Force Calculation to balance arm

AI Thread Summary
The discussion centers on calculating the force required to hold a pivoted arm using a pneumatic ram. The arm weighs 100 kg and has its center of gravity located 740 mm up the arm, with a torque calculation yielding 203.07 Nm at the pivot. Initial calculations suggested a force of 676N, but further analysis indicated a need for a higher force due to the angle of the ram. After reevaluating the problem using trigonometric functions, a final force requirement of 1054N was determined, which participants agreed appeared more accurate. The discussion emphasizes the importance of using correct torque calculations and the angles involved in the setup.
gavhowe
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Hello,

I am after a little help on an applicaition, I need to calculate the force require to hold a pivoted arm in a position using a pneumatic ram. Basically the arms C of G is 740mm up the arm, and the arm weights 100kg. Roughly 1000N working at an offset of 147mm. The arm is 16 degrees from vertical, with the ram at 62 degrees to the arm, so it is pulling to hold the arm in position.

I attach a diagram to give all the information

I had a friend take a look and he calculated 676N, I am not sure if this is correct. As I could not follow is method or workings. I attach them as well.

Thanks

Mr Howell
 

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The arm is rotating around the pivot at its lower end, right? The pivot at the end of ram's shaft is there just to allow the ram / arm angle to change freely.

Moments should be taken about the pivot at the base of the arm. I see a larger ram force being required than what you (and your friend) calculated.
 
yes the arm is free to rotate around the pivot at the base of it. The ram will slightly change the angle of the arm when it moves in and out.

My calculation were, to first calculate the torque at the pivot:

100kg x 9.81 x 0.147m = 203.07 Nm

so 203.07 / .06m = 3384.5N to support at the pivot point of 215mm up the arm. If the ram was at 90degrees to the arm.

So need to calculate the angle from 90 degrees to the arm to the real position of the ram.

90 - 62 = 28 degrees.

3384.5 / (cos 28) = 3833.18N

But I am not happy that my method is correct as it is vastly different end figure then my friends.
 
Calculating torques requires that you use the components of the forces that are perpendicular to the moment arm. So, for example, the torque about the pivot point due to gravity would use the component mg \; sin(16°).
 
Ok I have had another look at the problem and tried is a different way, please could some let me know if this is correct:

981N*0.74m*Sin16 = ? * 0.215m*Sin62

(981N*0.74m*Sin16) / (0.215m*Sin62)

= 1054N
 
gavhowe said:
Ok I have had another look at the problem and tried is a different way, please could some let me know if this is correct:

981N*0.74m*Sin16 = ? * 0.215m*Sin62

(981N*0.74m*Sin16) / (0.215m*Sin62)

= 1054N

Now THAT looks good!
 
thank you
 

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