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

Homework Help: Find Magnittude of the Force Supported by the bearing O

  1. Sep 15, 2012 #1
    The two light pulleys are fastened together and form an integral unit. They are prevented from turning about their bearing at O by a cable wound securely around the smaller pulley and fastened to point A. Calculate the magnitude R of the force supported by the bearing O for the applied 3.5-kN load.

    I have attached an image of the problem.

    1. The problem statement, all variables and given/known data

    2. Relevant equations
    ƩMO = 0
    ƩFx =0
    ƩFy =0

    3. The attempt at a solution

    First I found the angle between the positive x axis and the cable (AB)

    θ = arctan(130/285)
    θ = 24.5196°

    Then I found the moment about point to calculate the tension (T) in the cable
    ƩMO = 0
    0 = (3.5 kN)(200mm) -(130mm)T
    T = 5.3846 kN

    Then, knowing that the pivot point, O, has a force in the x direction (Rx) and a force in the y direction (Ry) I calculated these:

    ƩFx= 0
    0 = -Rx + Tcos(24.5196)
    Rx = 4.899 kN

    ƩFy = 0
    0 = -3.5 kN + Ry -Tsin(24.5196)
    Ry = 5.7305 kN

    Thus the magnitude R is:

    R = sqrt((4.899 kN)^2 +(5.7305 kN)^2)
    R = 7.53914 kN

    It says my answer is wrong and I can't see where my mistake is. Help would be appreciated.

    Attached Files:

  2. jcsd
  3. Sep 15, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's a significant figure error....too many places after the decimal point...try 7.5 kN
  4. Sep 16, 2012 #3
    7.5 kN didn't work either. I forget to include this in the image but the system has a tolerance to +/- 1 to the third significant digit.
  5. Sep 16, 2012 #4
    I found my mistake, the angle that I calculated using arctan(130/285) was wrong. It should have been arcsin(130/285) because 285mm is no the longest length of the triangle.

    arcsin(130/285) = 27.138

    Substituting 27.138 into my calculations I get fx = 4.7918kN and fy= 5.9561kN

    Taking the magnitude of these gives me a resultant of 7.644 kN which is the correct answer.
  6. Sep 17, 2012 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Sorry, how did I miss that??
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook