The x-component of the normal force is equal to 1706.45 N.
Why?
To solve the problem, and since there is no additional information, we can safely assume that the x-axis is parallalel to the hill surface and the y-axis is perpendicular to the x-axis. Knowing that, we can calculate the components of the normal force (or weight for this case), using the following formulas:
[tex]N_{x}=W*Sin(\alpha)=mg*Sin(\alpha)\\\\N_{y}=W*Cos(\alpha)=mg*Cos(\alpha)[/tex]
Now, using the given information, we have:
[tex]mass=m=1150Kg\\\alpha=8.70\°\\g=9.81\frac{m}{s^{2}}[/tex]
Calculating, we have:
[tex]N_{x}=mg*Sin(\alpha)[/tex]
[tex]N_{x}=1150Kg*9.81\frac{m}{s^{2}}*Sin(8.70\°)\\\\N_{x}=11281.5\frac{Kg.m}{s^{2} }*Sin(8.70\°)=1706.45\frac{Kg.m}{s^{2} }=1706.45.23N[/tex]
Hence, we have that the x-component of the normal force is equal to 1706.45 N.
Have a nice day!
