# Tourqe of a sign on the wall

• _wolfgang_

## 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

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

you never know the book might spell it that way to?!??