Question

a vertical force, p = 10 lb is applied to the ends of the 2 - ft cord ab and spring ac. the spring has an unstretched length of 2 ft. take k = 17 lb/ft. determine the angle θ for equilibrium. express your answer using three significant figures.

Explanation:

Step1: Analyze forces in x - direction

$\sum F_x = 0$. Let the tension in cord $AB$ be $T_{AB}$ and in spring $AC$ be $T_{AC}$. $T_{AB}\sin\theta=T_{AC}\sin\theta$. So $T_{AB} = T_{AC}=T$.

Step2: Analyze forces in y - direction

$\sum F_y = 0$. $P = 2T\cos\theta$.

Step3: Find the force in the spring

The elongation of the spring $\Delta x=2(\sec\theta - 1)$ (from right - triangle geometry). By Hooke's law, $T = k\Delta x=k\cdot2(\sec\theta - 1)$.

Step4: Substitute $T$ into the y - direction force equation

$P = 2k\cdot2(\sec\theta - 1)\cos\theta$. Given $P = 10$ lb and $k = 17$ lb/ft.
$10=4\times17(\sec\theta - 1)\cos\theta$. Since $\sec\theta=\frac{1}{\cos\theta}$, we have $10 = 4\times17(1 - \cos\theta)$.

Step5: Solve for $\cos\theta$

$10=68 - 68\cos\theta$.
$68\cos\theta=68 - 10$.
$68\cos\theta = 58$.
$\cos\theta=\frac{58}{68}\approx0.853$.

Step6: Solve for $\theta$

$\theta=\cos^{- 1}(0.853)\approx31.5^{\circ}$.

Answer:

$31.5^{\circ}$