Bar-Spring-Problem

Agrovolo Picture

Determine $\theta$ when the system reaches equilibrium. $\theta = 0$ when the spring is uncompressed.

Notice that

$$F_s = k\Delta x = k[l-lcos(\theta)]=kl[1-cos(\theta)]$$

Now, apply static equilibrium:

$$\Sigma F_x=0=B_x-F_s \Rightarrow B_x=F_s$$

$$\Sigma F_y=0=A_y-w \Rightarrow A_y=w$$

$$\Sigma M_A=0=w\frac{l}{2}cos(\theta)-F_slsin(\theta)$$

Substitute the expression that we found for $F_s$,

$$w\frac{1}{2}cos(\theta)=kl[1-cos(\theta)]sin(\theta)$$

$$\frac{w}{2kl}=[1-cos(\theta)]tan(\theta)$$

Now we can use graphical methods to solve, using the factor $\frac{2\pi}{360}$ to convert the x-axis to degrees:za

Agrovolo Picture