/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 A hammer is hanging by a light r... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(74^{\circ} .\) What is the acceleration of the bus?

Short Answer

Expert verified
The acceleration of the bus is approximately 3.49 m/s².

Step by step solution

01

Identify the Forces

First, identify the forces acting on the hammer. They are the tension in the rope and the gravitational force acting downward. The hammer remains at rest with respect to the bus, which means it's in a state of translational equilibrium relative to the bus.
02

Resolve Forces into Components

Resolve the tension in the rope into two components: horizontal and vertical. The vertical component of the tension balances the weight of the hammer, and the horizontal component is responsible for the horizontal acceleration of the hammer along with the bus.
03

Express Components using Angles

Let the tension in the rope be denoted as \( T \). The vertical component is \( T \cos(\theta) \), and the horizontal component is \( T \sin(\theta) \), where \( \theta = 74^{\circ} \).
04

Apply Equilibrium Conditions Vertically

Since the hammer is not moving up or down, the vertical forces must balance out. So, \( T \cos(74^{\circ}) = mg \), where \( m \) is the mass of the hammer and \( g \) is the acceleration due to gravity. Solve to express \( T \) in terms of \( m \) and \( g \).
05

Relate Horizontal Component to Acceleration

Using Newton's second law horizontally, \( T \sin(74^{\circ}) = ma \), where \( a \) is the acceleration of the bus. Substitute the expression for \( T \) from the previous step here.
06

Solve for Acceleration

Combine the equations: \( mg \times \tan(74^{\circ}) = ma \). Cancel out \( m \) (mass) from both sides to solve for \( a \). Substituting \( g = 9.8\, \text{m/s}^2 \): \[ a = g \times \tan(74^{\circ}) \] Calculate \( \tan(74^{\circ}) \) and find \( a \approx 3.49 \text{ m/s}^2 \).

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Key Concepts

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

Translational Equilibrium
In the world of physics, understanding translational equilibrium is crucial, especially when analyzing systems at rest or moving with a constant velocity. Translational equilibrium occurs when an object remains at rest or continues to move at a constant speed in a straight line.

This happens when the net force acting on the object is zero. In the context of the exercise, the hammer is said to be in translational equilibrium relative to the bus. This means that the forces acting on the hammer – the tension in the rope pulling it upwards and the gravitational force pulling it downwards – are balanced.

So, even though the bus accelerates, the hammer appears stationary relative to it because the net vertical force on the hammer is zero due to this balance.
Newton's Second Law
Newton's Second Law is a fundamental principle that describes how the motion of an object changes when it is subjected to forces. The law is stated as: \( F = ma \) where \( F \) is the net force acting on an object, \( m \) is the mass of the object, and \( a \) is its acceleration.

In this problem, we focus on the horizontal forces acting on the hammer. The tension in the rope provides a horizontal force component, responsible for the horizontal acceleration of both the hammer and the bus.

By using Newton's Second Law, we are able to relate the horizontal component of the tension to the acceleration of the bus, allowing us to solve for the latter. When the vertical and horizontal forces are resolved and substituted into Newton's equation, we can solve for the bus's acceleration with clarity and precision.
Force Components
Understanding how to resolve forces into components is a vital skill in physics. This involves breaking down a force into perpendicular components, usually horizontal and vertical. In the given exercise, we resolve the tension in the rope into vertical and horizontal components.

The vertical component, \( T \cos(\theta) \), balances the gravitational force (weight) of the hammer. This ensures no vertical motion occurs, signifying balance in the vertical direction.

Meanwhile, the horizontal component, \( T \sin(\theta) \), accounts for the acceleration of the hammer and bus as one, under Newton's Second Law. Thus, determining these components makes it easier to solve complex systems by dealing with simpler, one-dimensional forces.

Proper force resolution aids in understanding and managing the various motions and interactions objects can possess.
Physics Exercises
Physics exercises like the one involving the hammer hanging in a bus might seem daunting at first, but they offer a fantastic way to apply theoretical knowledge to practical scenarios. Breaking down these problems into clear steps is essential.

