a) 6.1 N
Given a force of magnitude F, the horizontal (x) component of the force can be found using the equation:
[tex]F_x = F cos \theta[/tex]
where
F is the magnitude of the force
[tex]\theta[/tex] is the angle between the direction of the force and the horizontal
In this problem, we have
F = 6.6 N
[tex]\theta=21.6^{\circ}[/tex]
Therefore, the horizontal component is
[tex]F_x = (6.6)(cos 21.6^{\circ})=6.1 N[/tex]
b) 2.4 N
The vertical (y) component of the force can be found using the equation:
[tex]F_y = F sin \theta[/tex]
where
F is the magnitude of the force
[tex]\theta[/tex] is the angle between the direction of the force and the horizontal
Here we have
F = 6.6 N
[tex]\theta=21.6^{\circ}[/tex]
So, the vertical component is
[tex]F_y = (6.6)(sin 21.6^{\circ})=2.4 N[/tex]
