kg while the other is 35kg. Find the acceleration of the 20kg mass and the tension of the cable.

    For the first two I know you have to like do T1x + T2y = T1 and T2x + T1y + F = T2 or something, but I can't remember the right formula for it. For the second one I don't know how to do it at all.

    Also, if anyone has any links on how to sum up masses that would be greatly appreciated.
  2. jcsd
  3. Mar 17, 2005 #2


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    I will do number 1 as an example:

    (we assume the sign is hanging at rest supported by the cables)
    cable1 is at 0 degree,
    cable2 is at 135 degrees above the horizontal.

    mass of sign = 10 kg, Force of weight = 10*9.81 = 98.1 Newtons

    so both of those cables must be supporting a combined 98.1 newtons of wieght.
    Tension in Cable1 in the Y direction + T in cable2 in the Y direction = -98.1 N
    Tension in Cable1 in the X direction + T in cable2 in the X direction = 0 N

    cable1 holds no weight in the Y direction, all of it is held by cable2 since it is the only cable with an upward component to its angle.
    F*sin(135) = 98.1, F = 138.7 N at an angle of 135 degrees,
    no being at 135 degrees, some of that force is in the X direction, that is where cable1 comes into play, its tention must cancel out cable2's X direction tention,
    Fx of cable2 = 138.7*cos(135) = -98.1 N, therefore cable1 must be supplying 98.1 N in the positive direction to cancel out the -98.1 N.

    UNderstand, you need to break down the cables so that first, they (together) will hold up the sign, and then so that the sign doesnt move either direction because of some imbalanced X direction force.
  4. Mar 17, 2005 #3
    Okay, I worked that out on paper and got it, but I'm having one heck of a time trying to figure out the second hanging sign problem. Do I need to find the two equations and substitute one in for the other? For the second one right now I have:

    T2= -T1cos170/cos35 and T1=4900-T2sin35

    Do I need to substitute one equation in the other to find the two tensions, or did I just do that totally wrong?
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