Problem 25 - College Physics

Question

\(\bullet$$\bullet\)The TGV, France's high-speed train, pulls out of the Lyons station and is accelerating uniformly to its cruising speed. Inside one of the cars, a 3.00 \(\mathrm{N}\) digital camera is hanging from the luggage compartment by a light, flexible strap that makes a \(12.0^{\circ}\) angle with the vertical. (a) Make a free-body diagram of this camera. (b) Apply Newton's second law to the camera, and find the acceleration of the train and the tension in the strap.

Step-by-Step Solution

Answer

The train's acceleration is approximately \(1.71\,\text{m/s}^2\), and the tension in the strap is approximately \(3.07\,\mathrm{N}\).

Step 1: Analyze the Scenario and Prepare the Free-Body Diagram (FBD)

The camera is suspended by a strap making an angle of \(12.0^{\circ}\) with the vertical while the train is accelerating horizontally. This suggests there are two forces acting on the camera: the gravitational force (\(3.00\,\mathrm{N}\)) acting downward and the tension \(T\) in the strap. The tension can be broken down into two components: vertical (\(T_y = T\cos\theta\)) and horizontal (\(T_x = T\sin\theta\)). The vertical component balances the weight, and the horizontal component is responsible for the train's acceleration.

Step 2: Set Up the Equations Using Newton's Second Law

For the vertical direction, there's no vertical acceleration, so:\[ T \cos\theta = mg \]For the horizontal direction, there is acceleration \(a\):\[ T \sin\theta = ma \]Where:- \( m \) is the mass of the camera.- \( g = 9.8\,\text{m/s}^2 \) is the acceleration due to gravity.- \( T \) is the tension in the strap.These equations can be used to solve for the mass \( m \) and the camera's acceleration.

Step 3: Solve for the Mass of the Camera

First, solve for the mass of the camera using the weight:\[ mg = 3.00\,\mathrm{N} \]\[ m = \frac{3.00\,\mathrm{N}}{9.8\,\text{m/s}^2} \]\[ m \approx 0.306\,\text{kg} \]

Step 4: Solve for Tension (T) in the Strap

Using the vertical component equation:\[ T \cos 12.0^\circ = 3.00\,\mathrm{N} \]\[ T = \frac{3.00\,\mathrm{N}}{\cos 12.0^\circ} \]\[ T \approx 3.07\,\mathrm{N} \]

Step 5: Solve for Acceleration of the Train

Using the horizontal component equation:\[ T \sin 12.0^\circ = ma \]Substitute for \(T\) and \(m\):\[ 3.07\,\mathrm{N} \sin 12.0^\circ = 0.306\,\text{kg} \cdot a \]\[ a \approx 1.71\,\text{m/s}^2 \]

Key Concepts

Free-Body Diagram

When analyzing forces acting on objects, it's useful to start with a free-body diagram. This diagram illustrates all forces involved and their directions. Here, we have a digital camera hanging in a train that is accelerating. The free-body diagram focuses on two crucial forces:

• The gravitational force of 3.00 N acting downwards, pulling the camera toward the Earth's center.
• The tension in the strap that holds the camera, which is at an angle of 12.0° with the vertical.

The tension force can be broken into two components:

• Vertical Component (\( T_y = T \cos \theta \)). This counterbalances the gravitational force.
• Horizontal Component (\( T_x = T \sin \theta \)). This component results from the train's acceleration.

The \( \theta \) symbol denotes the angle between the tension force and the vertical. Drawing these components accurately is critical to applying Newton's Second Law correctly in the next steps.

Tension in the Strap

The tension in the strap holds the camera steady as the train moves forward. This tension can be thought of as a "force holder" that counteracts gravitational pull and accounts for the forces caused by the train's movement.

• The vertical component of the tension (\( T \cos \theta \)) balances the camera's weight: \( T \cos 12.0^\circ = 3.00 \mathrm{N} \).
• The horizontal component (\( T \sin \theta \)) is what allows us to figure out the camera's horizontal motion: \( T \sin 12.0^\circ = ma \).

In this case, solving this system of equations gives the tension \( T \approx 3.07 \mathrm{N} \). Understanding the role of tension assists in determining how strong a strap must be to handle specific forces.

Horizontal and Vertical Components

Breaking forces into horizontal and vertical components simplifies the analysis of the forces acting on objects. By considering these components, we can apply Newton's Second Law separately to each direction.

• Vertical Components: In many cases, like this one, the vertical components are balanced. This means that the only vertical force we need to worry about is the weight being counteracted by an upward force, which is the vertical part of the tension here: \( T \cos \theta = mg \).
• Horizontal Components: In this problem, the horizontal component of tension is responsible for the acceleration: \( T \sin \theta = ma \). Solving for \( a \) gives the acceleration of the train, \( a \approx 1.71 \mathrm{m/s}^2 \).

Breaking down the forces allows for a targeted analysis that can apply other contexts and complexities, regardless of whether an object is stationary or in motion.