2 - 48. three forces act on the bracket. determine the magnitude and direction θ of f₁ so that the resultant…

Explanation:

Step1: Resolve forces in x - direction

$F_{Rx}=\sum F_{x}$, so $800\sin60^{\circ}=F_{1}\sin(60^{\circ}+\theta)+ 180\sin15^{\circ}$

Step2: Resolve forces in y - direction

$F_{Ry}=\sum F_{y}$, and since the resultant is along the positive x - axis, $F_{Ry} = 0$. So $800\cos60^{\circ}=200 + F_{1}\cos(60^{\circ}+\theta)-180\cos15^{\circ}$ We have a system of two equations with two unknowns ($F_{1}$ and $\theta$). First, from the y - direction equation: $400=200 + F_{1}\cos(60^{\circ}+\theta)-180\cos15^{\circ}$ $F_{1}\cos(60^{\circ}+\theta)=400 - 200+180\cos15^{\circ}$ $F_{1}\cos(60^{\circ}+\theta)=200 + 180\cos15^{\circ}$ From the x - direction equation: $800\sin60^{\circ}=F_{1}\sin(60^{\circ}+\theta)+180\sin15^{\circ}$ $F_{1}\sin(60^{\circ}+\theta)=800\sin60^{\circ}-180\sin15^{\circ}$ Then, $\tan(60^{\circ}+\theta)=\frac{800\sin60^{\circ}-180\sin15^{\circ}}{200 + 180\cos15^{\circ}}$ $\sin15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.259$, $\cos15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.966$, $\sin60^{\circ}=\frac{\sqrt{3}}{2}\approx0.866$ $\tan(60^{\circ}+\theta)=\frac{800\times0.866-180\times0.259}{200 + 180\times0.966}$ $\tan(60^{\circ}+\theta)=\frac{692.8-46.62}{200 + 173.88}$ $\tan(60^{\circ}+\theta)=\frac{646.18}{373.88}\approx1.728$ $60^{\circ}+\theta=\arctan(1.728)\approx59.9^{\circ}$ $\theta\approx - 0.1^{\circ}$ Substitute $\theta$ back into the y - direction equation to find $F_{1}$: $F_{1}\cos(60^{\circ}-0.1^{\circ})=200 + 180\times0.966$ $F_{1}\cos59.9^{\circ}=200+173.88$ $F_{1}=\frac{373.88}{\cos59.9^{\circ}}\approx747.7\ N$

Answer:

Magnitude of $F_{1}\approx748\ N$, $\theta\approx - 0.1^{\circ}$