/*! 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 80 A man seeking to set a world rec... [FREE SOLUTION] | 91Ó°ÊÓ

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A man seeking to set a world record wants to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to the airplane. The mass of the man is 85 kg, and the coefficient of static friction between his shoes and the runway is 0.77. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.

Short Answer

Expert verified
The greatest acceleration the man can give the airplane is 0.0059 m/s².

Step by step solution

01

Determine the Maximum Static Friction Force

The maximum static friction force (\( f_s \)) between the man's shoes and the ground is what enables him to pull the plane. This can be calculated using the formula \( f_s = \mu_s \times N \), where \( \mu_s \) is the coefficient of static friction (0.77) and \( N \) is the normal force. For the man standing on a flat surface, \( N = m \times g \), where \( m \) is the man's mass (85 kg) and \( g \) is the acceleration due to gravity (approximately 9.8 m/s²). So,\[f_s = 0.77 \times 85 \times 9.8 = 641.23 \text{ N}\]
02

Apply Newton's Second Law to the Airplane

For the airplane, apply Newton's second law, \( F = m \times a \), where \( F \) is the force acting on the plane, \( m \) is the mass of the airplane (109,000 kg), and \( a \) is the acceleration. Since the only horizontal force applied by the man is the static friction force determined in Step 1, we set \( F = f_s = 641.23 \text{ N} \). Thus,\[641.23 = 109,000 \times a\]
03

Solve for Maximum Acceleration

Solve the equation for \( a \) to find the greatest acceleration the man can impart to the plane:\[a = \frac{641.23}{109,000} = 0.005884 \text{ m/s}^2\]This is the maximum acceleration available given the friction limit between the man's shoes and ground.

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle that establishes a relationship between the motion of an object and the forces acting upon it. It states that the force applied on an object is equal to the mass of the object multiplied by the acceleration it experiences. This can be expressed with the formula:\[ F = m \times a \]where \( F \) is the force in newtons, \( m \) is the mass in kilograms, and \( a \) is the acceleration in meters per second squared.
When considering the man's interaction with the airplane in this exercise, Newton's Second Law helps us understand how a relatively small force, like static friction, can still impart an acceleration to the large mass of an airplane. By understanding this principle, we can determine the maximum potential acceleration that static friction allows, which is crucial for predicting the plane's movement along the runway.
Acceleration
Acceleration refers to the rate of change of velocity of an object. It is a vector quantity, meaning it has both magnitude and direction. When an object speeds up, slows down, or changes direction, it is accelerating.- In our exercise, the acceleration of the airplane is a direct result of the static frictional force being applied through the cable.- This is calculated using Newton's Second Law, where the static friction force is divided by the airplane's mass to find the acceleration (\( a \)).- The maximum acceleration the man can achieve is limited by the static friction and is calculated as: \[ a = \frac{f_s}{m} \] where \( f_s \) is the static friction force and \( m \) is the mass of the airplane.
Such calculations are important in predicting how quickly and effectively the airplane can be moved under the given conditions.
Normal Force
Normal force is the force exerted by a surface that supports the weight of an object resting on it. It acts perpendicular to the surface.- For the man attempting to pull the airplane, the normal force \( N \) is equal to his weight when standing on a flat surface. - It can be calculated using the formula: \[ N = m \times g \] where \( m \) is the man's mass and \( g \) is the acceleration due to gravity (9.8 m/s²).
This exercise demonstrates that the normal force is a key component in determining the maximum static friction force, which defines how much force he can exert on the airplane without slipping. By understanding the role of normal force, we can better appreciate its crucial part in facilitating the man's ability to move the plane.
Coefficient of Friction
The coefficient of friction is a dimensionless scalar value that describes the ratio of the force of friction between two bodies and the force pressing them together.- In this context, it represents the stickiness of the surface between the man's shoes and the runway.- It is denoted as \( \mu_s \) for static friction and is used in the formula to determine maximum static friction: \[ f_s = \mu_s \times N \] where \( f_s \) is the force of static friction and \( N \) is the normal force.
In our exercise, the given coefficient of static friction is 0.77. This value reveals that the man can exert a stronger force before slipping, as opposed to if a lower \( \mu_s \) was present. It's a crucial factor in ensuring that the man can pull the airplane without actually moving due to frictional resistance. Understanding this coefficient helps us grasp how different surfaces and materials interact and affect motion.

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Most popular questions from this chapter

The central ideas in this problem are reviewed in Multiple-Concept Example 9. One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and the magnitude of the force is slowly increased. When the force reaches 47.0 \(\mathrm{N}\), the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block so that it just begins to slide out from under the upper block?

The alarm at a fire station rings and an 86-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.0 m). Just before landing, his speed is 1.4 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

A train consists of 50 cars, each of which has a mass of \(6.8 \times 10^{3} \mathrm{kg} .\) The train has an acceleration of \(+8.0 \times 10^{-2} \mathrm{m} / \mathrm{s}^{2} .\) Ignore friction and determine the tension in the coupling (a) between the 30th and 31st cars and (b) between the 49th and 50th cars.

A girl is sledding down a slope that is inclined at \(30.0^{\circ}\) with respect to the horizontal. The wind is aiding the motion by providing a steady force of 105 \(\mathrm{N}\) that is parallel to the motion of the sled. The combined mass of the girl and the sled is 65.0 \(\mathrm{kg}\) , and the coefficient of kinetic friction between the snow and the runners of the sled is \(0.150 .\) How much time is required for the sled to travel down a \(175-\mathrm{m}\) slope, starting from rest?

A rescue helicopter is lifting a man (weight 822 N) from a capsized boat by means of a cable and harness. (a) What is the tension in the cable when the man is given an initial upward acceleration of 1.10 \(\mathrm{m} / \mathrm{s}^{2} ?\) (b) What is the tension during the remainder of the rescue when he is pulled upward at a constant velocity?

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