Answer to Question #201058 in Mechanics | Relativity for karen

Question

The supporting cables AB and AC are oriented so that the components of the 360-lb force along AB and AC are 185 lb and 200 lb, respectively. Determine the angles α and β

Answer 1

Since the cable setup is in equilibrium, the horizontal, as well as the vertical forces, should be in equilibrium.

From the given figure,

$F(AB)*sin(\alpha) = F(AC)*sin(\beta)$

Horizontal forces in equilibrium

$F(AB)*cos(\alpha) = F(AC)*cos(\beta)$

Vertical forces in equilibrium F(AB) and F(AC) are the forces in the cables AB and AC respectively. Balancing the horizontal forces,

$F = \sqrt{F_1^2+F_2^2-2F_1F_2cos\alpha}$

$cos\alpha = \large\frac{F_1^2+F_2^2-F^2}{2F_1F_2}$ = $\large\frac{185^2+200^2-360^2}{2*185*200}$ $=-0.74831 \to \alpha = 138^o$

$F(AB)*sin(\alpha) = F(AC)*sin(\beta)$

$\alpha + \beta = 138^o \to \alpha = 138^o-\beta$

$185cos\alpha=200cos\beta$

$185cos(138^o-\beta)=200cos\beta$

$185(cos138^ocos\beta-sin138^osin\beta)=200cos\beta$

$\beta=\large\frac{\ln \left(-\arccos \left(\frac{40\cos \left(β\right)}{37}\right)+2\pi n+β\right)}{\ln \left(138\right)}$