# Problem 2-1| Force Vectors| Engineering Mechanics: Statics| RC Hibbeler

#### Determine the magnitude of the resultant force $F_R=F_1+F_2$ and its direction, measured counterclockwise from the positive x axis.

SOLUTION:

Consider the free-body diagram below

The resultant force can be solved using the law of cosines. That is

$F_R=\sqrt{\left(250\right)^2+\left(375\right)^2-2\left(250\right)\left(375\right)cos\:75^{\circ} }=393.2=393\:lb$

The value of angle θ can be solved using sine law.

$\frac{393.2}{sin\:\left(75^{\circ} \right)}=\frac{250}{sin\:\theta }$

$\theta =37.89^{\circ}$

$\phi =360^{\circ} -45^{\circ} +37.89^{\circ} =353^{\circ}$

The resultant force has a magnitude of 393 lb and is located 353º measured counterclockwise from the positive x-axis.