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Tourqe of a sign on the wall

  • Thread starter _wolfgang_
  • Start date
  • #1
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Homework Statement



a uniform 2kg horizontal beam is 50cm long is bolted to a wall and supports a 5kg sign. calculate the tourqe produced by the combined weight of the beam and the sign about where it is mounted to the wall.

Homework Equations



tourqe = F x Perpendicular distance

The Attempt at a Solution



So im using the wall as the pivot point making the distance to the centre of mass of the beam 0.25m.

the tourqe of the beam= (9.8x2)x(0.25)
= 4.9Nm

The tourqe of the sign= (9.8x5)x(0.5)
=24.5Nm

And to get the total tourqe i have added these togther to get 29.4Nm

But my course book says the answer is 118Nm?? i know this is a pretty simple question but i cant pick up my error :confused:
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
Homework Helper
Gold Member
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476
looks like a book error, if you described the problem correctly.
 
  • #3
23
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the only thing i changed from the book is that it says a light rather than a sign.. but i think it may be a book error to. This book has many errors
 
  • #4
93
0
Well, one error is that it is a torque :biggrin:
 
  • #5
23
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you never know the book might spell it that way to??!?!!!??
 

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