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

$F=960N$ $\theta=45.2^{\circ}$
We are given that in the figure, that the magnitude of the resultant force is 500N. We are asked to determine the magnitude of the force, $F$, and its direction, measured counterclockwise from the positive x axis. We can treat the forces like a triangle and use the law of cosines: $c^2=a^2+b^2−2a∗b∗\cos C$ $c=\sqrt{a^2+b^2−2a∗b∗\cos C}$ $c=\sqrt{(500N)^2+(700N)^2−2∗700N∗500N∗\cos(90^{\circ}+15^{\circ})}$ $c=\sqrt{(500N)^2+(700N)^2−2∗700N∗500N∗\cos(105^{\circ})}$ $c=960N$ $F=960N$ Now we can use the law of sines to find the angle between $F$ and the x-axis. $\sin\alpha/700N=\sin105^{\circ}/960N$ $\alpha=\arcsin(\sin105^{\circ}/960N∗700N)$ $\alpha=44.8^{\circ}$ This is measured from the y-axis. We need to subtract this from 90 to get the angle from the x-axis, $\theta$. $\theta=90^{\circ}-44.8^{\circ}=45.2^{\circ}$