here is a side view of a wall. the weight is attached to an L-shaped bracket which has magnets firmly fixed to it. the bracket is then held in place on a metal wall.
considering the cantilever action, the top magnet has a certain amount of force trying to pull it away (sideways) from the wall. which orientation of the bracket will reduce that force?
this isn't homework - don't need to say any working just looking for a simple answer.
The moments and forces are the same for either case because you have changed the moment arms in either the top or bottom magnet.
*haven't
For example if the perpendicular distance from the line of action of the force of the weight is 5m, and the force is 5N, then the resulting moments at A (top in either case) and B (bottom) are the same. And since both magnets are essentially fixed connections along the wall, they will have equal moments.
*equal and opposite moments, the top one will always be forcef out of the wall
Let D be the length of the arms. I'm assuming the bracket has arms of equal lengths by the picture.
Left setup:
[math]
\tau = Dmg\\
[/math]
Right setup:
[math]
\tau = D\sqrt{2}mg\sin{\theta}
[/math]
Now consider that if the sides are of equal length, then theta will be 45 degrees and sine of theta will therefore be [math]\frac{\sqrt{2}}{2}[/math]
So the right side setup will be:
[math]
\tau = D\sqrt{2}\frac{\sqrt{2}}{2}mg = Dmg
[/math]
which is the same as on the left side. This is assuming the bracket arms of equal length. So either setup is the same.
I thought that to be the case at first but then I couldn't shake the feeling that the right-hand-side orientation would be more likely to "fail" first with the top magnet being more likely to be pulled away in this scenario.
Many thanks for the input.
In picture form
Ah yes, I had assumed the brackets wee of equal length without even thinking about it.
OP if you're reading this the best way to make the magnet on top not get muck force pulling it out would be to take case 1, and extend the vertical bracket upward, which would increase the distance between the magnets, lengthening the moment arm between A and B, which would decrease the force out of the wall at A
thanks for the math. helps to see it that way.
>I couldn't shake the feeling that the right-hand-side orientation would be more likely to "fail" first
In the real world where the weight might be knocked about some (bumped into) the bracket would tend to move more near the magnet near the bend in the L shape. If the magnet slides across the surface some, then it might tip over to the in or out of the picture. In the case to the left the contraption would still hang from the top magnet, at a 'bump' force where the right one would tip over, levered around the lower magnet and falling on the ground.
$x = 2$
Math mode on Veeky Forums is [ math ] and [ /math ]
Without the spaces
[math]
Didn't work.
My guess is that the right one is best because the force is more directly balanced between the two magnets.
seems to be the wrong angle noted or else it should be cosθ.
Why they are the same
I drew it for my own convenience, since if theta is 45 then the other angle will also be 45.
...
Think lever.
Where is the fulcrum?
What is the vector distance to the fulcrum?
While the forces are the same, the right hand side is better off from an engineering perspective. It swaps tensile stress for compressive stress which for all materials depicted (metal and plastic) is less straining.
Wrong, the pivot is still the under magnet.
Exactly the point I wanted to make. In real life, the top magnet on the left panel can provide much less attractive force to the wall compared to compression that the bottom magnet in the right panel can provide. Because of this, the right panel could potentially support a higher load.