## Engineering Mechanics: Statics & Dynamics (14th Edition)

$F_R=497N$ $\phi =155^{\circ}$
We are given that in the figure, $\theta=60^{\circ}$ and $F=450N$. We are asked to determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. We can treat the forces like a triangle and use the law of cosines: $c^2=a^2+b^2−2a∗b∗\cos C$ $c=\sqrt{a^2+b^2−2a∗b∗\cos C}$ $c=\sqrt{(700N)^2+(450N)^2−2∗700N∗450N∗\cos(60^{\circ}−15^{\circ})}$ $c=497N$ $F_R=497N$ Now we can use the law of sines to find the angle between $F$ and $F_r$, $\alpha$ $\sin\alpha*700N=\sin45^{\circ}497N$ $\alpha=\arcsin(sin45^{\circ}497N∗700N)$ $\alpha=95.19^{\circ}$ We will have to add this to θ to get the angle from the x-axis $ϕ=\alpha+\theta=95.19^{\circ}+60^{\circ}=155^{\circ}$