Question

2 - 48. three forces act on the bracket. determine the magnitude and direction θ of f₁ so that the resultant force is directed along the positive x axis and has a magnitude of 800 n.
f₂ = 200 n
f₃ = 180 n
solution
→σfₓ = σfₓ: 800 sin 60° = f₁ sin(60° + θ) - 12(180)

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