## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$F_x = 451~N$ $F_y = 315~N$
Paul is pulling with a force $F$ that is directed at an angle of $\theta = 35^{\circ}$ above the ground. We can find the horizontal component of the force; $F_x = F~cos(\theta)$ $F_x = (550~N)~cos(35^{\circ})$ $F_x = 451~N$ We can find the horizontal component of the force; $F_y = F~sin(\theta)$ $F_y = (550~N)~sin(35^{\circ})$ $F_y = 315~N$