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An elevator cab that weighs \(27.8 \mathrm{kN}\) moves upward. What is the tension in the cable if the cab's speed is (a) increasing at a rate of \(1.22 \mathrm{~m} / \mathrm{s}^{2}\) and (b) decreasing at a rate of \(1.22 \mathrm{~m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
(a) 31.1 kN when increasing; (b) 24.5 kN when decreasing.

Step by step solution

01

Understand the Forces Involved

The primary forces acting on the elevator are its weight (downward force due to gravity) and the tension in the cable (upward force). The weight of the elevator is given as \( W = 27.8 \text{ kN} \).
02

Convert Weight to Mass

First, we need to convert the weight from kilonewtons to newtons: \( 27.8 \text{ kN} = 27800 \text{ N} \). Using the relationship \( W = mg \), where \( g = 9.81 \text{ m/s}^2 \), we can find the mass: \( m = \frac{W}{g} = \frac{27800}{9.81} \approx 2834.35 \text{ kg} \).
03

Determine Net Force for Increasing Speed

When the speed is increasing, the net force \( F_{net} = ma \) acts upward. The upward force is the tension \( T \) in the cable, and the downward force is the weight \( W \). Thus, \( T - W = ma \), where \( a = 1.22 \text{ m/s}^2 \).
04

Calculate Tension for Increasing Speed

Rearranging the equation \( T - W = ma \) gives \( T = W + ma \). Substituting the values in, \( T = 27800 + (2834.35 \times 1.22) \approx 31126.7 \text{ N} \). Thus, the tension is approximately \( 31.1 \text{ kN} \).
05

Determine Net Force for Decreasing Speed

When the speed is decreasing, the acceleration is directing downward; thus, \( -ma \) is used for Newton's second law. The net force equation becomes \( T - W = -ma \).
06

Calculate Tension for Decreasing Speed

Rearrange the equation \( T - W = -ma \) to \( T = W - ma \). Substitute the values into this formula: \( T = 27800 - (2834.35 \times 1.22) \approx 24473.3 \text{ N} \). Thus, the tension is approximately \( 24.5 \text{ kN} \).

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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 in physics that relates the motion of an object to the forces acting upon it. The law is expressed with the formula \( F_{net} = ma \), where \( F_{net} \) represents the net force acting on an object, \( m \) is the mass of the object, and \( a \) is the acceleration of the object. In the context of an elevator problem, this means the difference between the tension in the cable and the weight of the elevator. Newton's Second Law helps us understand:
  • How much force is needed to move an object (e.g., the elevator cab).
  • How changes in force impact the acceleration of an object.
In our exercise, when the elevator speeds up, a net upward force is needed, which increases the tension in the cable. Conversely, when slowing down, the net force is downward, reducing cable tension. These calculations ensure the elevator safely operates within mechanical limits.
Tension in Cable
Tension in a cable is the pulling force exerted by a cable, string, or rope. It counters the weight of the object it's holding. In elevators, the tension in the cable is vital for control and safety. Key factors affecting tension include:
  • Weight of the object: The heavier the object, the more tension needed.
  • Acceleration: Changes in speed influence how much tension is required.
For the elevator problem, tension changes based on whether the elevator is accelerating upwards or downwards:
  • When accelerating up, an additional force is required, increasing tension.
  • When slowing down, less force is needed, so tension decreases.
This dynamic adjustment ensures the elevator remains stable as it starts or stops.
Force and Acceleration
The relationship between force and acceleration is crucial for understanding motion dynamics in physics. According to Newton's Second Law, force and acceleration are directly proportional. This means:
  • If more force is applied to an object, its acceleration increases, given the mass stays constant.
  • Conversely, less force results in deceleration or slower acceleration.
For an elevator, this relationship helps predict how quickly it will rise or fall given a certain input force (like the tension in the cable). In our scenario:
  • Increasing upwards speed requires additional force (hence higher tension).
  • Reducing downward speed means the tension can decrease as less force opposes gravity.
By mastering this, engineers ensure elevators operate efficiently and safely.
Weight and Mass Conversion
Understanding the difference between weight and mass is essential in physics, especially in problems involving forces. Here's how they differ:
  • Mass is a measure of the amount of matter in an object, measured in kilograms (kg).
  • Weight is the force exerted by gravity on an object, measured in newtons (N).
To convert weight to mass, use the formula \( m = \frac{W}{g} \), where \( W \) is weight and \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity). In the elevator exercise:
  • The given weight of the elevator (\(27.8 \text{ kN}\)) was converted to newtons (\(27800 \text{ N}\)) and then to mass, resulting in \(2834.35 \text{ kg}\).
This conversion is crucial for further calculations of force and tension, since calculations like those required by Newton's Second Law need mass to determine the effect of forces accurately.

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Most popular questions from this chapter

A \(2.0 \mathrm{~kg}\) particle moves along an \(x\) axis, being propelled by a variable force directed along that axis. Its position is given by \(x=\) \(3.0 \mathrm{~m}+(4.0 \mathrm{~m} / \mathrm{s}) t+c t^{2}-\left(2.0 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3}\), with \(x\) in meters and \(t\) in seconds. The factor \(c\) is a constant. At \(t=3.0 \mathrm{~s}\), the force on the particle has a magnitude of \(36 \mathrm{~N}\) and is in the negative direction of the axis. What is \(c\) ?

If the \(1 \mathrm{~kg}\) standard body is accelerated by only \(\vec{F}_{1}=\) \((3.0 \mathrm{~N}) \hat{\mathrm{i}}+(4.0 \mathrm{~N}) \hat{\mathrm{j}}\) and \(\vec{F}_{2}=(-2.0 \mathrm{~N}) \hat{\mathrm{i}}+(-6.0 \mathrm{~N}) \hat{\mathrm{j}}\), then what is \(\vec{F}_{\text {net }}(\) a \()\) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive \(x\) direction? What are the (d) magnitude and (e) angle of \(\vec{a}\) ?

A firefighter who weighs 712 N slides down a vertical pole with an acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2}\), directed downward. What are the (a) magnitude and (b) direction (up or down) of the vertical force on the firefighter from the pole and the (c) magnitude and (d) direction of the vertical force on the pole from the firefighter?

Three forces act on a particle that moves with unchanging velocity \(\vec{v}=(2 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(7 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). Two of the forces are \(\vec{F}_{1}=(2 \mathrm{~N}) \hat{\mathrm{i}}+\) \((3 \mathrm{~N}) \hat{\mathrm{j}}+(-2 \mathrm{~N}) \hat{\mathrm{k}}\) and \(\vec{F}_{2}=(-5 \mathrm{~N}) \hat{\mathrm{i}}+(8 \mathrm{~N}) \hat{\mathrm{j}}+(-2 \mathrm{~N}) \hat{\mathrm{k}}\). What is the third force?

A car traveling at \(53 \mathrm{~km} / \mathrm{h}\) hits a bridge abutment. A passenger in the car moves forward a distance of \(65 \mathrm{~cm}\) (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger's upper torso, which has a mass of \(41 \mathrm{~kg}\) ?

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