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Chapter 5: Problem 80

A hammer is hanging by a light rope from the ceiling of a bus. The ceiling of the bus is parallel to the roadway. The bus is traveling in a straight line on a horizontal street. You observe that the hammer hangs at rest with respect to the bus when the angle between the rope and the ceiling of the bus is \(67^{\circ} .\) What is the acceleration of the bus?

Short Answer

Expert verified
The acceleration of the bus is approximately \(23.1 \, \text{m/s}^2\).

Step by step solution

01

Understanding the Forces Involved

When the hammer is hanging at an angle of \(67^{\circ}\) with the ceiling, it suggests that there's a force component causing this hanging angle. The forces acting on the hammer are gravity (acting downward) and tension in the rope (directed along it). The horizontal component of the tension force must be providing the centripetal force needed for the bus's acceleration because the hammer is not moving sideways.
02

Analyze the Angle of the Rope

The angle \(\theta = 67^{\circ}\) formed by the rope with the vertical is crucial because it allows us to resolve the tension into components. The vertical component of tension balances the weight of the hammer (gravity), and the horizontal component provides the horizontal acceleration due to the bus's motion.
03

Set Up the Equations

Using Newton's second law, we can resolve the tension \(T\) into components: \( T \cos(\theta) = mg \) (vertical balance) and \( T \sin(\theta) = ma \) (horizontal acceleration), where \(m\) is the mass of the hammer, \(g\) is the acceleration due to gravity (9.8 m/s虏), and \(a\) is the acceleration of the bus.
04

Solve for Acceleration

From the two equations, we eliminate \(T\) by dividing \( T \sin(\theta) = ma \) by \( T \cos(\theta) = mg \), obtaining \( \tan(\theta) = \frac{a}{g} \). Solving for \(a\) gives \( a = g \tan(\theta) \). Substitute \(g = 9.8 \, \text{m/s}^2\) and \(\theta = 67^{\circ}\) (convert degrees to radians if necessary) into this equation.
05

Calculation

Calculate \( \tan(67^{\circ}) \) and multiply by \(9.8\), it comes to approximately \( 9.8 \times 2.355 \approx 23.1 \, \text{m/s}^2 \). Therefore, the acceleration of the bus is \(23.1 \, \text{m/s}^2\).

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Key Concepts

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

Centripetal Force
When the hammer inside the bus hangs at an angle, it indicates that there is a force acting towards the center of the circular path the hammer would take if the bus continued moving in a curved line. This force is known as the centripetal force. It is not a separate force but rather a result of the components of other forces acting together to keep the hammer at rest relative to the bus. - The tension in the rope provides the centripetal force needed, preventing sideways motion of the hammer. - Since the bus is accelerating, this centripetal force is essentially the horizontal component of the tension force. By understanding that centripetal force is the force that keeps an object moving in a curved path, we can see how it relates to other forces involved鈥攕uch as tension and gravity鈥攚hen resolving them into components.
Components of Forces
Forces acting on the hammer can be broken down into components to make complex problems easier to solve. When dealing with angled forces, like tension in a rope, it's helpful to use trigonometric functions to separate the force into its vertical and horizontal components. - The vertical component of the tension force counteracts the gravitational force pulling the hammer downwards. This means the hammer stays at a constant height without falling. - The horizontal component of the tension is responsible for the sideways or horizontal forces that we term as centripetal force in this scenario. By resolving forces into components, it becomes simpler to apply Newton's Second Law to different directions independently, which aids in understanding the motion and equilibrium of the system.
Tension in a Rope
The rope's tension is a crucial force in this situation, providing the necessary support and balance for the hammer. When a rope has tension, it means that the rope is being pulled tight between two forces, maintaining equilibrium. - Tension acts along the length of the rope, trying to pull equally on objects at both ends. - In this scenario, the rope's tension balances the downward pull of gravity through its vertical component. Meanwhile, the horizontal component of tension supplies the centripetal force needed for acceleration. This dual role of tension in both horizontal and vertical planes helps illustrate why understanding and calculating tension accurately is important in dynamics and system analysis.
Angle of Inclination
The angle of inclination is an important factor because it affects how forces are resolved into vertical and horizontal components. In this exercise, the angle between the rope and the vertical is specified as 67 degrees. - This angle plays a critical role because it determines the ratio of the horizontal to vertical force components. - Trigonometric functions like sine and cosine are used with the angle to calculate these components. - Through calculations, the angle helps us find how much of the tension contributes to balancing gravity versus contributing to the bus's acceleration. Understanding how angles affect force distributions can aid in solving a variety of problems involving motion and equilibrium, reinforcing the interconnectedness of geometry and physics.

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Most popular questions from this chapter

Two bicycle tires are set rolling with the same initial speed of 3.50 \(\mathrm{m} / \mathrm{s}\) on a long, straight road, and the distance each travels before its speed is reduced by half is measured. One tire is inflated to a pressure of 40 psi and goes 18.1 m; the other is at 105 psi and goes 92.9 \(\mathrm{m} .\) What is the coefficient of rolling friction \(\mu_{\mathrm{r}}\) for each? Assume that the net horizontal force is due to rolling friction only.

A horizontal wire holds a solid uniform ball of mass \(m\) in place on a tilted ramp that rises \(35.0^{\circ}\) above the horizontal. The surface of this ramp is perfectly smooth, and the wire is directed away from the center of the ball (Fig. P5.60). (a) Draw a free-body diagram for the ball. (b) How hard does the surface of the ramp push on the ball? (c) What is the tension in the wire?

One December identical twins Jena and Jackie are playing on a large merry-go- round (a disk mounted parallel to the ground, on a vertical axle through its center) in their school playground in northern Minnesota. Each twin has mass 30.0 \(\mathrm{kg}\) . The icy coating on the merry-go-round surface makes it frictionless. The merry-go-round revolves at a constant rate as the twins ride on it. Jena, sitting 1.80 \(\mathrm{m}\) from the center of the merry-go-round, must hold on to one of the metal posts attached to the merry-go-round with a horizontal force of 60.0 \(\mathrm{N}\) to keep from sliding off. Jackie is sitting at the edge, 3.60 \(\mathrm{m}\) from the center. (a) With what horizontal force must Jackie hold on to keep from falling off? (b) If Jackie falls off, what will be her horizontal velocity when she becomes airborne?

The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end (Fig. E5.46). Each arm supports a seat suspended from a cable 5.00 m long, the upper end of the cable being fastened to the arm at a point 3.00 \(\mathrm{m}\) from the central shaft. (a) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of \(30.0^{\circ}\) with the vertical. (b) Does the angle depend on the weight of the passenger for a given rate of revolution?

A 25.0 -kg box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is \(0.25,\) and the coefficient of static friction is \(0.35 .\) (a) As the angle \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 \(\mathrm{m}\) along the loading ramp?
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