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Archived Hanging a sign Torque and summation of forces

  1. Apr 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Please help me confirm my answer.

    A shop owner wants to hang a sign of mass 200 kg which is supported by which is supported by a uniform 155 N beam. What is the tension in the guy wires and the horizontal and vertical forces that the wall applies to the beam? The length of the beam is 400 cm.
    The diagram shows that the angle between the guy wire and the beam is 65 degrees. It also shows that the wall to center of mass for the sign is 350 cm. I hope that gives a good picture.

    2. Relevant equations

    Mb = mass of beam
    r = distance
    g = gravity
    Ms = mass of sign
    t = torque
    T = tension

    3. The attempt at a solution

    OK so sum the forces in each direction
    Fy = Fu +Tsin(θ) - Mbg - Msg = 0
    Fx = Fh - Tcos(θ) = 0
    We will return back to the equations after we torque about the beam where it meets the wall.
    Ʃt = Tr1 - Mbr2 +Msgr3
    r1 = 4tan(65) 4 is in meters it is the length of the beam. Tan(65) is to find the lever arm.
    r2 = 2 meters
    r3 = 3.5 meters this is from the wall to the signs center of mass. The diagram shows this.
    We solve for Tension and get

    T = (Mbr2 +Msgr3) / (4tan(65))
    T = (155N x 2) + (200kg x 9.8 m/s^2 x 3.5 m) /(4tan(65))
    T = 835 N

    Then put this back into the summation of forces to get the values for the Fh and Fu
    I got .. Fh = 353.2 using Fh = Tcos(65)
    And I got Fu = 1357 using Mb + Msg -Tsin(65)
     
  2. jcsd
  3. Feb 23, 2016 #2

    CWatters

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    Homework Helper

    I believe the missing diagram would looks like this (left hand side). Free Body diagram for the beam is shown on the right.

    Hanging Sign.jpg

    First thing to note is that in the problem statement the mass of the beam is given in Newtons and the mass of the sign is given in Kilograms so only the latter needs to be multiplied by g later. The OP spotted this.

    Then we should make some assumptions/definitions: For forces, up and right is positive. For torques anti-clockwise is positive.

    It's a statics problem, nothing is accelerating so the vertical forces sum to zero, the horizontal forces sum to zero and the torques sum to zero.

    First vertical and horizontal forces..
    Correct so far.
    I believe the OP missed out some brackets in that last line. The torque due to the mass of the sign and the mass of the beam should be in the same direction/same polarity so the equation should be..

    Ʃt = Tr1 - (Mbr2 +Msgr3) = 0

    The OP then states..
    However I believe r1 = 4Sin(65) not 4tan(65). Perhaps my drawing is incorrect but the problem statement does say that..
    So when I solve for Tension I get..

    T = (Mbr2 +Msgr3) / (4sin(65))
    T = ((155N x 2) + (200kg x 9.8 m/s^2 x 3.5 m)) /(4sin(65))
    T = 1978N

    Then substitute for T into equation (1) and (2) to get the values for the Fh and Fu

    From eqn (1)..
    Fh = Tcos(65)
    = 1978cos(65)
    = 836N

    From eqn (2)..
    Fu = Mb + Msg -Tsin(65)
    = 155 + (200*9.8) - 1978sin(65)
    = 155 + 1960 - 1793
    = 322N
    .
    The method the OP used is correct but I believe the OP may have made a mistake calculating r1.
     
    Last edited: Feb 23, 2016
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