problem 4: if θ = 30° and t = 6kn, determine the magnitude of the resultant force acting on the eyebolt and…

Explanation:

Step1: Resolve forces into components

Let the force $T = 6kN$ and the other force $F = 8kN$. The $x$-component of $T$ is $T_x=T\cos\theta$, and the $y$-component of $T$ is $T_y = T\sin\theta$. Given $\theta = 30^{\circ}$ and $T = 6kN$, we have $T_x=6\cos30^{\circ}=6\times\frac{\sqrt{3}}{2}=3\sqrt{3}kN$ and $T_y = 6\sin30^{\circ}=3kN$. The $x$-component of the $8 - kN$ force is $F_x=8\cos30^{\circ}=4\sqrt{3}kN$ and the $y$-component is $F_y=- 8\sin30^{\circ}=-4kN$. The total $x$-component of the resultant force $R_x=T_x + F_x=3\sqrt{3}+4\sqrt{3}=7\sqrt{3}kN$. The total $y$-component of the resultant force $R_y=T_y+F_y=3+( - 4)=-1kN$.

Step2: Calculate the magnitude of the resultant force

The magnitude of the resultant force $R$ is given by $R=\sqrt{R_x^{2}+R_y^{2}}$. Substitute $R_x = 7\sqrt{3}kN$ and $R_y=-1kN$ into the formula: [ \begin{align*} R&=\sqrt{(7\sqrt{3})^{2}+(-1)^{2}}\ &=\sqrt{147 + 1}\ &=\sqrt{148}\ &=2\sqrt{37}\approx 12.166kN \end{align*} ]

Step3: Calculate the direction of the resultant force

The direction $\alpha$ of the resultant force is given by $\tan\alpha=\frac{|R_y|}{|R_x|}$. $\tan\alpha=\frac{1}{7\sqrt{3}}=\frac{\sqrt{3}}{21}$. $\alpha=\arctan(\frac{\sqrt{3}}{21})\approx 4.79^{\circ}$ clock - wise from the positive $x$-axis.

Answer:

The magnitude of the resultant force is approximately $12.17kN$ and its direction is approximately $4.79^{\circ}$ clockwise from the positive $x$-axis.