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

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.\n\nsolution\n→f_{rx}=σf_x: 800 sin 60° = f₁ sin(60° + θ) - \\frac{12}{13}(180)\n+↑f_{ry}=σf_y: 800 cos 60° = f₁ cos(60° + θ) + 200+\\frac{5}{13}(180)\n60° + θ = 81.34°\nθ = 21.3°\nf₁ = 869 n

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.\n\nsolution\n→f_{rx}=σf_x: 800 sin 60° = f₁ sin(60° + θ) - \\frac{12}{13}(180)\n+↑f_{ry}=σf_y: 800 cos 60° = f₁ cos(60° + θ) + 200+\\frac{5}{13}(180)\n60° + θ = 81.34°\nθ = 21.3°\nf₁ = 869 n

Answer

Explanation:

Step1: Set up x - component equation

The resultant force $F_{Rx}$ along the x - axis is given by the sum of the x - components of the individual forces. The resultant force magnitude is $F_R = 800$ N and is directed along the positive x - axis. So, $F_{Rx}=F_R\sin60^{\circ}$. The x - component of $F_1$ is $F_1\sin(60^{\circ}+\theta)$, and the x - component of $F_3$ (where $F_3 = 180$ N and the ratio of its components is based on the right - triangle with sides 5 and 12) is $-\frac{12}{13}(180)$. So, $800\sin60^{\circ}=F_1\sin(60^{\circ}+\theta)-\frac{12}{13}(180)$.

Step2: Set up y - component equation

The y - component of the resultant force $F_{Ry}=F_R\cos60^{\circ}$. The y - component of $F_1$ is $F_1\cos(60^{\circ}+\theta)$, the y - component of $F_2$ is 200 N, and the y - component of $F_3$ is $\frac{5}{13}(180)$. So, $800\cos60^{\circ}=F_1\cos(60^{\circ}+\theta)+200 + \frac{5}{13}(180)$.

Step3: Solve the system of equations

We have a system of two equations with two unknowns $F_1$ and $\theta$. First, from the y - component equation: [ \begin{align*} 800\times\frac{1}{2}&=F_1\cos(60^{\circ}+\theta)+200+\frac{5}{13}\times180\ 400&=F_1\cos(60^{\circ}+\theta)+200 + \frac{900}{13}\ F_1\cos(60^{\circ}+\theta)&=400-(200+\frac{900}{13})\ F_1\cos(60^{\circ}+\theta)&=200-\frac{900}{13}=\frac{2600 - 900}{13}=\frac{1700}{13} \end{align*} ] From the x - component equation: [ \begin{align*} 800\times\frac{\sqrt{3}}{2}&=F_1\sin(60^{\circ}+\theta)-\frac{12}{13}\times180\ 400\sqrt{3}&=F_1\sin(60^{\circ}+\theta)-\frac{2160}{13}\ F_1\sin(60^{\circ}+\theta)&=400\sqrt{3}+\frac{2160}{13} \end{align*} ] Dividing the second equation by the first equation gives $\tan(60^{\circ}+\theta)=\frac{400\sqrt{3}+\frac{2160}{13}}{\frac{1700}{13}}$. Solving for $\theta$ gives $\theta = 21.3^{\circ}$. Substitute $\theta$ into either the x - or y - component equation to find $F_1$. Substituting into the y - component equation: [ \begin{align*} 400&=F_1\cos(60^{\circ}+ 21.3^{\circ})+200+\frac{900}{13}\ F_1\cos(81.34^{\circ})&=400-(200+\frac{900}{13})\ F_1&=\frac{400-(200+\frac{900}{13})}{\cos(81.34^{\circ})}\approx869 \end{align*} ]

Answer:

$\theta = 21.3^{\circ},F_1 = 869$ N