1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Torque and Equilibrium

  1. Dec 11, 2007 #1
    1. The problem statement, all variables and given/known data
    http://img404.imageshack.us/img404/248/sdaft7.png [Broken]

    3. The attempt at a solution
    The arrows on the diagram are drawn in, as are the labels (except for the angles).
    [tex]\sum F_{x} = P_{x} - Tcos30 = 0[/tex]
    [tex]T = \frac{P_{x}}{cos30}[/tex]
    [tex]\sum F_{y} = N_{p} - 8900N - 650g - Tsin30 = 0[/tex]
    [tex]N_{p} = 15274N + Tsin30[/tex]
    [tex]N_{p} = 15274N + P_{x}tan30[/tex]

    I think I'm missing forces, and probably making up a few. If anyone could check them over before I continue I'd really appreciate it.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Dec 11, 2007 #2
    you're off to a good start. make sure you have the weight of the box in kg. now you need a third equation, which comes from summing the torques about the pivot and setting that equal to 0.
  4. Dec 11, 2007 #3
    Are all the forces right though? Am I missing any?
    Here's the third equation:
    The angle between the strut and the tension from the cable is 16 degress: 180 - 30 - (180 - 46) = 16
    [tex]\sum\tau = Tlsin16 - \frac{1}{2}l650g(cos46) - 8900l(cos46) = 0[/tex]
    [tex]\frac{P_{x}}{cos30}lsin16 = 325lg(cos46) + 8900l(cos46)[/tex]
    [tex]\frac{P_{x}}{cos30}sin16 = 325g(cos46) + 8900(cos46)[/tex]
    [tex]Px = \frac{325g(cos46) + 8900(cos46)}{sin16} = 26381N[/tex]
    [tex]T = \frac{26381}{cos30} = 30462N[/tex]
    [tex]N_{p} = 15274 + P_{x}tan30 = 15274 + 26381(tan30) = 30505N[/tex]
    [tex]P_{tot} = 40330N[/tex]
    [tex]\phi = 49 degrees[/tex]
    Last edited: Dec 11, 2007
  5. Dec 11, 2007 #4
    i just noticed. in your summation of the forces in the y direction, you have the mass of the strut. mass is not a force, so it shouldnt be included in the summation of the forces. you have to find the weight of the strut. (i made a bad mistake by saying that the weight of the crate should be in kg. mass and weight are not the same thing). same deal in your summation of torques. torque is defined as the force times the moment arm, not the mass times moment arm. i think you should have it though. just play around with your three equations till you get it.
  6. Dec 11, 2007 #5
    When I listed 600g, I meant the g as in 9.81 not grams or kilograms. When I computed the answers, I multiplied things by 9.81 wherever g is. I use g so that I get a more accurate answer. Am I using all the right forces though? Tension in cable pulling down and to the left, normal force from pivot on the strut pushing up and to the right, weight of strut going down, and weight of crate going down. Is that all?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook