To avoid capture during a high speed car chase involving the Diplomatic Security Service, a saboteur releases an oil slick from their vehicle, then pulls a sudden U-turn. Having only a fraction of a second to react, DSS agent Luke Hobbes fires a grapple hook from the side of his car toward a nearby utility pole. Once the hook is anchored, the cable attached to it enables Hobbes to make a 8.52 m diameter U-turn in just 3.12 seconds. If the tension in the cable is the only horizontal force acting on the car, what must be the tension (in N) if Hobbes and his car have a combined mass of 1,993 kg?

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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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### Physics Problem Involving Tension and Circular Motion

**Scenario:**
To avoid capture during a high-speed car chase involving the Diplomatic Security Service, a saboteur releases an oil slick from their vehicle, then pulls a sudden U-turn.

**Details:**
Having only a fraction of a second to react, DSS agent Luke Hobbes fires a grapple hook from the side of his car toward a nearby utility pole. Once the hook is anchored, the cable attached to it enables Hobbes to make an 8.52-meter diameter U-turn in just 3.12 seconds.

**Problem Statement:**
If the tension in the cable is the only horizontal force acting on the car, what must be the tension (in Newtons, N) if Hobbes and his car have a combined mass of 1,993 kg?

To solve the problem, you need to apply the principles of circular motion and use the formula for the centripetal force, which is provided by the tension in the cable in this scenario.

**Step-by-Step Solution:**
1. **Determine the radius (r) of the U-turn:**
   - Diameter = 8.52 meters
   - Radius, \( r = \frac{Diameter}{2} = \frac{8.52 \, \text{m}}{2} = 4.26 \, \text{m} \)

2. **Calculate the velocity (v) of the car:**
   - The car completes a U-turn in 3.12 seconds, describing a circular path.
   - Circumference of the U-turn path, \( C = 2 \pi r = 2 \pi \times 4.26 \, \text{m} = 26.76 \, \text{m} \)
   - Velocity, \( v = \frac{C}{time} = \frac{26.76 \, \text{m}}{3.12 \, \text{s}} = 8.58 \, \text{m/s} \)

3. **Use the centripetal force formula:**
   - \( F_{centripetal} = \frac{mv^2}{r} \)
   - Where \( m = 1993 \, \text{kg} \), \( v = 8.58 \, \text{m/s} \), \( r = 4.26 \, \text
Transcribed Image Text:### Physics Problem Involving Tension and Circular Motion **Scenario:** To avoid capture during a high-speed car chase involving the Diplomatic Security Service, a saboteur releases an oil slick from their vehicle, then pulls a sudden U-turn. **Details:** Having only a fraction of a second to react, DSS agent Luke Hobbes fires a grapple hook from the side of his car toward a nearby utility pole. Once the hook is anchored, the cable attached to it enables Hobbes to make an 8.52-meter diameter U-turn in just 3.12 seconds. **Problem Statement:** If the tension in the cable is the only horizontal force acting on the car, what must be the tension (in Newtons, N) if Hobbes and his car have a combined mass of 1,993 kg? To solve the problem, you need to apply the principles of circular motion and use the formula for the centripetal force, which is provided by the tension in the cable in this scenario. **Step-by-Step Solution:** 1. **Determine the radius (r) of the U-turn:** - Diameter = 8.52 meters - Radius, \( r = \frac{Diameter}{2} = \frac{8.52 \, \text{m}}{2} = 4.26 \, \text{m} \) 2. **Calculate the velocity (v) of the car:** - The car completes a U-turn in 3.12 seconds, describing a circular path. - Circumference of the U-turn path, \( C = 2 \pi r = 2 \pi \times 4.26 \, \text{m} = 26.76 \, \text{m} \) - Velocity, \( v = \frac{C}{time} = \frac{26.76 \, \text{m}}{3.12 \, \text{s}} = 8.58 \, \text{m/s} \) 3. **Use the centripetal force formula:** - \( F_{centripetal} = \frac{mv^2}{r} \) - Where \( m = 1993 \, \text{kg} \), \( v = 8.58 \, \text{m/s} \), \( r = 4.26 \, \text
### Solution Explanation for Cable Tension Problem

When tackling problems involving the tension in a cable, it's crucial to follow a systematic approach. Here's how you can solve such problems, step by step:

#### Step 1: Identify Known and Unknown Quantities
- **Known Quantities**: Typically includes the weights of objects, dimensions, angles, and sometimes the lengths of the cables or ropes.
- **Unknown Quantities**: Usually, the unknown is the tension in the cable, but it can also include angles or positions in some scenarios.

#### Step 2: Drawing a Free-Body Diagram
- Draw a detailed diagram of the system, clearly indicating all forces acting on the objects and the cable.
- Mark the points where the cable is attached and the direction of tension force.

#### Step 3: Apply Principles and Concepts
- Utilize principles such as equilibrium conditions where the sum of forces in any direction equals zero (ΣF = 0) because the system is at rest or moving with constant velocity.
- In cases involving angles, resolve the forces into horizontal and vertical components.

#### Step 4: Set Up Equations
- Use equilibrium equations based on the free-body diagram to set up a system of equations. Typically, you will have:
  - The sum of horizontal forces equals zero.
  - The sum of vertical forces equals zero.
  
  For example:
  \[
  \sum F_x = 0 \quad \text{and} \quad \sum F_y = 0
  \]
  This might translate into equations involving the tension \(T\), angles \(\theta\), and weights \(W\).

#### Step 5: Solve the Equations
- Solve the system of equations to find the unknowns. This might involve algebraic manipulation or using trigonometric identities.

#### Step 6: Verify Your Solution
- Check that the calculated tensions satisfy all conditions of equilibrium.
- Ensure that your solution is physically reasonable (e.g., tension should not be negative).

### Example Equation Usage
If a weight \(W\) is suspended by two cables making angles \(\theta_1\) and \(\theta_2\) with the horizontal, the following equations might be used:
\[
T_1 \cos(\theta_1) + T_2 \cos(\theta_2) = 0
\]
\[
T_1 \sin(\theta_1) + T_2 \sin
Transcribed Image Text:### Solution Explanation for Cable Tension Problem When tackling problems involving the tension in a cable, it's crucial to follow a systematic approach. Here's how you can solve such problems, step by step: #### Step 1: Identify Known and Unknown Quantities - **Known Quantities**: Typically includes the weights of objects, dimensions, angles, and sometimes the lengths of the cables or ropes. - **Unknown Quantities**: Usually, the unknown is the tension in the cable, but it can also include angles or positions in some scenarios. #### Step 2: Drawing a Free-Body Diagram - Draw a detailed diagram of the system, clearly indicating all forces acting on the objects and the cable. - Mark the points where the cable is attached and the direction of tension force. #### Step 3: Apply Principles and Concepts - Utilize principles such as equilibrium conditions where the sum of forces in any direction equals zero (ΣF = 0) because the system is at rest or moving with constant velocity. - In cases involving angles, resolve the forces into horizontal and vertical components. #### Step 4: Set Up Equations - Use equilibrium equations based on the free-body diagram to set up a system of equations. Typically, you will have: - The sum of horizontal forces equals zero. - The sum of vertical forces equals zero. For example: \[ \sum F_x = 0 \quad \text{and} \quad \sum F_y = 0 \] This might translate into equations involving the tension \(T\), angles \(\theta\), and weights \(W\). #### Step 5: Solve the Equations - Solve the system of equations to find the unknowns. This might involve algebraic manipulation or using trigonometric identities. #### Step 6: Verify Your Solution - Check that the calculated tensions satisfy all conditions of equilibrium. - Ensure that your solution is physically reasonable (e.g., tension should not be negative). ### Example Equation Usage If a weight \(W\) is suspended by two cables making angles \(\theta_1\) and \(\theta_2\) with the horizontal, the following equations might be used: \[ T_1 \cos(\theta_1) + T_2 \cos(\theta_2) = 0 \] \[ T_1 \sin(\theta_1) + T_2 \sin
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