1. Not finding help here? Sign up for a free 30min 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!

Test question today

  1. Oct 19, 2007 #1
    1. The problem statement, all variables and given/known data
    The image is a crane.
    For the diagram given, find the tension in the cable and find the moment around point A.
    The length of the crane the dark bar is 12 m.

    http://img249.imageshack.us/img249/679/exqjm6.jpg


    2. Relevant equations
    sigma m = F *D


    3. The attempt at a solution

    I had this question on a test today and am very curious to know the answer.
    if someone strong in physics please solve it.

    I got 51 something Kilo Newton for the reaction at A.
    For the tension i got 46.08 Kilo newton i believe.

    thank you.

    EDIT: In my reaction/moment calculation I included the force of the cable. that being said i forgot to account for the y direction of the cable. should the cable have been accounted for?
     
    Last edited: Oct 19, 2007
  2. jcsd
  3. Oct 19, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    Here's what I get:

    let T be the force of tension..

    let B be the force in the bar...

    Bcos30 - Tcos20 = 0

    B = 1.085T

    Bsin30 - 35 - Tsin20 = 0

    plug in B = 1.085T

    I get T = 174.6N

    Then the torque about A:

    10kN*6cos30 + 35kN*12cos30 - Tsin20*12

    = 10kN*6cos30 + 35kN*12cos30 - 174.6sin20*12
    = -300.9 kNm

    so the reaction has to be +300.9 kNm ie 300.9 kNm clockwise

    I was a little thrown off by the problem... because I initially just tried to get T by making the torque about A to 0... then when I saw the question asking for the moment about A I changed...

    Not very confident with how i did this problem.
     
  4. Oct 19, 2007 #3
    If someone as intelligent as you are not confident then I am sure my class took a strong beating during this test.

    Btw the weight hanging is 25KN i should have wrote it neater.

    Yup totally blew the last page of this test. I even got the tension wrong..which was the easier question.

    He gave us 50 minutes to solve 4 questions. By the time i got to the 4th had very little time left.

    but listen in your moment calculation why didnt you account for the x component of the tensin in the string?
     
  5. Oct 19, 2007 #4

    learningphysics

    User Avatar
    Homework Helper

    Thanks I appreciate that. But in this question, I don't feel confident at all because I changed the way I solved the problem after reading the second part of the question. That's never a good sign. ;)

    The assumption I'm making here that there's a reaction force acting at A along the axis of the beam... along with a reaction moment at A. I don't feel good about this assumption that the net reactional force acts along the axis...

    yeah, I used 25kN... The 35kN... is the 25kN + the 10kN of the beam...

    I was using the component of tension perpendicular to the beam... ie Tsin10. then the torque due to that is -Tsin10*12 (minus sign indicates counteclockwise). The component parallel to the beam is Tcos10... but this creates 0 torque...

    You can also divide into the x (horizontal) and y(vertical)-components... magnitude of x-component of tension would be Tcos20. torque due to that is -Tcos20*(12sin30). The torque due to the y-component of tension is Tsin20*12cos30... net torque would be Tsin20*12cos30 - Tcos20*(12sin30) = Tsin(20-30) = Tsin(-10) = -Tsin10. comes out to the same answer...

    I would have blown this question on a test too.
     
  6. Oct 19, 2007 #5
    Go back and look at the moment about A and ask yourself "Is the boom arm rotating about A?" If not, what does that say about the moments?
     
  7. Oct 19, 2007 #6

    learningphysics

    User Avatar
    Homework Helper

    Oh... I was making this whole problem way more complicated than it was... for some reason I was fixated on the answer to the second part not being 0.

    so to get tension:

    10kN*6cos30 + 25kN*12cos30 - Tsin10*12 = 0

    T = 149.6 kN ?
     
    Last edited: Oct 19, 2007
  8. Oct 19, 2007 #7
    do you mean T cos 20? because the triangle of the given 10 degrees is not a right triangle.

    149.6N doesn't SEEM right because how could the tension be so low compared to the other forces?
     
    Last edited: Oct 19, 2007
  9. Oct 19, 2007 #8

    learningphysics

    User Avatar
    Homework Helper

    I blundered... it should be 149.6 kN.

    I'm taking the component of tension perpendicular to the arm... that's Tsin10
     
  10. Oct 20, 2007 #9
    Thanks ok I understand what you did there.

    Can someone clarify for what TVP45 meant about the boom bar please?

    Why should the moment/reaction be 0 again?
     
    Last edited: Oct 20, 2007
  11. Oct 20, 2007 #10
    Since the boom bar is not rotating about A, the moments about that point must sum to 0. That is a necessary condition for static equilibrium.

    That is a fairly simple equation to solve.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Test question today
  1. Test question (Replies: 5)

Loading...