A 640  N  person is riding on a 8.9  m radius Ferris Wheel. The ride rotates at a constant angular speed. When the person is at the bottom of the wheel, the normal force is 960  N
  . Determine the normal force on the person when they are at point A, which is 40  ∘
  to the left of the vertical.

We can approach this problem by analyzing the forces acting on the person in different positions on the Ferris wheel. Here's a breakdown of the situation:

### Given Data:
- Weight of the person, \( W = 640 \, \text{N} \) (which implies the person's mass \( m = \frac{W}{g} = \frac{640}{9.8} \, \text{kg} \)).
- Radius of the Ferris wheel, \( r = 8.9 \, \text{m} \).
- Normal force at the bottom, \( N_{\text{bottom}} = 960 \, \text{N} \).
- Angle of the person at point A is \( \theta = 40^\circ \) to the left of the vertical.

### Step 1: Analyzing forces at the bottom
At the bottom of the Ferris wheel, the forces acting on the person are:
- Weight \( W = 640 \, \text{N} \) acting downwards.
- The normal force \( N_{\text{bottom}} = 960 \, \text{N} \) acting upwards.

The net force towards the center of the Ferris wheel provides the centripetal force \( F_c \). Therefore, the equation of motion at the bottom is:
\[
N_{\text{bottom}} - W = F_c \quad \text{(1)}
\]
Substitute the values:
\[
960 - 640 = F_c
\]
\[
F_c = 320 \, \text{N}
\]

### Step 2: Angular velocity and centripetal acceleration
The centripetal force \( F_c \) is related to the mass \( m \) and centripetal acceleration \( a_c \) by:
\[
F_c = m a_c = m \frac{v^2}{r}
\]
First, calculate the mass \( m \):
\[
m = \frac{W}{g} = \frac{640}{9.8} = 65.31 \, \text{kg}
\]
Now, the centripetal acceleration \( a_c \):
\[
a_c = \frac{F_c}{m} = \frac{320}{65.31} = 4.9 \, \text{m/s}^2
\]
This is the acceleration towards the center of the Ferris wheel.

### Step 3: Forces at point A (40° to the left of the vertical)
At point A, the person experiences two forces:
1. Weight \( W = 640 \, \text{N} \), which can be split into components:
   - \( W_{\parallel} = W \sin(\theta) = 640 \sin(40^\circ) \).
   - \( W_{\perp} = W \cos(\theta) = 640 \cos(40^\circ) \).
2. Normal force \( N_A \), which acts perpendicular to the Ferris wheel’s surface.

At point A, the centripetal force is the net inward force, so we have:
\[
N_A + W \sin(\theta) = F_c
\]
Solving for \( N_A \):
\[
N_A = F_c - W \sin(\theta)
\]
Substitute the known values:
\[
N_A = 320 - 640 \sin(40^\circ)
\]
Using \( \sin(40^\circ) \approx 0.6428 \):
\[
N_A = 320 - 640 \times 0.6428 = 320 - 410 = -90 \, \text{N}
\]

The negative sign indicates that the normal force at point A would need to act inward (toward the center) rather than outward.

### Final Answer:
The normal force on the person at point A is \( \boxed{90 \, \text{N}} \) directed inward.

---

Would you like further explanation or more details on any part? Here are some follow-up questions:

1. What is the role of centripetal force in circular motion?
2. How does the normal force change at different points on a Ferris wheel?
3. Can we find the speed of the Ferris wheel using the given forces?
4. How does the angle \( \theta \) affect the components of weight?
5. How would the problem change if the Ferris wheel increased its speed?

**Tip:** In circular motion, the net inward force is always the centripetal force, regardless of whether the force is provided by gravity, tension, or normal force.

A 640 N person is riding on a 8.9 m radius Ferris Wheel. The ride rotates at a constant angular speed. When the person is at the bottom of the wheel, the normal force is 960 N. Determine the normal force on the person when they are at point A, which is 40° to the left of the vertical.

Determine the normal force on a 640 N person riding a Ferris wheel with an 8.9 m radius at 40 degrees to the left of the vertical. This problem involves circular motion, centripetal force, and trigonometric analysis, suitable for students in Grades 11-12.

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