# Calculating the force and/or torque required to pivot an arm from one end

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 $L$, pulley diameter $D$, and maximum weight $W$ 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 ($F$) 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. Thanks 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 $F = ma$, and $T = Fd$.

So am I doing this right?

Let's assume 225 kg weight to be lifted, and an arm length of 6 m, and a 30 cm pulley diameter.

\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?

$T = 13243.5\text{ Nm}$ and $d = 0.15\text{ m}$, so $N = 88290 N$

So in this example I would need a cylinder that would be able to generate 88,290 N pull force (or roughly 20,000 lb pull), and a cable that could also withstand more than 20,000 lb strain?

• Think about it like this: You have two levers, the difference in force is the ratio of the lengths. 6/.15=40. That matches your numbers.
– hazzey
Feb 1 '17 at 19:25
• @hazzey thanks for that. I feel more confident in the calculations now. :) Feb 1 '17 at 19:52

My calculation got the same answer as you. Here's what I did.

You mentioned that the arm has some weight so I added that force to your diagram. Also, I labeled the center of the elbow joint as point 'E'. Assumptions

• The pully is mounted on a frictionless pivot (friction will reduce the amount of force required)
• The arm is perfectly rigid (if the arm bends, the geometry changes so the forces will change)
• The system is at rest (rotating motion means acclerations which need forces)

Because nothing is in motion, we can solve for the force using statics (https://en.wikipedia.org/wiki/Statics). Summing of the moments (torques) acting on the blue arm around point E lets us solve for the force with just one equation.

$\Sigma M_E=0$

$F\dfrac{D}{2}-ca-WL=0$

$F=\dfrac{2}{D}(ca+WL)$

You can see from the equation that the resultant force is sensitive to the pully diameter. As the diameter tends towards zero, the force tends towards infinity.

Like you already calculated above, the force from the weight is:

$W=(225kg)(9.81\frac{m}{s^2})=2,207N$

If we set the arm weight to zero $(a=0)$ then

$F=\dfrac{2}{0.3m}(2207N)(6m)$

$F=88290N$

Side note:

Overhead cranes usually require a high factor of safety in their design. I found an OSHA reference which says use 5:1.

OSHA 1926.753(c)(1)(i)(C)