# 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 $$L_1$$?

• 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. – daDib Apr 14 at 5:02