how to calculate the holding force

You would also need to specify the size of contact surface where the ball is gripped, and the coefficient of friction between the two materials.

okay, lets say the ball is d 30 mm, contact surface is something like 40% of the ball surface. material ball steel, socket brass. i don't know how much the friction pressure is but ball is pressed into socket by stainless steel cap.

will that do?

I am wondering if the strongest one of these http://www.tmcmagnetics.com/ball_joint_assembly.html would do the job too?

No it won't nearly hold it

Your torque is 1.2 Kg X 26" (that's 28.54 COS(25°))

=31.2 Kg.inches (sorry about mixed units)

The "socket" is 1" wide.

Even if you treat one side of the "cup" as the fulcrum and the other side the point where the force needs to be applied: you would need a force of 31.2 Kg to hold it there.

The holding force of 20Kg in the spec. is what it would support if you held the socket upside down and dangled a weight from the rod.

I have no idea how these magnetic ball joints are constructed, but could it be possible to use more than one magnetic socket? Maybe the the whole joint has to be reconstructed then?

The inside of it looked quite complicated on a patent sheet..

But it still sounds doable

I THINK ITS ZERO BEING A BALL JOINT WHICH WILL NOT ALLOW ANY INTERMEDIATE DEGREE ORIENTATION EXCEPT THAT AT POSITION DUE TO GRAVITY..

did you draw a free body diagram with all the forces on it?

considering that the ball joint is holding the rod at position, the force reaction there should equal the weight of the rod and the torque reaction should be 1.2kg*g*1.45/2*cos(25°) considering that the 25° are measured form the horizontal.

However, ball joints don't generate torque reactions, so, how do you keep the rod at 25°?

I havent drawn a fb diagram, but the angle 25 is from the vertical. the rod should stay in the angle due to the holding force of the joint or friction pressure is perhaps what it is called. The joint will have a max angle at 25 too so there the rod will lean on the edge of the cap of the joint. (24.9 will be the actual max then...)

should it be sin(25) instead then?

Yes it's sin(25°) instead.

So you're down to about 14.5 kg force applied at the edge.

I still don't think the units you are looking at will be strong enough: depends on the coefficient of friction.

How much money time and design flexibility have you got? Could you adapt a magswitch?

It could perhaps be possible with a magswitch, those are new to me, but the switch itself seems a bit to big for my design. and they are maybe quite expensive? regarding money I try to keep the costs as low as possible

Is this about that floor lamp?

I've had to deal with a lot of speakers big and small that were supported by all manner of ball joints..

well at least they were supposed to be.

it only takes a bit of grit to secure a ball joint, but it only takes a bit of slip to fail .. the length of rod is a problem.

good luck

yes it is..and it is quite tricky stuff

I originally visualized this as 25° from horizontal. With the clarification, the rod's mass of 1.2kg may be considered as concentrated at 0.725m on a 65° angle. This results in a moment (torque) of 1.2 x 0.725 x cos65 ≈ 0.37 kg-m. (I am omitting the conversion to N-m, but will do the same on the next calc.)

Assuming that the clamping force will be applied to the ball in the most advantageous direction, we need 0.37/0.030 ≈ 12.3kg to resist movement. The actual clamping force required will be 12.3/coeff. fric. Taking a wild guess of 0.40 (which depends on materials and surface finish), you would need ~ 31kg of clamping effect.

This seems to be achievable. Even if you double it, you should be able to swing the rod to other positions. If much lower, though, the clamping force will need to be higher.

Thanks for your calculation!

These magnetic ball joints (the biggest) http://www.tmcmagnetics.com/ball_joint_assembly.html has a holding force of 20 kg, if it was possible to use two sockets on one ball, do you know if it then would be covering the need of 31 kg clamping effect?

You're welcome!

If the coefficient of friction is higher than my crude estimate, one of these magnets might suffice. I'm not sure that two would fully double the clamping force, but I suspect it would come close. Time for a trial run, I think. Good luck!