What is the Torque Produced by a Beam and Sign on a Wall?

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The discussion revolves around calculating the torque produced by a 2kg beam and a 5kg sign mounted on a wall. The user correctly identifies the torque formula as torque = F x perpendicular distance and calculates the torque for both the beam and the sign. The calculated torque for the beam is 4.9Nm, and for the sign, it is 24.5Nm, leading to a total of 29.4Nm. However, the course book states the answer is 118Nm, prompting confusion and speculation about potential errors in the book. The user questions whether the discrepancy is due to a misunderstanding or an actual error in the textbook.
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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 I am 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 can't pick up my error :confused:
 
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looks like a book error, if you described the problem correctly.
 
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
 
Well, one error is that it is a torque :biggrin:
 
you never know the book might spell it that way to?!??
 
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