
#1
Apr2013, 11:00 AM

P: 3

This problem should be an introductory physics problem, but I am not sure if I am going about it the right way. This Link is the total system that has been created, and I have attached an isolated image to clarify what I am about to ask, so please refer to that.
I have a pointmass (m) attached to a two arm system, noted as L and R. The upper arm is attached to a servo that has a range of 135 deg (0 deg is straight up in this instance). I know that the max torque on the upper arm will be when it is fully extended to the horizontal. I am drawing 2 free body diagrams for each arm, for maximum torque, where the lower arm is fixed (but free to rotate) about the upper arm. The lower arm will never reach an angle of more than 90 deg relative to the upper arm. The problem I am running into, is that this isolated system is replicated 3 times, all supporting the mass m. Do I divide the mass by 3 for each system or is this problem much harder than that? 



#2
Apr2013, 12:32 PM

Sci Advisor
PF Gold
P: 11,401

If your arrangement is symmetrical then the weight force is equally shared between all three supports but if it is not symmetrical then the shares will depend upon the three angles (ref vertical) of the ropes. The total vertical force will be equal to the weight and the proportions are given by the angles so that gives you enough equations to work out the situation.
Of course, the tensions are not simply weight/3 because of the angles involved. BTW, where is 'vertical' in your diagram? 



#3
Apr2013, 02:31 PM

P: 3

Thanks for the quick reply.
That's about what I concluded. Vertical, would be straight up (pi/2). Let me know if I went about this the wrong way, here are my thoughts, assuming the entire weight was pulling perpendicular in the isolated system: m(total) = .746kg L = .585m R = .485m F = m*g = .746kg * 9.8 = 7.31 t = R*F = 3.545Nm So, if the entire weight, pulling down at 90deg from the end of the upper arm were experienced, it would be a total of 3.55Nm, which should NEVER happen, because the weight is distributed. So, a mild approximation, would be to divide it by 3, assuming that all upper arms were at a 90 deg angle to the horizontal. At that point, we'd have t/3 = 3.55Nm/3 = 1.183Nm on each arm. Now, the servo's I am looking at produce 333oz/in stall torque. I'm estimating 75% of that as dynamic torque, for approximately 250oz/in, which convert to 7.09kg/.0254m. So, with an upper arm length of .485m, the servo would be able to lift a maximum of about .370kg, slightly less than half of the total mass. I would assume then, that if the state where 1/2 of the total weight was ever directly under the outermost edge of one arm directed at the horizontal, it would not lift. However, during that same state, the other two arms would be in a position such that they support some of the load. Am I safe to estimate this way? 



#4
Apr2013, 04:15 PM

P: 3

Torque with two arms
Looking back, the upper arm weight will mostly be supported by the other arms, so in effect, I would think that the total amount of torque necessary should be a fraction of what is previously estimated. Can anyone support the idea that I should, in fact, be well above the required amount of rotational force if I have 250oz/in of torque to lift the specified amount of weight?



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