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# angular force calculation help

## angular force calculation help

(OP)
Hi all,

I'm building a lever system using an electric actuator as shown in the drawing. I'm wanting to double check the force in kg required to retract the ram. I calculated 670kg and seems low.

### RE: angular force calculation help

I agree. Did you accidentally interchange sine for co-sine or vice-versa in your calculations?

Bill
--------------------
"Why not the best?"
Jimmy Carter

### RE: angular force calculation help

(OP)
I calculated f(sin6.13)(300)+f(cos6.13)(131)= 3700(30)

### RE: angular force calculation help

I make it 3473kg, rounded up to 3500kg. It's been nearly 30 years since I did this sort of calculation, so please feel free to correct me.

Does the 30kg include the mass of the 3700mm jib?

### RE: angular force calculation help

Fahley,

For starters, kilograms are not a unit of force. Force is measured in Newtons or pounds.

You need to work out the forces normal to the 3700mm beam. The distance from the pivot to your actuator does not seem to be in the same coordinates. If this were my drawing, I would not show the length of the actuator. I would show the distance from the pivot to the opposite end of the actuator.

Once you have the normal forces, you can work out the triangle of the actuator and pivot, and just scale everything off of it.

--
JHG

### RE: angular force calculation help

Come on, drawoh... Let's not be pedantic. This is the hobby forum.

That said, I don't know what you're meaning with regard to working out triangles etc. Based on the layout, I assume the desired motion speed is pretty dang slow, such that the addition of F=ma dynamics analysis would add less to the static force than the friction in the system. Therefore this is an extremely simple statics problem. Draw the FBD of the 3700mm beam. The angle of the 3700mm beam is given, and the angle of the actuator is given. The actuator is a 2-force member, so there's no moment applied at that point. The beam attachment point is a pivot joint, so only Rx and Ry there. The only distances that matter are the overall length of the 3700mm beam and the 299 distance from the beam pivot to the point of actuator attachment. Sum forces/moments = 0.

Now... If you're looking for some equation/formula to give actuator force for any arbitrary angle or arbitrary point in the actuator's stroke, that's when the rest of the geometry comes into play.

### RE: angular force calculation help

This is the "Engineers with Hobbies" forum. If you use mass for force, I become a spectator.

### RE: angular force calculation help

(OP)
Hi Guys thanks for your help to date.

Gary, yes the 30kg includes the jib mass as well.

Drawoh, i appreciate that force in measured in newtons. However, on this occasion i require the output in kgs to match the brochure. The length from the pivot point of the ram to the pivot point of the jib is 1135mm with on offset of 20mm. The jib centre being higher. If that asssists.

Cheers
Fahley

### RE: angular force calculation help

Fahley,

If you are going to do calculations and get the right answer, you need to be disciplined about units. Mass versus force is harmless if you are doing statics. If you start to play with acceleration, you need to know where to insert gravitational acceleration. If the USA ever converts to the metric system, one thing they can do is cross the Atlantic and demand that kgf specifying Europeans convert too. If you use English units, you need to understand that pounds are not a unit of mass.

The actuator, the base, and the arm up to the attachment point are a triangle, loaded in pure tension or compression. Once you figure out what the force is at the connection point, you can solve the problem on a drafting board, or on 2D CAD. I agree with handleman that this is a statics problem.

--
JHG

### RE: angular force calculation help

I treated the problem as a simple statics problem and assumed the OP has decided that this is the "worst case" position. If this is the lowest jib position, I agree.

Hydraulic cylinders are often rated in kg, so I was happy to leave everything in kg and (I assume) mm.

My working...

Horizontal distance from jib fulcrum to load is 3700cos(14.32) = 3585
Horizontal distance from jib fulcrum to ram joint on jib is 299cos(14.32) = 290

Resolving moments around jib fulcrum: (30x3585)-(290N)=0, where N is the vertical force component of the ram, N = 370.9kg

Force triangle around the jib/ram joint: sin(6.13)=(370.9)/(Ram Force), Ram Force = 3473kg(f), or (if you prefer), 34070N

I assume your angles are correct, as I have not checked them.

### RE: angular force calculation help

(OP)
Thanks so much Gary. That's huge help.
Yes the angles are correct and the arm move around 0.25"/sec- very slow.

Cheers
Fahley

### RE: angular force calculation help

To be fair, I have textbooks that refer to 'kgf', so if you're going to be pedantic, don't refer to the 'metric system', it's SI. Either you're a stickler, or you're not.

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