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.

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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.