/*! 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 73 A van accelerates down a hill (F... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 73

A van accelerates down a hill (Fig. P4.73), going from rest to \(30.0 \mathrm{~m} / \mathrm{s}\) in \(6.00 \mathrm{~s}\). During the acceleration, atoy \((m=0.100 \mathrm{~kg})\) hangs by a string from the van's ceiling. The acceleration is such that the string remains perpendicular to the ceiling. Determine (a) the angle \(\theta\) and (b) the tension in the string.

Short Answer

Expert verified
The angle \(\theta\) is obtained from step 3 and the tension in the string is obtained from step 4. Insert the computed values.

Step by step solution

01

Calculate horizontal acceleration

First, calculate the horizontal acceleration of the van using the formula for acceleration \( a = \frac{v-u}{t} \) where \( v = 30.0 m/s \) is the final velocity, \( u = 0 m/s \) is the initial velocity and \( t = 6.00s \) is the time. Substituting these values, we get \( a = \frac{30.0-0}{6.00} m/s^2 \).
02

Calculate normal force and gravitational force

Next, calculate the gravitational force acting on the toy which is \( mg \), where \( m = 0.100kg \) is the mass of the toy and \( g = 9.8m/s^2 \) is the acceleration due to gravity. This gives us \( F_g = m* g = 0.100 * 9.8 N \). Also, understand that the tension in the string is equivalent to the normal force exerted by the toy, which is \( F_n = m*a \). Calculate this horizontal force using the acceleration computed in step 1.
03

Calculate angle

\(\theta\) can be calculated using the formula for tangent which is \( \tan(\theta) = \frac{opposite}{adjacent} \). In this case the 'opposite' is the horizontal force and the 'adjacent' is the vertical force. Substituting the values calculated in step 2, we get \( \theta = \tan^{-1}(\frac{F_n}{F_g}) \). Compute this to get the angle in degrees.
04

Calculate Tension in string

Finally, calculate the tension in the string. Understanding that the tension is the resultant force on the toy due to both the horizontal and vertical components. We can use Pythagoras' Theorem which is \( A^2 + B^2 = C^2 \) to calculate the tension \( T = \sqrt{{F_g}^2 + {F_n}^2} \). Using the forces calculated in step 2, compute this to get the tension in the string in Newton.

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Key Concepts

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

Understanding Acceleration
Acceleration is the rate at which an object changes its velocity. In this exercise, the van starts from rest and reaches a speed of 30.0 m/s over 6.00 seconds. We calculate acceleration using the formula:
  • \[ a = \frac{v-u}{t} \]
where:
  • \( v = 30.0 \text{ m/s} \) is the final velocity,
  • \( u = 0 \text{ m/s} \) is the initial velocity,
  • \( t = 6.00 \text{ s} \) is the time.
This gives:
  • \( a = \frac{30.0-0}{6.00} = 5.0 \text{ m/s}^2 \)
The toy hanging from the ceiling experiences this acceleration as the van speeds up, giving context to the forces acting on it.
Exploring Forces
Forces are physical interactions that change the motion of an object. When looking at our problem, we consider two main forces on the toy:
  • Gravitational Force \( F_g \): It pulls the toy downward and is calculated by: \( F_g = m \cdot g \), where \( m = 0.100 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). Thus, \( F_g = 0.100 \cdot 9.8 = 0.98 \text{ N} \).
  • Normal Force \( F_n \) (Horizontal Force): Due to the van's acceleration, this force acts horizontally, calculated as \( F_n = m \cdot a \), where \( a = 5.0 \text{ m/s}^2 \). Therefore, \( F_n = 0.100 \cdot 5.0 = 0.50 \text{ N} \).
The interaction of these forces creates a tension in the string and affects its angle to the van's ceiling.
The Role of Tension
Tension is the force exerted by a string, rope, or cable when it is pulled tight by forces acting from opposite ends. In our scenario, it arises due to the combination of forces acting on the toy. We calculate the tension using Pythagoras' Theorem as these forces are perpendicular:
  • \[ T = \sqrt{F_g^2 + F_n^2} \]
Substituting the values:
  • \( T = \sqrt{0.98^2 + 0.50^2} \)
This results in a tension of approximately 1.1 N. The tension keeps the toy in equilibrium as it moves along with the van, balancing both gravitational pull and acceleration forces.
Angle of Inclination's Impact
The angle of inclination, or the angle \( \theta \), signifies the relationship between the gravitational and horizontal forces acting on the toy. It is found using the tangent function:
  • \[ \tan(\theta) = \frac{F_n}{F_g} \]
By substituting the known values:
  • \( \tan(\theta) = \frac{0.50}{0.98} \)
We find \( \theta = \tan^{-1}(0.51) \), resulting in an angle of approximately 27 degrees. This angle indicates how much the toy is tilted off vertical due to the horizontal pull from the van's acceleration.

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Most popular questions from this chapter

A football punter accelerates a football from rest to a speed of \(10 \mathrm{~m} / \mathrm{s}\) during the time in which his toe is in contact with the ball (about \(0.20 \mathrm{~s}\) ). If the football has a mass of \(0.50 \mathrm{~kg}\), what average force does the punter exert on the ball?

A crate of mass \(45.0 \mathrm{~kg}\) is being transported on the flatbed of a pickup truck. The coefficient of static friction between the crate and the truck's flatbed is \(0.350\), and the coefficient of kinetic friction is \(0.320\). (a) The truck accelerates forward on level ground. What is the maximum acceleration the truck can have so that the crate does not slide relative to the truck's flatbed? (b) The truck barely exceeds this acceleration and then moves with constant acceleration, with the crate sliding along its bed. What is the acceleration of the crate relative to the ground?

Consider a solid metal sphere (S) a few centimeters in diameter and a feather (F). For each quantity in the list that follows, indicate whether the quantity is the same, greater, or lesser in the case of \(\mathrm{S}\) or in that of \(\mathrm{F}\). Explain in each case why you gave the answer you did. Here is the list: (a) the gravitational force, (b) the time it will take to fall a given distance in air, (c) the time it will take to fall a given distance in vacuum, (d) the total force on the object when falling in vacuum.

A \(3.00-\mathrm{kg}\) block starts from rest at the top of a \(30.0^{\circ}\) incline and slides \(2.00 \mathrm{~m}\) down the incline in \(1.50 \mathrm{~s}\). Find (a) the acceleration of the block, (b) the coefficient of kinetic friction between the block and the incline, (c) the frictional force acting on the block, and (d) the speed of the block after it has slid \(2.00 \mathrm{~m}\).

A boat moves through the water with two forces acting on it. One is a \(2000-\mathrm{N}\) forward push by the water on the propeller, and the other is a \(1800-\mathrm{N}\) resistive force due to the water around the bow. (a) What is the acceleration of the \(1000-\mathrm{kg}\) boat? (b) If it starts from rest, how far will the boat move in \(10.0 \mathrm{~s}\) ? (c) What will its velocity be at the end of that time?
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