# Chapter 3 - Vectors and Coordinates Systems - Exercises and Problems - Page 79: 38

$F_x = 451~N$ $F_y = 315~N$

#### Work Step by Step

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; $F_y = F~sin(\theta)$ $F_y = (550~N)~sin(35^{\circ})$ $F_y = 315~N$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.