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!

Torque on crane arm and load limit

  1. Jan 6, 2007 #1

    i had this problem: the arm of a crane is 15.0 m long and makes an angle of 20.0 degrees with the horizontal. assume that the maximum load for the crane is limited by the amount of torque the load produces around the base of the arm.
    a) what is the mximum torque the crane can withstand if the maximum load is 450 N?
    b) what is the maximum load for this crane at an angle of 40.0 degrees.

    my answer/work: a) max=450 N t=?
    (6300)(sin 20)
    2154.73 N m​

    b) t=fd(sin 40)
    divide by 15d(sin 40) on both sides

    is this correct? are my answers and work correct? please let me know by commenting on this post. thank you:smile:
  2. jcsd
  3. Jan 6, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Firstly, note that you require cos20, not sin20 (check your diagram!)
    I'm not sure how you've gone from the 2nd to the 3rd line. It should read t=fd=450*cos(20)*15

    Once again, you have the wrong angle. However, your method is correct, apart from the bit in bold.. you should only divide by 15 sin(40)! (although, I must reiterate; this is not the correct angle)
  4. Jan 7, 2007 #3
    first off i want to thank you for taking the time to try to help me out.
    secondly so the answer for part a should be 6342.93 N m?? why do you use cos instead of sin???
    thirdly i do not understand what you are talking about in part b. what angle would i use??? am i suppose to use cos instead of sin???

  5. Jan 7, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    You're welcome.
    I've not done the calculation, but if you've used the right angle then this should be correct. If you imagine the diagram, you have the angle between the crane and the horizontal. Now the load will be hanging vertically downwards, and we want the angle between this and the normal to the crane. This angle is equal to the angle the crane makes with the horizontal (try it for yourself, using simple trig). Thus, since we are resolving through this angle, we require cos(20).
    Sorry, that wasn't very clear- it was late last night when I typed the response! We require cos(angle) for the same reasoning as above.
  6. Jan 7, 2007 #5
    thank you again that made a lot more sense and i was able to get an answer. thank you!!!!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Torque on crane arm and load limit
  1. Torque in a Crane (Replies: 5)

  2. Moment of load on arm. (Replies: 8)