Recent content by RoboRaptor

I assumed that was the case because I thought Normal Force had the same magnitude as the force that is being applied perpendicular to the road, in this case, the y-component of the weight. Our teacher taught us a way to do free-body diagrams on inclined ramps which I'm not sure if it's the norm...

Hey BvU! I just realized how I wrote 40° instead of 14°, that was just a silly mistake on my part 🤦‍♂️. However, this changes the other 50° angle to 76°. Here's a more detailed image of my work. It's a bit messy but I hope you can understand it. Not sure I understand the question. Basically...

First I figured out the normal force being exerted on the car using the equation above. Cos(40°)*(1050*9.8) = 7883N Next, I tried to find out the horizontal component of the normal force by doing: Cos(50) * 7883 = 5067N I figured out the angle by using certain geometrical properties. Next, I...