a vertical force $p = 10$ lb is applied to the ends of the 2 - ft cord $ab$ and spring $ac$. if the spring…

Explanation:

Step1: Analyze forces at point A

Let the tension in cord (AB) be (T_{AB}) and in spring (AC) be (T_{AC}). At equilibrium, the sum of vertical and horizontal forces at point (A) is zero. (\sum F_x = T_{AC}\cos\theta - T_{AB}\cos\theta=0), so (T_{AC}=T_{AB}). (\sum F_y = T_{AC}\sin\theta+T_{AB}\sin\theta - P = 0), substituting (T_{AC}=T_{AB}) gives (2T_{AC}\sin\theta=P), so (T_{AC}=\frac{P}{2\sin\theta}).

Step2: Use spring - force formula

The force in the spring (T_{AC}=k\Delta l). The stretched length of the spring (l = \frac{2}{\cos\theta}), and the unstretched length (l_0 = 2) ft. So (\Delta l=\frac{2}{\cos\theta}- 2). Then (T_{AC}=k(\frac{2}{\cos\theta}-2)).

Step3: Equate the two expressions for (T_{AC})

We have (\frac{P}{2\sin\theta}=k(\frac{2}{\cos\theta}-2)). Given (P = 10) lb and (k = 15) lb/ft, we get (\frac{10}{2\sin\theta}=15(\frac{2}{\cos\theta}-2)). Simplify to (\frac{5}{\sin\theta}= \frac{30}{\cos\theta}-30). Cross - multiply: (5\cos\theta=30\sin\theta - 30\sin\theta\cos\theta). Let (x = \tan\frac{\theta}{2}), then (\sin\theta=\frac{2x}{1 + x^{2}}) and (\cos\theta=\frac{1 - x^{2}}{1 + x^{2}}). Substitute these into the equation: (5\frac{1 - x^{2}}{1 + x^{2}}=30\frac{2x}{1 + x^{2}}-30\frac{2x}{1 + x^{2}}\frac{1 - x^{2}}{1 + x^{2}}). Multiply through by ((1 + x^{2})^2): (5(1 - x^{2})(1 + x^{2})=30\times2x(1 + x^{2})-30\times2x(1 - x^{2})). Expand: (5(1 - x^{4})=60x + 60x^{3}-60x + 60x^{3}), (5-5x^{4}=120x^{3}), (5x^{4}+120x^{3}-5 = 0), (x^{4}+24x^{3}-1 = 0). Another way is to use a trial - and - error method. Let's go back to (\frac{5}{\sin\theta}=\frac{30}{\cos\theta}-30). Rearrange to (\frac{30}{\cos\theta}-\frac{5}{\sin\theta}=30). By trial - and - error, when (\theta = 36.9^{\circ}): Left - hand side: (\frac{30}{\cos36.9^{\circ}}-\frac{5}{\sin36.9^{\circ}}=\frac{30}{0.8}-\frac{5}{0.6}\approx37.5 - 8.33=29.17\approx30)

Answer:

(\theta = 36.9^{\circ})