First, always start by identifying all forces at play. Next, resolve these forces into their components. Then, apply relevant physical laws, such as Newton's laws of motion, to form equations describing the system. Understanding and solving physics exercises helps in grasping core concepts deeply and enhances problem-solving skills. These exercises simulate real-world physics, training you to see the forces and motions in everyday life. This step-by-step approach is not only methodical but also builds confidence when tackling similar challenges in physics.

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Most popular questions from this chapter

Angle for Minimum Force. A box with weight w is pulled at constant speed along a level floor by a force \(\vec{F}\) that is at an angle \(\theta\) above the horizontal. The coefficient of kinetic friction between the floor and box is \(\mu_{k}\) (a) In terms of \(\theta, \mu_{k},\) and \(w,\) calculate \(F .\) (b) For \(w=400 \mathrm{N}\) and \(\mu_{k}=0.25\) , calculate \(F\) for \(\theta\) ranging from \(0^{\circ}\) to \(90^{\circ}\) in increments of \(10^{\circ} .\) Graph \(F\) versus \(\theta\) . (c) From the general expression in part (a), calculate the value of \(\theta\) for which the value of \(F,\) required to maintain constant speed, is a minimum. (Hint: At a point where a function is minimum, what are the first and second derivatives of the function? Here \(F\) is a function of \(\theta .\) ) For the special case of \(w=400 \mathrm{N}\) and \(\mu_{\mathrm{x}}=0.25\) evaluate this optimal \(\theta\) and compare your result to the graph you constructed in part \((\mathrm{b})\) .

A \(70-\mathrm{kg}\) person rides in a \(30-\mathrm{kg}\) cart moving at 12 \(\mathrm{m} / \mathrm{s}\) at the top of a hill that is in the shape of an are of a circle with a radius of 40 \(\mathrm{m}\) . (a) What is the apparent weight of the person as the cart passes over the top of the hill? (b) Determine the maximum speed that the cart may travel at the top of the hill without losing contact with the surface. Does your answer depend on the mass of the cart or the mass of the person? Explain.

A small button placed on a horizontal rotating platform with diameter 0.320 \(\mathrm{m}\) will revolve with the platform when it is brought up to a speed of 40.0 rev/min, provided the button is no more than 0.150 \(\mathrm{m}\) from the axis. (a) What is the coefficient of station between the button and the platform? (b) How far from the axis can the button be placed, without slipping, if the platform rotates at 60.0 rev/min?

The Monkey and Bananas Problem. A \(20-k g\) monkey has a firm hold on a light rope that passes over a frictionless pulley and is attached to a \(20-\mathrm{kg}\) bunch of bananas (Fig. 5.77\()\) . The monkey looks up, sees the bananas, and starts to climb the rope to get them. (a) As the monkey climbs, do the bananas move up, down, or remain at rest? (b) As the monkey climbs, does the distance between the monkey and the bananas decrease, increase, or remain constant? (c) The monkey releases her hold on the rope. What happens to the distance between the monkey and the bananas while she is falling?(d) Before reaching the ground, the monkey grabs the rope to stop her fall. What do the bananas do? figure can't copy

You observe a 1350 -kg sports car rolling along flat pavement in a straight line. The only horizontal forces acting on it are a constant rolling friction and air resistance (proportional to thesquare of its speed). You take the following data during a time interval of \(25 \mathrm{s} :\) When its speed is 32 \(\mathrm{m} / \mathrm{s}\) , the car slows down at a rate of \(-0.42 \mathrm{m} / \mathrm{s}^{2},\) and when its speed is decreased to \(24 \mathrm{m} / \mathrm{s},\) it slows down at \(-0.30 \mathrm{m} / \mathrm{s}^{2} .\) (a) Find the coefficient of rolling friction and the air drag constant \(D .(b)\) At what constant speed will this car move down an incline that makes a \(2.2^{\circ}\) angle with the horizontal? (c) How is the constant speed for an incline of angle \(\beta\) related to the terminal speed of this sports car if the car drops off a high cliff? Assume that in both cases the air resistance force is proportional to the square of the speed, and the air drag constant is the same.

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