Vector Mechanics Dynamics 9th Edition Beer Johnston Solution 1 -

The y-component of $F_2$ is: $F_{2y} = F_2 \sin 60^\circ = 150 \sin 60^\circ = 129.90 \text{ N}$

The y-component of $F_1$ is: $F_{1y} = F_1 \sin 30^\circ = 100 \sin 30^\circ = 50 \text{ N}$

The preceding example displays one method of supporting students as they build foundational knowledge with "Vector Mechanics for Engineers: Dynamics 9th Edition Beer Johnston Solution 1" . The y-component of $F_2$ is: $F_{2y} = F_2

The x-component of the resultant force $R$ is: $R_x = F_{1x} + F_{2x} = 86.60 + 75 = 161.60 \text{ N}$

Problem 1.1 in Chapter 1 of the book asks to determine the magnitude of the resultant of two forces applied to a particle. The magnitude of the resultant force $R$ is:

The x-component of $F_2$ is: $F_{2x} = F_2 \cos 60^\circ = 150 \cos 60^\circ = 75 \text{ N}$

The magnitude of the resultant force $R$ is $242.11 \text{ N}$. The y-component of $F_2$ is: $F_{2y} = F_2

The magnitude of the resultant force $R$ is: $R = \sqrt{R_x^2 + R_y^2} = \sqrt{(161.60)^2 + (179.90)^2} = 242.11 \text{ N}$