20) why do you think its convenient to use parallel/perpendicular instead of x/y for problems involving inclines? justify your response.

21) now lets consider the normal force. if the angle of inclination is 0°, what would the magnitude of the normal force be equal to? what about if the angle of inclination was 90°? justify your answer.

22) conceptually, what do you think the equation would be for $f_{g\perp}$ in terms of $\theta$? does that fit your answer to question 21?

challenge: use your fbd you drew earlier to help make an equation to determine $f_{g\parallel}$ in terms of $\theta$ and $f_g$ using algebra and geometry. while youre at it, determine an equation for $f_{g\perp}$ in terms of $f_g$ and $\theta$.

Question

20) why do you think its convenient to use parallel/perpendicular instead of x/y for problems involving inclines? justify your response.

21) now lets consider the normal force. if the angle of inclination is 0°, what would the magnitude of the normal force be equal to? what about if the angle of inclination was 90°? justify your answer.

22) conceptually, what do you think the equation would be for $f_{g\perp}$ in terms of $\theta$? does that fit your answer to question 21?

challenge: use your fbd you drew earlier to help make an equation to determine $f_{g\parallel}$ in terms of $\theta$ and $f_g$ using algebra and geometry. while youre at it, determine an equation for $f_{g\perp}$ in terms of $f_g$ and $\theta$.

Accepted Answer

### Question 20
# Brief Explanations:
When dealing with incline problems, forces like gravity can be resolved into components parallel and perpendicular to the incline. Using parallel/perpendicular (to the incline) instead of x/y (standard coordinate system) simplifies force analysis. The normal force is perpendicular to the incline, and the component of gravity along the incline causes motion. Resolving forces along these directions (parallel to the incline's surface and perpendicular to it) makes it easier to apply Newton's laws, as the normal force has no component parallel to the incline (so only gravity's parallel component and friction act along that direction), and the perpendicular direction has the normal force balancing gravity's perpendicular component. This alignment reduces the number of force components to consider and simplifies equations (e.g., the perpendicular component of gravity is $ F_{g\perp} = F_g \cos\theta $ and parallel is $ F_{g\parallel} = F_g \sin\theta $, which are straightforward for incline analysis).
# Answer:
Using parallel/perpendicular to the incline simplifies force analysis by aligning with the incline's geometry, reducing force components to consider (normal force is perpendicular, gravity's components parallel/perpendicular are easy to define), and making Newton's laws easier to apply (e.g., perpendicular forces balance, parallel forces cause motion).

### Question 21
# Brief Explanations:
The normal force ($ F_N $) on an incline is equal to the perpendicular component of gravity ($ F_{g\perp} $) when there's no other perpendicular force. The formula for $ F_{g\perp} $ is $ F_g \cos\theta $, where $ \theta $ is the angle of inclination.
- For $ \theta = 0^\circ $: The incline is horizontal. $ \cos(0^\circ) = 1 $, so $ F_{g\perp} = F_g \times 1 = F_g $. Thus, the normal force equals the weight of the object (since the object is on a horizontal surface, normal force balances gravity).
- For $ \theta = 90^\circ $: The incline is vertical. $ \cos(90^\circ) = 0 $, so $ F_{g\perp} = F_g \times 0 = 0 $. Thus, the normal force is zero because the object is in free - fall (or sliding vertically, with no normal force from the vertical surface as there's no perpendicular contact force).
# Answer:
- At $ \theta = 0^\circ $ (horizontal incline), the magnitude of the normal force is equal to the weight of the object ($ F_N = F_g $) because $ F_{g\perp}=F_g\cos(0^\circ) = F_g $, and the normal force balances this perpendicular component.
- At $ \theta = 90^\circ $ (vertical incline), the magnitude of the normal force is $ 0 $ because $ F_{g\perp}=F_g\cos(90^\circ)=0 $, so there's no perpendicular component of gravity to balance, and the object is in a state where the surface can't exert a normal force (or the object is falling freely).

### Question 22
# Brief Explanations:
Conceptually, the perpendicular component of gravity ($ F_{g\perp} $) on an incline should be related to the weight ($ F_g $) and the cosine of the angle of inclination ($ \theta $). From the right - triangle formed by resolving gravity into components (where $ F_g $ is the hypotenuse, $ F_{g\perp} $ is the adjacent side to $ \theta $), the equation is $ F_{g\perp}=F_g\cos\theta $.
To check with Question 21:
- When $ \theta = 0^\circ $, $ \cos(0^\circ)=1 $, so $ F_{g\perp}=F_g\times1 = F_g $, which matches the normal force (equal to weight) at $ \theta = 0^\circ $.
- When $ \theta = 90^\circ $, $ \cos(90^\circ)=0 $, so $ F_{g\perp}=F_g\times0 = 0 $, which matches the normal force (zero) at $ \theta = 90^\circ $. So the equation fits the answers to Question 21.
# Answer:
The equation for $ F_{g\perp} $ in terms of $ \theta $ is $ F_{g\perp}=F_g\cos\theta $. Yes, it fits the answer to Question 21: at $ \theta = 0^\circ $, $ F_{g\perp}=F_g $ (normal force equals weight); at $ \theta = 90^\circ $, $ F_{g\perp}=0 $ (normal force is zero), which matches the normal force values from Question 21.

### Challenge
# Explanation:
## Step 1: Recall the Free - Body Diagram (FBD)
The FBD for an object on an incline includes gravity ($ F_g $) acting vertically downward, the normal force ($ F_N $) perpendicular to the incline, and (if present) friction parallel to the incline. We resolve $ F_g $ into two components: one parallel ($ F_{g\parallel} $) to the incline and one perpendicular ($ F_{g\perp} $) to the incline.

## Step 2: Use Geometry to Relate Angles
The angle between $ F_g $ (vertical) and $ F_{g\perp} $ (perpendicular to the incline) is equal to the angle of inclination $ \theta $. This is because the incline angle $ \theta $ is equal to the angle between the vertical (direction of $ F_g $) and the perpendicular to the incline (direction of $ F_{g\perp} $) (alternate interior angles in the right - triangle formed by the incline, vertical, and perpendicular to incline).

## Step 3: Derive $ F_{g\parallel} $
In the right - triangle formed by $ F_g $, $ F_{g\parallel} $, and $ F_{g\perp} $, $ F_{g\parallel} $ is the side opposite to the angle $ \theta $. Using the definition of sine in a right - triangle ($ \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} $), where the hypotenuse is $ F_g $ and the opposite side is $ F_{g\parallel} $, we get:
$ \sin\theta=\frac{F_{g\parallel}}{F_g} $
Multiplying both sides by $ F_g $ gives $ F_{g\parallel}=F_g\sin\theta $.

## Step 4: Derive $ F_{g\perp} $
For $ F_{g\perp} $, it is the side adjacent to the angle $ \theta $ in the right - triangle. Using the definition of cosine in a right - triangle ($ \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} $), where the adjacent side is $ F_{g\perp} $ and the hypotenuse is $ F_g $, we get:
$ \cos\theta=\frac{F_{g\perp}}{F_g} $
Multiplying both sides by $ F_g $ gives $ F_{g\perp}=F_g\cos\theta $.

# Answer:
The equation for $ F_{g\parallel} $ in terms of $ \theta $ and $ F_g $ is $ \boldsymbol{F_{g\parallel}=F_g\sin\theta} $.
The equation for $ F_{g\perp} $ in terms of $ F_g $ and $ \theta $ is $ \boldsymbol{F_{g\perp}=F_g\cos\theta} $.

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QUESTION IMAGE

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Explanation:

Question 20

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Question 21