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Chapter 5: Problem 25

\(\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.

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

Solve for the Mass of the Camera

First, solve for the mass of the camera using the weight:\[ mg = 3.00\,\mathrm{N} \]\[ m = \frac{3.00\,\mathrm{N}}{9.8\,\text{m/s}^2} \]\[ m \approx 0.306\,\text{kg} \]
04

Solve for Tension (T) in the Strap

Using the vertical component equation:\[ T \cos 12.0^\circ = 3.00\,\mathrm{N} \]\[ T = \frac{3.00\,\mathrm{N}}{\cos 12.0^\circ} \]\[ T \approx 3.07\,\mathrm{N} \]
05

Solve for Acceleration of the Train

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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掳 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.

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Most popular questions from this chapter

\(\bullet$$\bullet\) A winch is used to drag a 375 \(\mathrm{N}\) crate up a ramp at a con- stant speed of 75 \(\mathrm{cm} / \mathrm{s}\) by means of a rope that pulls parallel to the surface of the ramp. The rope slopes upward at \(33^{\circ}\) above the horizontal, and the coefficient of kinetic friction between the ramp and the crate is 0.25 . (a) What is the tension in the rope? (b) If the rope were suddenly to snap, what would be the accel- eration of the crate immediately after the rope broke?

\(\bullet$$\bullet\) A man pushes on a piano of mass 180 \(\mathrm{kg}\) so that it slides at a constant velocity of 12.0 \(\mathrm{cm} / \mathrm{s}\) down a ramp that is inclined at \(11.0^{\circ}\) above the horizontal. No appreciable friction is acting on the piano. Calculate the magnitude and direction of this push (a) if the man pushes parallel to the incline, (b) if the man pushes the piano up the plane instead, also at 12.0 \(\mathrm{cm} / \mathrm{s}\) paral- lel to the incline, and (c) if the man pushes horizontally, but still with a speed of 12.0 \(\mathrm{cm} / \mathrm{s}\) .

the task of designing an accelerometer to be used inside a rocket ship in outer space. Your equipment consists of a very light spring that is \(15.0 \mathrm{~cm}\) long when no forces act to stretch or compress it, \(\begin{array}{lllll}\text { plus } & \text { a } & 1.10 & \text { kg } & \text { weight. }\end{array}\) be attached to a friction- free tabletop, while the \(1.10 \mathrm{~kg}\) weight is attached to the other end, as shown in Figure 5.67 . (Such a spring-type accelerometer system was actually used in the ill-fated Genesis Mission, which collected particles of the solar wind. Unfortunately, because it was installed backward, it did not measure the acceleration correctly during the craft's descent to earth. As a result, the parachute failed to open and the capsule crashed on Sept. \(8,2004 .)\) (a) What should be the force constant of the spring so that it will stretch by \(1.10 \mathrm{~cm}\) when the rocket accelerates forward at \(2.50 \mathrm{~m} / \mathrm{s}^{2}\) ? Start with a free-body diagram of the weight. (b) What is the acceleration (magnitude and direction) of the rocket if the spring is compressed by \(2.30 \mathrm{~cm} ?\)

\(\bullet$$\bullet\) Mars Exploration Rover landings. In January 2004 the Mars Exploration Rover spacecraft landed on the surface of the Red Planet, where the acceleration due to gravity is 0.379 what it is on earth. The descent of this 827 kg vehicle occurred in several stages, three of which are outlined here. In Stage I, friction with the Martian atmosphere reduced the speed from \(19,300 \mathrm{km} / \mathrm{h}\) to 1600 \(\mathrm{km} / \mathrm{h}\) in a 4.0 min interval. In Stage II, a parachute reduced the speed from 1600 \(\mathrm{km} / \mathrm{h}\) to 321 \(\mathrm{km} / \mathrm{h}\) in \(94 \mathrm{s},\) and in Stage III, which lasted 2.5 \(\mathrm{s}\) , retrorockets fired to reduce the speed from 321 \(\mathrm{km} / \mathrm{h}\) to zero. As part of your solu- tion to this problem, make a free-body diagram of the rocket during each stage. Assuming constant acceleration, find the force exerted on the spacecraft (a) by the atmosphere during Stage I, (b) by the parachute during Stage II, and (c) by the retrorockets during Stage III.

\(\bullet$$\bullet\) The coefficients of static and kinetic friction between a 476 \(\mathrm{N}\) crate and the warehouse floor are 0.615 and 0.420 , respectively. A worker gradually increases his horizontal push against this crate until it just begins to move and from then on maintains that same maximum push. What is the acceleration of the crate after it has begun to move? Start with a free-body diagram of the crate.
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