# Weight on a fulcrum style problem - Personal Interest

1. Aug 9, 2010

### c4nc3r

Just to preface, this isn't something that I've been given as homework, I'm not currently enrolled in any schooling, however as a matter of personal interest I wanted to work out the solution to the problem without a guess-check-improve method. I have an idea, and it seems too obvious to be true but still.

1. The problem statement, all variables and given/known data

I'm building a coffee table which will consist of a sheet of glass resting on top of the block of an engine (basically, a rectangular prism on its end). I want to find out the highest weight that can be put on the very edge of the table without the glass falling off. My basic MS Paint drawing is below, with the red arrow being the weight point.

I don't yet know the weight of the glass, as I'm unsure of the size of the sheet, I'm more interested in finding an equation to use when I do know all the variables.

2. Relevant equations

Not sure yet

3. The attempt at a solution

From the thinking I've been doing, it originally seemed to be simply a case of balancing the weight of the left and right hand sides, with the pivot point being the side of the block closest to the weight (see attached image)

For example, the image below would be perfectly balanced and therefore wouldn't fall;

But then the more I think about it the more that seems to be too simple a breakdown, as the further from the pivot point the weight is the more force it has. So that's where I'm stumped, which is frustrating as I know I should know the answer.

I'm not sure if my explanation has made sense to anyone, if you need clarification don't hesitate to ask, and likewise if this is in the wrong place then let me know and I'll do what I can to move it.

Thanks guys
Josh
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 9, 2010

### arkofnoah

this is quite a simple problem involving torque.

assuming that the sheet of glass is uniform, the centre of gravity is at the centre of the glass sheet (to the right of the left edge of the block), and it will produce a clockwise moment about the pivot point.

moment = force x perpendicular distance to pivot.

in order to balance out the moment, the anti-clockwise moment produced by the weight must be equal to that generated by the weight of the glass sheet acting on its centre of mass.

Last edited: Aug 10, 2010
3. Aug 9, 2010

### c4nc3r

But as the block in the center is solid, wouldn't that mean that is the center of the rotation? Or am I missing something in your explanation?

4. Aug 10, 2010

### arkofnoah

Actually i meant that the pivot point is the left edge of the block, not the center of it.

5. Aug 10, 2010

### thrill3rnit3

I'm sorry, it was kinda late at night when I posted that message [ which I deleted ], and I didn't realize what we're talking about here.

Anyways, yes, the pivot point is the left edge of the center block because that's where the glass rotates.