The century-old ascensores in Valparaiso, Chile, are small cable cars that go up and down the steep hillsides. 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. What is the tension in the connecting cable? The tension in the cable to the motor is 820N and the angle is 35 degrees from horizontal

To solve this problem, we need to calculate the tension in the **connecting cable**, which connects the two cars in the ascensores system. Given that the system is moving at constant speed, this means the net force on the system is zero, implying that the forces due to gravity, friction, and tension are balanced.

We are given:
- Mass of each car with passengers: \( m = 1700 \, \text{kg} \)
- Coefficient of rolling friction: \( \mu_r = 0.030 \)
- Tension in the cable to the motor: \( T_{\text{motor}} = 820 \, \text{N} \)
- Angle of the incline: \( \theta = 35^\circ \)

### Step 1: Calculate the force of friction

The force of rolling friction \( F_{\text{friction}} \) is given by:
\[
F_{\text{friction}} = \mu_r \cdot N
\]
where \( N \) is the normal force. The normal force on the car is the perpendicular component of the gravitational force, which is:
\[
N = m \cdot g \cdot \cos(\theta)
\]
where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

Thus, the friction force is:
\[
F_{\text{friction}} = \mu_r \cdot m \cdot g \cdot \cos(\theta)
\]

Substituting the known values:
\[
F_{\text{friction}} = 0.030 \cdot 1700 \cdot 9.8 \cdot \cos(35^\circ)
\]
\[
F_{\text{friction}} \approx 0.030 \cdot 1700 \cdot 9.8 \cdot 0.8192
\]
\[
F_{\text{friction}} \approx 412.57 \, \text{N}
\]

### Step 2: Calculate the component of the weight parallel to the incline

The gravitational force acting parallel to the incline is:
\[
F_{\text{gravity, parallel}} = m \cdot g \cdot \sin(\theta)
\]
Substituting the known values:
\[
F_{\text{gravity, parallel}} = 1700 \cdot 9.8 \cdot \sin(35^\circ)
\]
\[
F_{\text{gravity, parallel}} \approx 1700 \cdot 9.8 \cdot 0.5736
\]
\[
F_{\text{gravity, parallel}} \approx 9576.28 \, \text{N}
\]

### Step 3: Analyze the forces on the system

Since the cars are moving at a constant speed, the net force must be zero. Therefore, the tension in the connecting cable \( T_{\text{connect}} \) must balance the difference between the gravitational force pulling one car down and the friction force opposing the motion. 

For each car, the forces along the incline are:
- **Car going up**: The tension in the connecting cable must overcome the friction and gravity pulling it down the incline.
- **Car going down**: The tension in the connecting cable is reduced by the force of gravity pulling it down.

So, the net force equation for the system can be written as:
\[
T_{\text{connect}} = F_{\text{gravity, parallel}} - F_{\text{friction}}
\]
Substituting the values:
\[
T_{\text{connect}} = 9576.28 - 412.57
\]
\[
T_{\text{connect}} \approx 9163.71 \, \text{N}
\]

### Conclusion:
The tension in the **connecting cable** is approximately \( \mathbf{9164 \, N} \).

Would you like further details or explanations on any step? 

Here are some related questions to deepen understanding:
1. How does the angle of the incline affect the tension in the cable?
2. What would happen if the friction coefficient were doubled?
3. How would increasing the mass of the cars affect the tension?
4. What other forces might act on the cars if they were not moving at constant speed?
5. How does the motor's tension relate to the tension in the connecting cable?

**Tip**: Remember, when analyzing forces, always consider both the forces parallel and perpendicular to the incline.

The century-old ascensores in Valparaiso, Chile, are small cable cars that go up and down the steep hillsides. 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. What is the tension in the connecting cable? The tension in the cable to the motor is 820N and the angle is 35 degrees from horizontal.

Learn how to calculate the tension in the connecting cable of Valparaiso's ascensores. This problem involves Newtonian mechanics, including forces on an incline, friction, and tension. Given data includes a mass of 1700 kg, a coefficient of rolling friction of 0.030, and a 35-degree incline. The tension in the motor cable is 820 N.

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