Another wonderful Torque problem

  • Thread starter Thread starter flynostrich08
  • Start date Start date
  • Tags Tags
    Torque
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 1K views
flynostrich08
Messages
20
Reaction score
0

Homework Statement


A 50 kg sign is hanging at the end of a 3 m long beam. The beam is held in place by a hinge at the wall and a wire attached to the wall and the end of the beam at 30 degrees. The beam has a uniform mass of 20 kg. Find the tension in the wire.


Homework Equations





The Attempt at a Solution


i pretty much drew the picture and free body diagrams and realized torque net= 0 = torque sign + torque of cable= torque beam.

i found torque of sign=490 torque of beam= 98 and torque of cable=T(tension force)sin150 i solved for T and got 1176.

I need to know if i am right because i feel like i need to solve a net force problem and is the answer 1176 or -1176?
Thanks for anyone's help
 
Physics news on Phys.org
flynostrich08 said:
i found torque of sign=490 torque of beam= 98 and torque of cable=T(tension force)sin150 i solved for T and got 1176.
Your final number looks okay to me. :approve:
I need to know if i am right because i feel like i need to solve a net force problem and is the answer 1176 or -1176?
Thanks for anyone's help

In general, the direction of forces involved on rigid bodies might depend on how you set up your free body diagram. In this particular case the problem involves a "tension," which has its own meaning. It's understood that a positive tension means that the wire/string/cable is being pulled apart. When something gets pulled apart, that thing has positive tension. When it is pushed together (if possible, such as for a rigid rod) it has negative "tension". But like I mentioned before, the direction of force might depend on what part of the free body diagram you're working with. A straight, tense string pulls in one direction on one side of the string, and in the other direction on the opposite side of the string (yet it only has one positive "tension" -- you just need to look at the tension direction differently at different points in the FBD).