Answer:
72.0 N
Explanation:
The horizontal component of the net force is given by:
[tex]F_x = F cos \theta[/tex]
where
F is the magnitude of the net force
[tex]\theta[/tex] is the angle between the net force and the horizontal.
In this problem, we know:
- the horizontal component, [tex]F_x = 68 N[/tex]
- the angle, [tex]\theta=19.2^{\circ}[/tex]
Therefore, we can re-arrange the equation to find the magnitude of the net force:
[tex]F=\frac{F_x}{cos \theta}=\frac{68 N}{cos 19.2^{\circ}}=72.0 N[/tex]
