# Energy analysis of an unwinding cable drum and attached lever raising a weight [closed] I have been circling the drain on this question for a couple years, and have not properly asked the question before. This is not a homework problem - I have created the attached diagram and conceived the problem statement. I was hoping someone could evaluate my work and tell me where I am going wrong. When I calculate the energy input by pulling the cable with tension $$W_1$$ over a distance $$L_1$$, it does not match the energy required to lift weight $$W_2$$ to the height $$L_2$$. I know there is something wrong with how I am approaching the problem, but I can't figure out where my misstep is. I have been working on a couple inventions that center on this fundamental problem and I can't seem to get it right.

Problem Statement

A cable is partially wound around a drum of radius $$R_1$$ and fastened to its surface. The opposite end of this cable hangs freely. The drum rotates around a central axis, and is rigidly connected to a straight lever arm. A weight $$W_2$$ acts on the lever arm. The weight $$W_2$$ can slide freely so that when the arm rotates, the weight maintains a constant horizontal distance $$R_{2,x}$$ from the drum axis.

The arm begins in a horizontal orientation, a vertical tension $$W_1$$ is applied to the free end of the cable and the cable is pulled down a distance $$L_1$$. As a result, the drum rotates through an angle $$a$$ and the weight $$W_2$$ is raised to a height $$L_2$$.

Analysis

Ignore stretching of the cable, and the inertia of the rigid body system (drum and lever arm). At all positions of the lever arm where $$(0^\circ < a < 90^\circ)$$, the weight W2 should generate a constant torque T2 about the drum's axis. This can be shown by: $$T_2=R\cdot W_{2,\bot}$$ $$T_2=\frac{R_{2,x}}{\cos(a)}W_2\cdot \cos(a)=W_2*R_{2,x}$$ The equilibrium tension in the cable should then be: $$W_1=\frac{T_2}{R_1}=W_2\frac{R_{2,x}}{R_1}$$ Since the torque generated by weight $$W_2$$ is constant, I assume that the equilibrium tension $$W_1$$ should also be constant. Therefore, by pulling the cable vertically down a distance $$L_1$$, the energy input into the system should be: $$E_1=W_1\cdot L_1$$ The angle $$a$$ that the lever rotates to, without slipping or stretching of the cable, should be: $$a=\frac{L_1}{R_1}$$ This rotation should lift the weight $$W_2$$ to a height of: $$L_2=R_{2,x}\tan(a)$$ This action should require an energy of: $$E_2=W_2\cdot L_2$$

In Summary

By pulling on the free end of the cable with tension $$W_1$$ and moving it a distance $$L_1$$ vertically, the weight $$W_2$$ will raise to a height of $$L_2$$. The input and output energies should be equal, but are calculated as: $$E_1=W_1\cdot L_1=\frac{W_2\cdot R_{2,x}\cdot L_1}{R_1}$$ $$E_2=W_2\cdot L_2=W_2\cdot R_{2,x}\cdot \tan(\frac{L_1}{R_1})$$

Substitution

$$W_2=1~lb,~~R_{2,x}=10~in,~~R_1=3~in,~~L_1=3~in$$ $$E_1=\frac{(1~lb)(10~in)(3~in)}{3~in}=10~in\cdot lbf$$ $$E_2=(1~lb)(10~in)tan(\frac{(3~in)}{(3~in)})=15.57~in\cdot lbf$$

I did find that performing a similar analysis with a see-saw type setup (placing a pivot at the center of a straight bar with vertical forces maintained at specified horizontal distances in each direction) produces a correct result with balanced energy. So there should be some problem with how I am calculating the height $$L_2$$ in the drum and lever setup. However it seems trivial to determine the angle $$a$$: without slipping or stretching of the cable, shouldn't the drum should rotate through an arc-distance equal to $$L_1$$?

The torque generated by W2 is not constant. The higher the arm, the greater the torque. Imagine the extreme case of the arm being almost vertical. W2 must travel a much longer vertical distance to cause a the same angular rotation of the drum as when the arm is horizontal.

Here's a more detailed diagram of the connection between W2 and the arm. There are two other forces too which all sum to give zero at equlibrium. That means there's some horizontal force on the arm as well as W2, adding to the torque. When the arm is horizontal, this extra force is zero so the torque is simply R2x * W2.

Another way to look at it, the mechanism which keeps R2x constant applies a horizontal force, and thus a torque to the arm. I think you've neglected this in your analysis. • I took another look based on your response and posted and image to imgur. This time, I found the reaction torque to be $t= \frac{W \cdot R_x}{cos^2(a)}$. If this is correct, it also explains why calculations workout with a symmetric setup (i.e. vertical forces acting on either side of a central pivot). The geometric symmetry cancels out the cosine functions, and you're left with a typical fulcrum advantage. Could you confirm if I am correct? This would mean that, even though the torque is increasing in the arm, the mechanical advantage is constant. Apr 14, 2021 at 5:02
• I confirm your formula t = W R_x / cos^2(a) Apr 14, 2021 at 18:57
• Thank you for the help - means a lot! Working alone has me going in circles without external feedback. I really appreciate your time. Apr 14, 2021 at 19:00
• I don't think the mechanical advantage is constant though, by definition, because W2/W1 is not constant and that's what mechanical advantage is. Apr 14, 2021 at 19:05
• The way I was seeing it - W * rx is a constant (should have been small 'r' in my response). The initial force multiplied by the horizontal distance Rx. Another piece that pushed me to that conclusion is that the velocities are constant, relying on the tightly coupled torque/velocity parameters. Apr 14, 2021 at 19:10