I'm trying to design a two-stage "arm" which articulates from an attached base. The arm uses a pulley and cable at the "elbow" and a hydraulic cylinder mounted on the lower stage to lift the upper stage (similar to a boom truck).
For a given arm length L$L$, pulley diameter D$D$, and maximum weight W$W$ at the end of the arm (obviously the arm itself has some weight too), how can I determine the maximum required hydraulic cylinder pull force (F$F$) required to lift the arm from a horizontal orientation?
If there's a formula where I can plug in the values, I can experiment with different arm lengths, pulley circumferences, and weights to hopefully find a sweet spot in relation to available hydraulic cylinders and pulley sizes.
Here's a rough drawing. ThanksThanks for any pointers, links, etc.
UPDATE:
So I found this document which clearly expresses the formulas used to calculate forces required to lift a weight on an arm.
From this document, I understand that F = m x a$F = ma$, and T = F x d$T = Fd$.
So am I doing this right? Let's
Let's assume 225kg225 kg weight to be lifted, and an arm length of 6m6 m, and a 30cm30 cm pulley diameter.
F = 225kg x 9.81m/s2 = 2,207.25 N T = 2,207.25 N x 6m = 13,243.5 N-m$$\begin{alignat}{2} F &= 225 \times 9.81 &&= 2207.25\text{ N} \\ T &= 2207.25 \times 6 &&= 13243.5\text{ Nm} \end{alignat}$$
So pulley radius actually becomes the length of a lever, right? So the same torque would need to be generated by applying a linear force to a 15cm lever?
T = 13,243.5 N-m,$T = 13243.5\text{ Nm}$ and d = 0.15m$d = 0.15\text{ m}$, so N = 88,290$N = 88290 N$
So in this example I would need a cylinder that would be able to generate 88,290 N N pull force (or roughly 20,000 lb lb pull), and a cable that could also withstand more than 20,000lb000 lb strain?
Am I doing this right?
