Neodymium magnets are powerful and give the ability to adjust the position as they can be pulled or slide.
The principle is simple, under the bench we glue a steel plate of 0.8mm thickness and the servo holder is attracted by magnets to the plate.
How strong magnets do we need? To answer this, i use a primary design of the servo holder, the Magnetic force calculator, the forces analysis described in my post Design a servo switch machine to control a model railroad turnout [excel].
The force F between the wire and the throwbar creates a moment in movement plane. That moment M=F*H must be carried by a pair of magnets. The magnets pair has also to resist the F force and not slide. So, if we call the distance between the two magnets dm, Fm1, Fm2 the forces on them and Gs the weight of the servo and servo holder, then:
Fm1 = M/dm + Gs/2 and Fm1 = -M/dm + Gs/2
Fmag > Fm1
The sign (+) means a pulling force and the (-) a push force.
The magnet's pull force Fmag (the case 2 in the magnetic force calculator) must be bigger than Fm1.
For the force Fr resisting the slide force F we have to make an assumption about the value of the static friction coeficient between the magnet and the steel plate. The steel on steel static friction coeficient has values from 0,15 to 0,7. Assuming the static friction coeficient equal to 0,25 we obtain:
Fr = 0,25 * (Fmag * 2 - Gs) > F => Fmag > 2*F + 0,5*Gs
Now we can choose the magnet with pull force that satisfies the underlined conditions with a reserve (let's say 100%) to cover eventually uncertainties.
Example
F= 1,0 N ; H= 60 mm ; Gs= 0,5N ; dm=20 mm
M=F*H= 1*60= 60 Nmm
Fm1= 60/20 + 0,5/2= 3,25 N
Fmag > 3,25 N
Fmag > 2*1,0 + 0,5*0,5= 2,25 N
We have to choose magnets with pull force 2*Fmag each, according to the case 2 in the magnetic force calculator.
The third magnet is used to stabilize the connection, but its favorable participation is neglected.
Note that bigger distance dm leads to smaller magnets.
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