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The century-old ascensores in Valparaiso, Chile, are small cable cars that go up and down the steep hillsides. As (Figure 1) shows, one car ascends as the other descends. The cars use a two-cable arrangement to compensate for friction; one cable passing around a large pulley connects the cars, the second is pulled by a small motor. Suppose the mass of each car (with passengers) is 1700 kg, the coefficient of rolling friction is 0.030, and the cars move at constant speed. 

This image illustrates a cable car system, where one car ascends as the other descends. The cars are connected via a pulley, and a motor drives them up a slope inclined at 35 degrees. The problem gives the mass of each car as 1700 kg, the coefficient of rolling friction as 0.030, and it states that the cars are moving at constant speed, implying the net force on each car is zero.

To find the forces involved, we can break the problem into the following steps:

### 1. **Calculate the gravitational force acting on the cars along the incline**:
- The component of gravitational force along the incline is given by:
  \[
  F_{\text{gravity}} = m g \sin \theta
  \]
  where:
  - \( m = 1700 \, \text{kg} \) (mass of the car),
  - \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity),
  - \( \theta = 35^\circ \) (angle of incline).

  Substituting the values:
  \[
  F_{\text{gravity}} = 1700 \times 9.81 \times \sin 35^\circ
  \]

### 2. **Calculate the force due to rolling friction**:
- The frictional force is given by:
  \[
  F_{\text{friction}} = \mu_r \times N
  \]
  where:
  - \( \mu_r = 0.030 \) (coefficient of rolling friction),
  - \( N = m g \cos \theta \) is the normal force acting on the car, perpendicular to the incline.

  Thus:
  \[
  F_{\text{friction}} = 0.030 \times 1700 \times 9.81 \times \cos 35^\circ
  \]

### 3. **Determine the total force required to move each car**:
- Since the cars move at constant speed, the motor only needs to overcome the sum of the gravitational force and the rolling friction:
  \[
  F_{\text{total}} = F_{\text{gravity}} + F_{\text{friction}}
  \]

### 4. **Net force between cars**:
Since one car is moving up and the other is moving down, the tension in the cable connecting them helps counterbalance the forces, reducing the motor's work.

Let's compute the forces step-by-step.

#### Step 1: Gravitational Force
\[
F_{\text{gravity}} = 1700 \times 9.81 \times \sin 35^\circ = 1700 \times 9.81 \times 0.5736 = 9570.77 \, \text{N}
\]

#### Step 2: Frictional Force
\[
F_{\text{friction}} = 0.030 \times 1700 \times 9.81 \times \cos 35^\circ = 0.030 \times 1700 \times 9.81 \times 0.8192 = 409.46 \, \text{N}
\]

#### Step 3: Total Force
\[
F_{\text{total}} = 9570.77 + 409.46 = 9979.23 \, \text{N}
\]

Thus, the motor must apply a force of approximately **9979.23 N** to keep the system in motion at constant speed.

Would you like further clarification or additional details? 

Here are 5 related questions:
1. How does the coefficient of friction affect the total force required by the motor?
2. What changes in the force calculations if the angle of incline increases?
3. How does the mass of the cars influence the system's equilibrium?
4. What role does the pulley system play in reducing the motor's workload?
5. How would the system behave if the speed were not constant (e.g., accelerating)?

**Tip:** When solving inclined plane problems, always decompose the gravitational force into parallel and perpendicular components relative to the surface for easier calculations.

The century-old ascensores in Valparaiso, Chile, are small cable cars that go up and down the steep hillsides. As (Figure 1) shows, one car ascends as the other descends. The cars use a two-cable arrangement to compensate for friction; one cable passing around a large pulley connects the cars, the second is pulled by a small motor. Suppose the mass of each car (with passengers) is 1700 kg, the coefficient of rolling friction is 0.030, and the cars move at constant speed.

Explore the forces acting on a cable car system moving on an inclined plane in Valparaiso, Chile. With a mass of 1700 kg, a rolling friction coefficient of 0.030, and an incline of 35 degrees, learn how to calculate the total force required for constant motion. Suitable for advanced physics students in Grades 11-12.

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