problem 1.2 - the resultant of the two forces acting on the screw is vertical and pointing down. determine…

Explanation:

Step1: Resolve forces horizontally

Since the resultant is vertical, the sum of horizontal - components of the two forces is zero. Let the 40 - lb force be $F_1$ and the 60 - lb force be $F_2$. The horizontal component of $F_1$ is $F_{1x}=40\cos30^{\circ}$, and the horizontal component of $F_2$ is $F_{2x} = 60\cos\theta$. So, $40\cos30^{\circ}-60\cos\theta = 0$. $40\times\frac{\sqrt{3}}{2}-60\cos\theta=0$ $20\sqrt{3}-60\cos\theta = 0$ $\cos\theta=\frac{20\sqrt{3}}{60}=\frac{\sqrt{3}}{3}$ $\theta=\cos^{- 1}(\frac{\sqrt{3}}{3})\approx54.7^{\circ}$

Step2: Resolve forces vertically

The vertical component of $F_1$ is $F_{1y}=-40\sin30^{\circ}$, and the vertical component of $F_2$ is $F_{2y}=-60\sin\theta$. The magnitude of the resultant $R$ is the sum of the vertical components of the two forces. $F_{1y}=-40\times\frac{1}{2}=-20$ lb. $\sin\theta=\sqrt{1 - \cos^{2}\theta}=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}$ $F_{2y}=-60\times\sqrt{\frac{2}{3}}=- 20\sqrt{6}$ lb. $R = F_{1y}+F_{2y}=-20-20\sqrt{6}\approx - 20(1 + \sqrt{6})\approx-68.99$ lb. The magnitude of the resultant is $|R|\approx69.0$ lb.

Answer:

$\theta\approx54.7^{\circ}$, magnitude of the resultant $\approx69.0$ lb