The y-component of $F_2$ is: $F_{2y} = F_2 \sin 60^\circ = 150 \sin 60^\circ = 129.90 \text{ N}$
The magnitude of the resultant force $R$ is $242.11 \text{ N}$. The y-component of $F_2$ is: $F_{2y} = F_2
The x-component of the resultant force $R$ is: $R_x = F_{1x} + F_{2x} = 86.60 + 75 = 161.60 \text{ N}$ The y-component of $F_2$ is: $F_{2y} = F_2
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 $F_1$ is: $F_{1x} = F_1 \cos 30^\circ = 100 \cos 30^\circ = 86.60 \text{ N}$
To find the resultant force $R$, resolve the forces $F_1$ and $F_2$ into their x and y components.