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Chapter 6: Problem 25

A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of \(591 \mathrm{N}\). As the elevator later stops, the scale reading is \(391 \mathrm{N}\). Assume the magnitude of the acceleration is the same during starting and stopping, and determine (a) the weight of the person, (b) the person's mass, and (c) the acceleration of the elevator.

Short Answer

Expert verified
The weight of the person is \(591N\), the mass is approximately \(60.3 kg\), and the acceleration of the elevator is approximately \(3.3 m/s^{2}\).

Step by step solution

01

Calculate the weight of the person

The weight of the person when the elevator is stationary is equal to the constant scale reading, which is \(591N\). Weight is the force on an object due to gravity. When the elevator is not accelerating, all the force on the person is due to gravity (weight). Therefore, the weight of the person is \(591N\).
02

Calculate the person's mass

We know that weight \(W\) is equal to the mass \(m\) times gravitational acceleration \(g\), expressed as \(W = mg\). We can rearrange this to solve for mass: \(m = \frac{W}{g}\). Knowing that the gravitational acceleration is approximately \(9.8 m/s^{2}\), we can substitute the values into the formula to find the mass: \(m = \frac{591N}{9.8 m/s^{2}} \approx 60.3 kg\).
03

Calculate the acceleration of the elevator

The scale reading decreased when the elevator stopped, indicating it underwent acceleration in the opposite direction of the gravitational force. The change in force, which equals to \(591N - 391N = 200N\), is due to this acceleration. From Newton's second law (force equals mass times acceleration, \(F = ma\)), we can calculate for acceleration as \(a = \frac{F}{m}\). Therefore, substituting the values, we get \(a = \frac{200N}{60.3 kg} \approx 3.3 m/s^{2}\).

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Key Concepts

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

elevator physics
Understanding elevator physics involves grasping how forces interact when an elevator accelerates or decelerates. When you stand on a scale in an elevator, the scale measures the normal force, which can change as the elevator moves.
  • If the elevator accelerates upward, you feel heavier because the normal force increases to counteract both gravity and the upward acceleration.
  • Conversely, if the elevator accelerates downward, you feel lighter because the normal force decreases.
  • When the elevator is at rest or moves at constant velocity, the scale shows your actual weight鈥攖he force due to gravity.

In our exercise, the scale initially reads 591 N, indicating your weight while the elevator is stationary. As the elevator stops, the scale shows 391 N due to downward acceleration, meaning the normal force is lower. This reading variations are practical examples of how elevator physics influences apparent weight.
gravitational force
Gravitational force is the attraction that Earth exerts on objects, pulling them toward its center. It is a key component in understanding how forces work in the universe and is essential to the calculations in our exercise.
The gravitational force acting on an object near the Earth's surface is given by the equation: \[ F_g = mg \] where
  • \( F_g \) is the gravitational force,
  • \( m \) is the mass of the object, and
  • \( g \) is the acceleration due to gravity (approximately 9.8 m/s虏).

For a 60.3 kg person (as calculated from the initially stable reading of 591 N), the gravitational force is simply their weight. When the elevator moves, changes in scale readings result from the acceleration adding or subtracting from this gravitational force. Gravitational force remains constant, but the apparent force varies with elevator acceleration.
Newtonian mechanics
Newtonian mechanics, or classical mechanics, provides the foundation for analyzing forces and motion in our everyday world. One of its central principles is Newton's Second Law, expressed as \[ F = ma. \] This law states that the force acting on an object equals its mass times its acceleration.
In this exercise, when the person is on a scale in a moving elevator, Newton's Second Law helps us understand the relationship between their weight (the gravitational force) and the additional forces due to elevator motion.
During the elevator's start and stop phases, the scale shows different readings because of acceleration forces that combine with the static gravitational force:
  • Upward acceleration increases the net force, creating a higher reading.
  • Downward acceleration decreases the net force, leading to a lower reading.

By applying the equation \( F = ma \), we calculate the elevator's acceleration from changes in the normal force as indicated by the scale, further highlighting the practicality of Newtonian mechanics.

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Most popular questions from this chapter

A \(50.0-\mathrm{kg}\) parachutist jumps from an airplane and falls to Earth with a drag force proportional to the square of the speed, \(R=C v^{2} .\) Take \(C=0.200 \mathrm{kg} / \mathrm{m}\) (with the parachute closed) and \(C=20.0 \mathrm{kg} / \mathrm{m}\) (with the chute open). (a) Determine the terminal speed of the parachutist in both configurations, before and after the chute is opened. (b) Set up a numerical analysis of the motion and compute the speed and position as functions of time, assuming the jumper begins the descent at \(1000 \mathrm{m}\) above the ground and is in free fall for 10.0 s before opening the parachute. (Suggestion: When the parachute opens, a sudden large acceleration takes place; a smaller time step may be necessary in this region.)

A crate of eggs is located in the middle of the flat bed of a pickup truck as the truck negotiates an unbanked curve in the road. The curve may be regarded as an arc of a circle of radius \(35.0 \mathrm{m}\). If the coefficient of static friction between crate and truck is \(0.600,\) how fast can the truck be moving without the crate sliding?

A small block is at rest on the floor at the front of a rail. road boxcar that has length \(\ell .\) The coefficient of kinetic friction between the floor of the car and the block is \(\mu_{k}\) The car, originally at rest, begins to move with acceleration a. The block slides back horizontally until it hits the back wall of the car. At that moment, what is its speed (a) relative to the car? (b) relative to Earth?

Consider an object on which the net force is a resistive force proportional to the square of its speed. For example, assume that the resistive force acting on a speed skater is \(f=-k m v^{2},\) where \(k\) is a constant and \(m\) is the skater's mass. The skater crosses the finish line of a straight-line race with speed \(u_{0}\) and then slows down by coasting on his skates. Show that the skater's speed at any time \(t\) after crossing the finish line is \(v(t)=u_{0} /\left(1+k t u_{0}\right) .\) This problem also provides the background for the two following problems.

The pilot of an airplane executes a constant-speed loop-theloop maneuver in a vertical circle. The speed of the airplane is \(300 \mathrm{mi} / \mathrm{h},\) and the radius of the circle is \(1200 \mathrm{ft}\). (a) What is the pilot's apparent weight at the lowest point if his true weight is 160 lb? (b) What is his apparent weight at the highest point? (c) What If? Describe how the pilot could experience weightlessness if both the radius and the speed can be varied. (Note: His apparent weight is equal to the magnitude of the force exerted by the seat on his body.)
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