Question

the gusset plate is subjected to the forces of three members as shown in (figure 1). the forces are concurrent at point o. take f = 2 kn. determine the magnitude of the tension force in member c for equilibrium. express your answer to three significant figures and include the appropriate units. determine the angle θ for equilibrium. express your answer in degrees to three significant figures.

Explanation:

Step1: Set up equilibrium equations

For equilibrium, $\sum F_x = 0$ and $\sum F_y=0$. Let the force in member $C$ be $T$. Resolve force $F = 2$ kN into components. $F_x=F\cos30^{\circ}$ and $F_y = F\sin30^{\circ}$.
$\sum F_x=F\cos30^{\circ}-T\cos\theta = 0$ and $\sum F_y=9 - F\sin30^{\circ}-T\sin\theta=0$. Given $F = 2$ kN, $F_x = 2\cos30^{\circ}=\sqrt{3}$ kN and $F_y=2\sin30^{\circ}=1$ kN.

Step2: Solve for $T$ from $x -$ component equation

From $\sum F_x = 0$, we have $T\cos\theta=\sqrt{3}$.
From $\sum F_y = 0$, we have $9 - 1-T\sin\theta=0$, so $T\sin\theta=8$.

Step3: Square and add the two - equations

$(T\cos\theta)^2+(T\sin\theta)^2 = (\sqrt{3})^2 + 8^2$. Using $\sin^{2}\theta+\cos^{2}\theta = 1$, we get $T^{2}=3 + 64=67$.

Step4: Find the magnitude of $T$

$T=\sqrt{67}\approx 8.19$ kN.

Step5: Find the angle $\theta$

Since $T\cos\theta=\sqrt{3}$ and $T=\sqrt{67}$, then $\cos\theta=\frac{\sqrt{3}}{\sqrt{67}}$. $\theta=\cos^{-1}(\frac{\sqrt{3}}{\sqrt{67}})\approx 77.4^{\circ}$

Answer:

Magnitude of tension force in member $C$: $8.19$ kN
Angle $\theta$: $77.4^{\circ}$