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Chapter 4: Problem 24

The upward normal force exerted by the floor is 620 \(\mathrm{N}\) on an elevator passenger who weighs 650 \(\mathrm{N}\) . What are the reaction forces to these two forces? Is the passenger accelerating? If so, what are the magnitude and direction of the acceleration?

Short Answer

Expert verified
Reaction forces are -620 N (down) for the normal force and -650 N (up) for the weight. The passenger is accelerating downward at 0.45 m/s虏.

Step by step solution

01

Identify Reaction Forces

According to Newton's Third Law of Motion, every action has an equal and opposite reaction. The reaction to the upward normal force of 620 \( \mathrm{N} \) exerted by the floor on the passenger is an equal and opposite force of 620 \( \mathrm{N} \) exerted by the passenger on the floor, directed downward. Similarly, the reaction to the gravitational force (weight) of 650 \( \mathrm{N} \) exerted by the Earth on the passenger is an equal and opposite force of 650 \( \mathrm{N} \) exerted by the passenger on the Earth, directed upward.
02

Determine Net Force on the Passenger

The net force acting on the passenger is the difference between the upward normal force and the downward gravitational force. Thus, the net force can be calculated as: \[ F_{\text{net}} = F_{\text{normal}} - F_{\text{weight}} = 620 \, \mathrm{N} - 650 \, \mathrm{N} = -30 \, \mathrm{N} \]. The negative sign indicates that this net force is directed downward.
03

Calculate Acceleration Using Newton's Second Law

To find the acceleration, apply Newton's second law: \( F_{\text{net}} = m \cdot a \), where \( m \) is the mass of the passenger and \( a \) is the acceleration. We first calculate the mass using \( F_{\text{weight}} = m \cdot g \), with \( g = 9.8 \, \mathrm{m/s^2} \): \[ m = \frac{650 \, \mathrm{N}}{9.8 \, \mathrm{m/s^2}} = 66.33 \, \mathrm{kg} \]. Substituting this value back into the equation: \[ -30 \, \mathrm{N} = 66.33 \, \mathrm{kg} \cdot a \], we solve for \( a \): \[ a = \frac{-30 \, \mathrm{N}}{66.33 \, \mathrm{kg}} \approx -0.45 \, \mathrm{m/s^2} \]. The negative sign indicates that the acceleration is downward.

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Key Concepts

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

Normal Force
Normal force acts perpendicular to the surface of contact between two objects. It "pushes back" against the object. In this exercise, the floor exerts a normal force of 620 N on the elevator passenger. This force supports the passenger against gravity.

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. Thus, the passenger simultaneously exerts a force of 620 N downwards on the floor.

Normal force adjusts so that objects don't pass through each other. Situations with "extra" normal force, like elevators, help understand forces better: **stronger floors prevent passengers from falling through**. It's not only crucial for elevators, but also when you're parked on a ramp or walking up a hill.
Gravitational Force
Gravitational force is the attraction between two masses. For our passenger in the elevator, this force is pulling them down towards Earth with 650 N. It acts downward, always aiming towards the center of Earth.

Much like the reaction of the normal force, the gravitational force also abides by the equal and opposite principle. The passenger also pulls on Earth with a force of 650 N, but directed upward.

This might seem odd because we don't feel Earth being tugged up, but this concept is important to understand the interaction between all masses. Remember, Earth is huge so its response isn't noticeable when compared to the passenger.
Acceleration
Acceleration is the change in velocity. It tells how quickly an object is speeding up or slowing down. When examining forces, we use Newton's Second Law: \( F_{\text{net}} = m \cdot a \).

In our problem, the net force is calculated by subtracting the gravitational force from the normal force. Here, the net force is \(-30\, \mathrm{N}\), indicating a downward direction.

To find acceleration, we first determine the passenger's mass. Using the formula for weight, \( 650 \, \mathrm{N} = m \cdot 9.8 \, \mathrm{m/s^2} \), we get around 66.33 kg. Then, we solve the acceleration: \( a = \frac{-30 \, \mathrm{N}}{66.33 \, \mathrm{kg}} \approx -0.45 \, \mathrm{m/s^2} \). This means the passenger is accelerating downward, slowly deviating from their initial speed. This kind of analysis is particularly beneficial when considering scenarios like elevators starting or stopping.

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Most popular questions from this chapter

An astronaut is tethered by a strong cable to a spacecraft. The astronaut and her spacesuit have a total mass of 105 \(\mathrm{kg}\) , while the mass of the cable is negligible. The mass of the spacecraft is \(9.05 \times 10^{4} \mathrm{kg}\) . The spacecraft is from any large astronomical bodies, so we can ignore the gravitational forces on it and the astronaut. We also assume that both the spacecraft and the astronaut are initially at rest in an inertial reference frame. The astronaut then pulls on the cable with a force of 80.0 \(\mathrm{N}\) . (a) What force bodies, so we can ignore the gravitational forces on it and the astronaut. We also assume that both the spacecraft and the astronaut are initially at rest in an inertial reference frame. The astronaut then pulls on the cable with a force of 80.0 \(\mathrm{N}\) . (a) What force does the cable exert on the astronaut? (b) Since \(\Sigma \overrightarrow{\boldsymbol{F}}=m \overrightarrow{\mathbf{a}},\) how can a "massless" \((m=0)\) cable exert a force? (c) What is the astronaut's acceleration? (d) What force does the cable exert on the spacecraft? (e) What is the acceleration of the spacecraft?

A student with mass 45 \(\mathrm{kg}\) jumps off a high diving board. Using \(6.0 \times 10^{24} \mathrm{kg}\) for the mass of the earth, what is the acceleration of the earth toward her as she accelerates toward the earth with an acceleration of 9.8 \(\mathrm{m} / \mathrm{s}^{2}\) ? Assume that the net force on the earth is the force of gravity she exerts on it.

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An object with mass \(m\) initially at rest is acted on by a force \(\vec{F}=k_{1} \hat{z}+k_{2} t^{3}\) , where \(k_{1}\) and \(k_{2}\) are constants. Calculate the velocity \(\vec{v}(t)\) of the object as a function of time.

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