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!

Homework Help: Torque and Tension of two weights

  1. Oct 21, 2004 #1
    Guys - I'm back again :rofl: . I hate to have so many questions, but I really appreaciate all the help you guys give me. For this problem, I am having some problems with the clockwise/counterclockwise Torque rule. Anyways, let me give you the problem and work.

    Two weights attached to a uniform beam of mass 26 kg are supported in a horizontal position by a pin and cable as shown in the figure (I have placed the figure in the attachment doc. 1). The acceleration of gravity is 9.8m/s^2. What is the tension in the cable which supports the beam? Answer in units of kN.

    To solve this problem, I knew I had to apply the rule that clockwise torques are equal to counterclockwise torques.

    Clockwise torques: Weight of the bar, Weight 1, and Weight 2.

    Counterclockwise torques: Tension

    Here is where my work gets shady :confused: . For the tensions, I broke the tension line up into components Ty and Tx. Ty = Tsin (theta). Tx = Tcos (theta). Do I take both of these for the counterclockwise torques in the equation? When I set the problem up, I only used the Ty value. Is this incorrect?

    Anyways here is the final problem set up as I had it with only one of the Tension components:

    [Weight of the bar X (length/2)] + [Weight1 X distance from weight] + [Weight2 X length] = Tsin (theta) X length

    In numerical values: 26(5.6/2) + 32(3.1) + 37(5.6) = Tsin (39) X 5.6

    Obviously this answer is wrong, but my question is....is the correct equation this:

    [Weight of the bar X (length/2)] + [Weight1 X distance from weight] + [Weight2 X length] = [Tsin (theta) X length] + [Tcos (theta) X length]

    In numerical values: 26(5.6/2) + 32(3.1) + 37(5.6) = [Tsin (39) X 5.6] + [Tcos (39) X 5.6]

    OR is it something else entirely? Thanks for all the help guys!

    Attached Files:

  2. jcsd
  3. Oct 21, 2004 #2


    User Avatar

    the last equation should be, for the x-component of T, the distance is 0, not 5.6. Only y-component of T has the distance 5.6 . So u just cancel out x-component cuz it's direction goes through the point O.
  4. Oct 22, 2004 #3
    Would that not leave me with the exact same initial equation that I started off with? I used that equation and came up with the wrong answer. Either I made a careless mistake (likely), I'm not understanding you, or there is something else wrong with the problem.
  5. Oct 24, 2004 #4
    bump....Still need help :grumpy:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook