0

When discussing torques on a beam-balance (or see-saw) system, most textbooks focus on the requirements for the equilibrium of the system while being horizontal. However, I got interested in the following problem, but it seems like I have arrived at contradictory results:

Let two objects, with weights $F_1$ and $F_2$, be situated at opposite sides of the pivot at $d_1$ and $d_2$ distances from the pivot respectively. The pivot is at the center of gravity of the see-saw. Suppose that the system is not initially balanced at the horizontal position. We want to find out the angle at which this system rotates (towards the side with greater torque, of course) to arrive at equilibrium.

This is my workout: enter image description here

As you can see, I have considered the perpendicular components (with respect to the see-saw) of $F_1$ and $F_2$, and used them to find the torques while being tilted at an angle $\theta$. Since the tilted system is considered to be at equilibrium, the two torques (one clockwise and the other anticlockwise) must be equal. But, the $cos(\theta)$ cancels out and we arrive at a contradiction. Where is the problem?

I have found a similar (but not same) version of my problem at: Angular equilibrium on a see-saw

However, the setup of that system is a bit different. There, the pivot (center of gravity of the see-saw) is suspended and can rotate about a fixed point. The answer to that post seems reasonable to me, but my situation here is not exactly the same as that.

Any solutions or insights into my problem will be much appreciated. Thanks!

Faiyaz
  • 35

1 Answers1

0

Suppose that the system is not initially balanced at the horizontal position. We want to find out the angle at which this system rotates (towards the side with greater torque, of course) to arrive at equilibrium.

If the system is not initially balanced at the horizontal position, there will be a net torque acting on the see-saw giving it angular acceleration. It will continue to have angular acceleration (i.e., be in disequilibrium) as long as there is a net torque.

The only time the net torque can subsequently be zero is if the see-saw gets into the vertical position, i.e., $\theta =90^0$. (For that to happen the ground below would need to be removed!). If it could do so, at that instant its angular acceleration becomes zero and it has maximum angular velocity (kind of like a simple pendulum at the bottom of its swing).

That said, if the system is initially balanced horizontally, it can be re-positioned to any angle $\theta$ and it will still be balanced. Likewise, if the see-saw is initially balanced at some angle $\theta$, it can be re-positioned to any other angle, including $\theta=0$, and still be balanced. The reason for both is that although the torque applied by each of the weights changes with the angle $\theta$, the net torque remains zero.

Hope this helps.

Bob D
  • 81,786