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 ±f_{rx}=σf_{x}: 800 sin 60°=f₁ sin(60° + θ)-\frac{12}{13}(180)

Explanation:

Step1: Resolve forces in x - direction

The resultant force in the x - direction $F_{Rx}=\sum F_{x}$. We know $F_{Rx} = 800\ N$. The x - component of $F_2$ is $0$ since it is along the y - axis. The x - component of $F_3$ is $-\frac{12}{13}(180)$ and the x - component of $F_1$ is $F_1\sin(60^{\circ}+\theta)$. So, $800 = F_1\sin(60^{\circ}+\theta)-\frac{12}{13}(180)$.

Step2: Resolve forces in y - direction

Since the resultant force is along the positive x - axis, the sum of the y - components of the forces is zero. The y - component of $F_2$ is $200\ N$, the y - component of $F_3$ is $\frac{5}{13}(180)$ and the y - component of $F_1$ is $F_1\cos(60^{\circ}+\theta)$. So, $0 = 200+\frac{5}{13}(180)-F_1\cos(60^{\circ}+\theta)$. We can rewrite it as $F_1\cos(60^{\circ}+\theta)=200 + \frac{5}{13}(180)=200 + \frac{900}{13}=\frac{2600 + 900}{13}=\frac{3500}{13}$.

Step3: Use the identity $\tan(A + B)$

We know that $\tan(60^{\circ}+\theta)=\frac{F_1\sin(60^{\circ}+\theta)}{F_1\cos(60^{\circ}+\theta)}$. From the x - direction equation $F_1\sin(60^{\circ}+\theta)=800+\frac{12}{13}(180)=800+\frac{2160}{13}=\frac{10400+2160}{13}=\frac{12560}{13}$. Then $\tan(60^{\circ}+\theta)=\frac{\frac{12560}{13}}{\frac{3500}{13}}=\frac{12560}{3500}=\frac{628}{175}$. So, $60^{\circ}+\theta=\arctan(\frac{628}{175})\approx74.3^{\circ}$, and $\theta\approx74.3^{\circ}- 60^{\circ}=14.3^{\circ}$.

Step4: Find the magnitude of $F_1$

Using $F_1\cos(60^{\circ}+\theta)=\frac{3500}{13}$, and $\cos(60^{\circ}+\theta)=\cos(74.3^{\circ})\approx0.27$. Then $F_1=\frac{\frac{3500}{13}}{\cos(74.3^{\circ})}=\frac{\frac{3500}{13}}{0.27}\approx981\ N$.

Answer:

The magnitude of $F_1$ is approximately $981\ N$ and the direction $\theta$ is approximately $14.3^{\circ}$.