Coupled pendulum question about equations of motion [closed]

I am working on problem number 2.3 of the Franklin, Powell, Naemi book Feedback Control of Dynamic Systems. The problem uses the simple coupled pendulum system below, where the two pendulum masses are joined by a spring that is attached $$3/4$$ of the length of the masses.

A similar problem was addressed in this question:

Coupled pendulums at half height

Below are the equations of motion from the solution to the problem. The variable $$l$$ refers to the length of the pendulum cable. Note that I copied the result exactly from the solution manual--though the notation seems a bit odd with the repeated $$3/4$$:

$$ml^2\ddot{\theta_1} = -mgl\sin{\theta_1} - \frac{3}{4}kl(\sin{\theta_1} - \sin{\theta_2})\cos{\theta_1}\frac{3}{4}l$$

$$ml^2\ddot{\theta_2} = -mgl\sin{\theta_2} + \frac{3}{4}kl(\sin{\theta_1} - \sin{\theta_2})\cos{\theta_2}\frac{3}{4}l.$$

My question is, I do not understand why we need the $$\frac{3}{4}l \cos{\theta_i}$$ factor in the second term of both equations. I am not clear on what the meaning or intuition is behind that factor. The $$\sin{\theta_1} - \sin{\theta_2}$$ factor makes sense because the spring is pulling in the opposite direction of the restoring force. But I am just not clear on what that $$\cos{\theta}$$ factor is doing? Is the attachment of the spring pulling in or rectracting the mass of the pendulum or something?

1 Answer

If you use the full force in a formula for torque, then remember that the corresponding distance must be the lever arm relative to the pivot. The force is horizontal, so you need the vertical distance from the pivot.

• Oh I get it now. Yeah so it is coming from the torque formula, and force of the spring is acting like a torque. That was the little piece I was missing. Thanks so much for the explanation. Mar 8 at 19:44