Torque of Two Arm System with Point Mass m

[amainejr]

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 point-mass (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?

 

[sophiecentaur]

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?

 

[amainejr]

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?

 

[amainejr]

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?

 

[PJC01]

I can provide some guidance for approaching this problem. First, it is important to clearly define the variables and parameters involved in the system. In this case, the point mass (m) and the two arm system (L and R) are the main components, with the upper arm being attached to a servo that has a range of 135 degrees. The upper arm is also connected to the lower arm, which is fixed but free to rotate around the upper arm.

Next, it is important to consider the forces acting on the system. In this case, there are two main forces to consider: the weight of the point mass and the torque applied by the servo on the upper arm. The weight of the point mass will always act downwards, while the torque from the servo will vary depending on the angle of the upper arm.

To determine the maximum torque on the upper arm, we can use the equation T = F x d, where T is torque, F is the force applied, and d is the distance from the point of rotation. In this case, the distance from the point of rotation to the point mass will vary depending on the angle of the upper arm. Therefore, the maximum torque will occur when the upper arm is fully extended to the horizontal, as this will give the maximum distance from the point of rotation to the point mass.

To address the issue of the replicated system, we can consider each system separately and then combine the results. Since the lower arm will never reach an angle of more than 90 degrees relative to the upper arm, we can ignore the replicated systems and focus on the maximum torque of one system. However, if the replicated systems are not identical, then it may be necessary to consider them separately and add the results together.

In summary, to solve this problem, you will need to consider the forces acting on the system and use the equation for torque to determine the maximum torque on the upper arm. You can then repeat this process for each replicated system and combine the results if necessary. It is also important to clearly define the variables and parameters involved in the problem to ensure a correct solution.