the resultant of the two forces acting on the screw eye is known to be vertical. determine the angle θ and…

Explanation:

Step1: Resolve forces horizontally

Since the resultant is vertical, the sum of horizontal - components of the two forces is zero. Let the force of magnitude $F_1 = 40$ lb and $F_2=60$ lb. The horizontal - component of $F_1$ is $F_{1x}=- 40\sin30^{\circ}$ and the horizontal - component of $F_2$ is $F_{2x}=60\sin\theta$. So, $-40\sin30^{\circ}+60\sin\theta = 0$. $-40\times\frac{1}{2}+60\sin\theta = 0$ $- 20+60\sin\theta = 0$ $60\sin\theta=20$ $\sin\theta=\frac{1}{3}$ $\theta=\sin^{- 1}(\frac{1}{3})\approx19.5^{\circ}$

Step2: Resolve forces vertically

The vertical - component of $F_1$ is $F_{1y}=-40\cos30^{\circ}$ and the vertical - component of $F_2$ is $F_{2y}=-60\cos\theta$. The magnitude of the resultant $R$ is the sum of the vertical - components of the two forces. $F_{1y}=-40\cos30^{\circ}=-40\times\frac{\sqrt{3}}{2}=-20\sqrt{3}$ lb $\cos\theta=\sqrt{1 - \sin^{2}\theta}=\sqrt{1-\left(\frac{1}{3}\right)^{2}}=\frac{2\sqrt{2}}{3}$ $F_{2y}=-60\times\frac{2\sqrt{2}}{3}=-40\sqrt{2}$ lb $R = |F_{1y}+F_{2y}|=| - 20\sqrt{3}-40\sqrt{2}|$ $R=20\sqrt{3}+40\sqrt{2}\approx20\times1.732 + 40\times1.414=34.64+56.56 = 91.2$ lb

Answer:

$\theta\approx19.5^{\circ}$, $R\approx91.2$ lb