Explanation
Step 1
angle of the inclined plane
to find the angle, we can use the sin function
[tex]\sin\theta=\frac{opposite\text{ side}}{hypotenuse}[/tex]so
a)let
[tex]\begin{gathered} hypotenuse(the\text{ longest side\rparen=6.8} \\ opposite\text{ side\lparen the one in front of the angle\rparen=3.05} \end{gathered}[/tex]b) replace and solve for the angle
[tex]\begin{gathered} \sin\theta=\frac{opposite\text{ side}}{hypotenuse} \\ sin\theta=\frac{3.05}{6.8} \\ \theta=\sin^{-1}(\frac{3.05}{6.8}) \\ \theta=26.65° \end{gathered}[/tex]therefore
the angle is 26.65 degrees
Step 2
Magnitude of the normal force
a) Diagram
Free body diagram
now ,as the object is at rest, the sum of the forces acting on it must equals zero , hence
for y -axis
[tex]\begin{gathered} Normal\text{ force-mgcos}\theta=0 \\ \end{gathered}[/tex]replace and solve for normal force
[tex]\begin{gathered} Normal\text{ force-mgcos}\theta=0 \\ Normal\text{ force=mg cos}\theta \\ Normal\text{ force=15.5 kg*9.8}\frac{m}{s^2}cos26.65 \\ Normal\text{ force=131.38 Newtons} \end{gathered}[/tex]so, the Normal force is 131.38 Newtons
Step 3
finally, the component of the force of gravity
to know this, let's set the equation in the x-axis
so
[tex]\begin{gathered} Ff-mg*sin\theta=0 \\ Ff=mgsin\theta \\ (mg\text{ sin}\theta)\rightarrow component\text{ of the force of gravity in the plane } \\ so \\ component=\text{ 15.5kg*9.8}\frac{m}{s^2}sin26.65 \\ component\text{ =68.13 Newtons} \end{gathered}[/tex]so, the component of the force of gravity along the plane is 68.13 Newtons
I hope this helps you
