Static Equilibrium

Yes, there’s a section about statics under dynamics. Whatcha gonna do about it? Static equilibrium is the analysis of forces when nothing’s actually moving. Certainly, forces are often at play even when a system is not accelerating (usually, it’s entirely still). A frequent case is that of a…thing…suspended by two (or more!) ropes at certain angles, as shown in the diagram.

[Diagram]

Given that diagram, the mass $ m$ of the “thing” and the angles $ \theta_1$ and $ \theta_2$ of the ropes, we want to find the force of tension $ \mathbf{F}_{t1}$ and $ \mathbf{F}_{t2}$ in ropes $ 1$ and $ 2$, respectively.

There are three forces at play here: $ \mathbf{F}_{t1}$ and $ \mathbf{F}_{t2}$ as well as $ \mathbf{F}_g$. The essence of the problem is that $ \sum\mathbf{F}=0$, meaning that $ \sum\mathbf{F}_x=0$ and $ \sum\mathbf{F}_y=0$. (We’ll define our coordinate system just like the Cartesian system; that is, the $ x$ and $ y$ directions are the horizontal and vertical ones, respectively, and the top and right directions are each positive.) There are two forces at play in the $ x$ direction, since $ \mathbf{F}_g$ has no horizontal component:

  • $ \mathbf{F}_{t1x} =-\mathbf{F}_{t1}\cos\theta_1$
  • $ \mathbf{F}_{t2x} = \mathbf{F}_{t2}\cos\theta_2$

which sum to zero: $\displaystyle -\mathbf{F}_{t1}\cos\theta_1 + \mathbf{F}_{t2}\cos\theta_2 = 0$ the first of two equations we’ll need to solve a two-variable system.

There are three forces at play in the $ y$ direction:

  • $ \mathbf{F}_{t1y} = \mathbf{F}_{t1}\sin\theta_1$
  • $ \mathbf{F}_{t2y} = \mathbf{F}_{t2}\sin\theta_2$
  • $ \mathbf{F}_g = -m\mathbf{a}_g$

and those three must also sum to zero: $\displaystyle \mathbf{F}_{t1}\sin\theta_1 + \mathbf{F}_{t2}\sin\theta_2—m\mathbf{a}_g = 0$ so our system is the two above equations in a system which we solve using our handy-dandy TI-89s, getting $\displaystyle \mathbf{F}_{t1} = \frac{m\mathbf{a}_g}{\sin\theta_1 + \sin\theta_2}$ and $\displaystyle \mathbf{F}_{t2} = \frac{m\mathbf{a}_g\cos\theta_1}{\cos\theta_2\left({\sin\theta_1+\sin\theta_2}\right)}$ (and that’s it.)

As an aside, there should be some assorted magnitude signs ($ \Vert\mathbf{blah}\Vert$) sprinkled across some of the equations, but I neglected them partly out of laziness and partly because I feel they’d do no more than detract from the clarity.

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