Questions: Problem 6: (7% of Assignment Value)
Circular turns of radius r in a race track are often banked at an angle θ to allow the cars to achieve higher speeds around the turns. Assume friction is not present, and use the coordinate system specified.
- Part (a) v
Find the y-component of the normal force FN on a car going around the turn in terms of the angle θ and the magnitude of the normal vector FN.
FK, y=FN cos(θ)
Correct!
Part (b)
Find the x-component of the normal force FN on a car going around the turn in terms of the angle θ and the magnitude of the normal vector FN.
FN x x=
Transcript text: Problem 6: ( $7 \%$ of Assignment Value)
Circular turns of radius $r$ in a race track are often banked at an angle $\theta$ to allow the cars to achieve higher speeds around the turns. Assume friction is not present, and use the coordinate system specified.
- Part (a) $\boldsymbol{v}$
Find the $y$-component of the normal force $F_{\mathrm{N}}$ on a car going around the turn in terms of the angle $\theta$ and the magnitude of the normal vector $F_{\mathrm{N}}$.
\[
F_{\mathrm{K}, \mathrm{y}}=\mathrm{F}_{\mathrm{N}} \cos (\theta)
\]
$\checkmark$ Correct!
Part (b)
Find the $x$-component of the normal force $F_{\mathrm{N}}$ on a car going around the turn in terms of the angle $\theta$ and the magnitude of the normal vector $F_{\mathrm{N}}$.
\[
F_{\mathrm{N} x \mathrm{x}}=
\]
Solution
Solution Steps
Step 1: Understanding the Problem
We need to find the x-component of the normal force \( F_N \) on a car going around a turn in terms of the angle \( \theta \) and the magnitude of the normal vector \( F_N \).
Step 2: Analyzing the Components of the Normal Force
The normal force \( F_N \) can be broken down into its x and y components using trigonometric functions. The y-component is given by \( F_{N_y} = F_N \cos(\theta) \).
Step 3: Finding the x-component of the Normal Force
To find the x-component of the normal force, we use the sine function:
\[ F_{N_x} = F_N \sin(\theta) \]