Problem 20 - University Physics with Modern Physics

Question

A 550 -N physics student stands on a bathroom scale in an 850 -kg (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 450 \(\mathrm{N}\) . (a) Find the acceleration of the elevator (magnitude and direction). (b) What is the acceleration if the scale reads 670 \(\mathrm{N}\) ? (c) If the scale reads zero, should the student worry? Explain. (d) What is the tension in the cable in parts (a) and (c)?

Step-by-Step Solution

Answer

(a) Acceleration is -1.78 m/s² (downward). (b) Acceleration is 2.14 m/s² (upward). (c) Yes, free fall is dangerous; the elevator is in free fall. (d) Tensions are 6817 N (a) and 0 N (c).

Step 1: Understand the Forces

The forces acting on the student are the gravitational force \( F_g = mg \) and the normal force \( N \), which corresponds to the scale reading. The net force on the student determines the acceleration of the elevator.

Step 2: Apply Newton's Second Law for Part (a)

For part (a), when the scale reads \( N = 450 \ \text{N} \), use Newton's second law to find the acceleration:\[ \begin{align*}F_{net} &= N - mg, \m &= \frac{550}{9.8} = 56.12 \ \text{kg}, \450 - 550 &= 56.12 imes a, \-100 &= 56.12a, \a &= \frac{-100}{56.12} \approx -1.78 \ \text{m/s}^2.\end{align*}\]The negative sign indicates the elevator is accelerating downwards.

Step 3: Determine Acceleration for Part (b)

For part (b), when the scale reads \( N = 670 \ \text{N} \):\[ \begin{align*}F_{net} &= N - mg, \670 - 550 &= 56.12 imes a, \120 &= 56.12a, \a &= \frac{120}{56.12} \approx 2.14 \ \text{m/s}^2.\end{align*}\]The positive acceleration indicates the elevator is moving upwards.

Step 4: Analyze Condition for Part (c)

In part (c), the scale reads 0 N. This means the normal force \( N = 0 \). Therefore:\[ F_{net} = 0 - mg = 56.12 imes a \Rightarrow a = -9.8 \ \text{m/s}^2. \]This acceleration equals the acceleration due to gravity, indicating the elevator is in free fall, which is a dangerous situation.

Step 5: Calculate the Tension for Parts (a) and (c)

For part (a), the tension in the cable is:\[ T = mg + ma = 850 \times 9.8 + 850 \times (-1.78) = 850 \times 8.02 = 6817 \ \text{N}. \]For part (c), the tension is zero because the elevator is falling freely along with the student.

Key Concepts

Acceleration Calculation

To find the acceleration of the elevator, we can use Newton's Second Law of Motion. This fundamental principle states that the force acting on an object is equal to the mass of the object multiplied by its acceleration: \[ F = ma \]. In this exercise, we are given the scale reading, which acts as the normal force \( N \), and the gravitational force acting on our physics student, \( F_g = mg \). The net force on the student is the difference between the normal force and the gravitational force.
When the scale reads less than the gravitational force exerted on the student, this means the elevator is accelerating downwards. Conversely, if the scale reading is higher, the elevator is accelerating upwards, as seen in part (b).

• If the scale reads 450 N, the elevator's net force is negative, indicating a downward acceleration: \[ a = \frac{-100}{56.12} \approx -1.78 \ \text{m/s}^2 \].
• For a scale reading of 670 N, the net force is positive, meaning an upward acceleration: \[ a = \frac{120}{56.12} \approx 2.14 \ \text{m/s}^2 \].

Understanding this balance of forces helps determine the direction and magnitude of acceleration.

Gravitational Force

Gravitational force is a fundamental concept, especially in this context where it impacts the entire system in the elevator. It is the attractive force that the Earth exerts on objects, which can be calculated with the equation \( F_g = mg \). Here, \( m \) is the mass of the object, and \( g \) is the gravitational acceleration, approximately \( 9.8 \ \text{m/s}^2 \) on Earth.
In the given exercise, the gravitational force acting on the student is \( F_g = 550 \ \text{N} \). This force operates in the negative y-direction (downwards) as considered in most physics problems.

• Role of Gravitational Force: It contributes to the net force calculation when determining the elevator's acceleration.
• In a free-fall situation, like when the scale reads zero, gravitational force equals the net force, resulting in the student and elevator accelerating downward at \( g \), which is \( -9.8 \ \text{m/s}^2 \).

Gravitational force is a constant that can be crucial in analyzing motion, especially in free-fall scenarios.

Tension in Cable

Tension in the cable is a key factor when discussing elevators, as it must counteract both the gravitational force of the elevator's mass and any additional forces due to its acceleration. In our exercise, tension is the force transmitted through the cable as it supports the moving elevator.
To calculate tension, we combine both the gravitational force and the force due to the elevator's acceleration:

• Scenario A (scale reads 450 N): The tension is given by \[ T = mg + ma \], resulting in a reduced tension due to the downward motion of the elevator: \[ T = 850 \times 8.02 = 6817 \ \text{N} \].
• Free Fall (scale reads 0 N): Here, the elevator is essentially falling under gravity, meaning tension becomes zero because the cable is not exerting an upward force.

Understanding tension in cables is crucial for ensuring the structural integrity and safe operation of elevators, highlighting the importance of careful calculations in engineering.