Answer:
262.5 N, 454.7 N
Explanation:
The component of the weight parallel to the inclined plane is given by:
[tex]F_{par}=mg sin \theta[/tex]
where
(mg) is the weight of the trunk
[tex]\theta[/tex] is the angle of the ramp
In this problem, [tex](mg)=525 N[/tex] while [tex]\theta=30.0^{\circ}[/tex], so t he component of the weight parallel to the inclined plane is
[tex]F_{par}=(525 N)( sin 30.0^{\circ})=262.5 N[/tex]
Instead, the component of the weight normal to the plane is given by:
[tex]F_{nor}=mg cos \theta =(525 N)(cos 30.0^{\circ})=454.7 N[/tex]
