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Chapter 3: Problem 59

A 700 -N man stands on a scale on the floor of an elevator. The scale records the force it exerts on whatever is on it. What is the scale reading if the elevator has an acceleration of (a) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) up? (b) \(1.8 \mathrm{~m} / \mathrm{s}^{2}\) down? (c) \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) down?

Short Answer

Expert verified
(a) 828.6 N, (b) 571.4 N, (c) 0 N.

Step by step solution

01

Understand the Problem

The scale measures the apparent weight of the man, which is the normal force exerted by the scale. When the elevator accelerates, this apparent weight changes depending on whether the elevator accelerates up or down.
02

Calculate Apparent Weight During Upward Acceleration

When the elevator accelerates upwards, the apparent weight increases. Use the equation for apparent weight: \( F_{\text{apparent}} = F_{\text{gravity}} + F_{\text{acceleration}} \), where \( F_{\text{gravity}} = mg \) and \( F_{\text{acceleration}} = ma \). With a man's weight of 700 N, we find his mass as \( m = \frac{700 \text{ N}}{9.8 \text{ m/s}^2} \approx 71.43 \text{ kg}\). The apparent weight becomes: \( 700\text{ N} + 71.43\text{ kg} \times 1.8\text{ m/s}^2 = 828.6 \text{ N} \).
03

Calculate Apparent Weight During Downward Acceleration

When the elevator accelerates downwards, the apparent weight decreases. Use the same equation \( F_{\text{apparent}} = mg - ma \). This results in \( 700 \text{ N} - 71.43 \text{ kg} \times 1.8 \text{ m/s}^2 = 571.4 \text{ N} \).
04

Calculate Apparent Weight During Free Fall

If the elevator accelerates downwards at \( 9.8 \text{ m/s}^2 \), the elevator is in free fall, and the apparent weight is zero, as the force exerted by the scale is equal to the gravitational force, but acts in the opposite direction, cancelling it out. Hence, the reading is 0 N.

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Key Concepts

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

Apparent Weight
When you stand on a scale, it typically measures your normal weight, which is the gravitational force pulling down on you. However, this can change when you're in a moving elevator. The scale actually measures what is called "apparent weight."

Apparent weight is the normal force exerted by the scale and is influenced by external factors like acceleration. It can be greater or lesser than your actual weight depending on how the elevator is moving.

  • When the elevator accelerates upwards, the apparent weight feels heavier.
  • When it accelerates downwards, you feel lighter as if your weight has reduced.

Understanding apparent weight helps us fathom sensations we experience in elevators and even amusement park rides. It is an excellent practical application of physics concepts.
Elevator Motion
Elevator motion significantly impacts apparent weight. When you step into an elevator and it starts moving, the changes in your apparent weight are due to the elevator's motion.

Let's look at different scenarios:
  • Upward Acceleration: The elevator starts moving upward, increasing the apparent weight. You feel a stronger push of the floor against your feet because the floor must push harder to overcome gravity and the acceleration.
  • Downward Acceleration: This reduces your apparent weight. You might feel a bit lighter because the gravitational pull becomes more significant compared to the normal force exerted by the scale.
  • Free Fall: In this case, the elevator accelerates downwards at the same rate as gravitational acceleration. You experience weightlessness since there's no net force acting on you except gravity.

The changes due to elevator motion can be explained entirely by Newton's second law, correlating with how forces interact.
Newton's Second Law
Newton's second law gives us a clear framework to understand elevator physics. The law states that the force exerted on an object is equal to the mass of the object times its acceleration: \[ F = ma \].

Using this law, we can find out why the scale readings change:
  • As the elevator moves up, the acceleration (positive) increases, adding to the gravitational force and thus increasing the apparent weight.

  • If the elevator moves down, the downward acceleration subtracts from the gravitational force, reducing the apparent weight.

The force you feel (or the scale reads) depends on the direction and magnitude of acceleration. Newton's laws thus help us compute changes in apparent weight based on acceleration.
Acceleration Effects
Physics problems like these are a prime example of how acceleration affects everyday experiences. Acceleration doesn't just change speed; it influences the forces we feel.

  • Upward acceleration causes an increase in apparent weight due to the added force lifting against gravity.

  • Downward acceleration results in a reduction of felt weight, making "weightless" experiences possible.

  • In free-fall acceleration (e.g., at \(9.8 ext{ m/s}^2 \) downward), you experience full weightlessness as both you and the elevator move solely under gravity's influence.
Understanding how acceleration affects physical sensations can bring clarity to more complex physics phenomena and help in future problem-solving scenarios.

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Most popular questions from this chapter

Compute algebraically the resultant of the following coplanar forces: \(100 \mathrm{~N}\) at \(30^{\circ}, 141.4 \mathrm{~N}\) at \(45^{\circ}\), and \(100 \mathrm{~N}\) at \(240^{\circ}\). Check your result graphically.

Consider an essentially spherical homogeneous celestial body of mass \(M\). The acceleration due to gravity in its vicinity beyond its surface at a distance \(R\) from its center is \(g_{R}\). Show that $$ g_{R}=\frac{G M}{R^{2}} $$ Notice that the acceleration drops off as \(1 / R^{2}\). Imagine an object of mass \(m\) at a distance \(R\) from the center of our celestial body. Its weight is \(F_{W}=m g_{R}\), but that's also the gravitation force on it due to the mass \(M\), that is, \(F_{W}=F_{G} .\) Hence, $$ m g_{R}=G \frac{m M}{R^{2}} $$

Typically, a bullet leaves a standard 45-caliber pistol (5.0-in. barrel) at a speed of \(262 \mathrm{~m} / \mathrm{s}\). If it takes \(1 \mathrm{~ms}\) to traverse the barrel, determine the average acceleration experienced by the \(16.2\) -g bullet within the gun, and then compute the average force exerted on it.

The only force acting on a \(5.0\) -kg object has components \(F_{x}=20 \mathrm{~N}\) and \(F_{y}=30 \mathrm{~N}\). Find the acceleration of the object. Use \(\sum F_{x}=m a_{x}\) and \(\sum F_{y}=m a_{y}\) to obtain $$ \begin{array}{l} a_{x}=\frac{\sum F_{x}}{m}=\frac{20 \mathrm{~N}}{5.0 \mathrm{~kg}}=4.0 \mathrm{~m} / \mathrm{s}^{2} \\ a_{y}=\frac{\sum F_{y}}{m}=\frac{30 \mathrm{~N}}{5.0 \mathrm{~kg}}=6.0 \mathrm{~m} / \mathrm{s}^{2} \end{array} $$ These components of the acceleration are shown in Fig. \(3-10 .\) From the figure, $$ a=\sqrt{(4.0)^{2}+(6.0)^{2}} \mathrm{~m} / \mathrm{s}^{2}=7.2 \mathrm{~m} / \mathrm{s}^{2} $$ and \(\theta=\arctan (6.0 / 4.0)=56^{\circ}\).

Find the weight on the surface of the Earth of a body whose mass is (a) \(3.00 \mathrm{~kg}\), and (b) \(200 \mathrm{~g}\). The general relation between mass \(m\) and weight \(F_{W}\) is \(F_{W}=m g\). In this relation, \(m\) must be in kilograms, \(g\) in meters per second squared, and \(F_{W}\) in newtons. On Earth, \(g=9.81 \mathrm{~m} / \mathrm{s} 2\). The acceleration due to gravity varies from place to place in the universe. (a) \(F_{W}=(3.00 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=29.4 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}=29.4 \mathrm{~N}\) (b) \(F_{W}=(0.200 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=1.96 \mathrm{~N}\)
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