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A lamp hangs vertically from a cord in a descending elevator that decelerates at \(2.4 \mathrm{~m} / \mathrm{s}^{2}\). (a) If the tension in the cord is \(89 \mathrm{~N}\), what is the lamp's mass? (b) What is the cord's tension when the elevator ascends with an upward acceleration of \(2.4 \mathrm{~m} / \mathrm{s}^{2}\) ?

Short Answer

Expert verified
(a) The mass of the lamp is approximately 12.03 kg. (b) The tension in the cord is approximately 146.77 N.

Step by step solution

01

Understand Forces Involved

In both scenarios, the forces acting on the lamp are its weight due to gravity and the tension in the cord. The weight of the lamp is the gravitational force acting downward, which can be calculated using the formula \(F_g = mg\), where \(m\) is mass and \(g\) is gravitational acceleration (\(9.8 \, \text{m/s}^2\)). The net force on the lamp is determined by the elevator's acceleration and tension provided by the cord.
02

Write Equation for Decelerating Elevator

For a descending elevator that is decelerating, the net upward force is \(T - mg = ma\), where \(a = 2.4 \, \text{m/s}^2\) is the deceleration. Since the elevator's direction is opposite to the gravitational force, the net force equation becomes \(T - mg = m(-a)\). Rearrange this to \(T = mg - ma\).
03

Solve for Mass

Given that the tension \(T = 89 \, \text{N}\) and \(a = 2.4 \, \text{m/s}^2\), substitute the values and rearrange the equation \(T = mg - ma\) to find the mass. \[m = \frac{T}{g-a} = \frac{89}{9.8-2.4} \, \text{kg} = \frac{89}{7.4} \, \text{kg} \]Calculate \(m\).
04

Calculate Mass

Perform the calculation from Step 3:\[m = \frac{89}{7.4} \approx 12.03 \, \text{kg}\]This is the mass of the lamp when the elevator is decelerating.
05

Write Equation for Ascending Elevator

For the elevator ascending with an acceleration, the tension must provide a net upward force equal to the sum of the force due to acceleration and the weight of the lamp. This can be expressed as: \(T = ma + mg\).
06

Calculate Tension During Ascend

Now using \(m = 12.03 \, \text{kg}\) from the previous calculation and \(a = 2.4 \, \text{m/s}^2\), substitute into the formula: \[T = 12.03 \times 2.4 + 12.03 \times 9.8 \] Calculate \(T\).
07

Complete Calculation for Tension

Compute the tension:\[T = 12.03 \times 2.4 + 12.03 \times 9.8 = 28.872 + 117.894 \approx 146.766 \, \text{N}\]Thus, the tension in the cord when the elevator accelerates upwards is approximately \(146.77 \, \text{N}\).

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Key Concepts

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

Force and Acceleration
Understanding the relationship between force and acceleration is crucial when analyzing motion, particularly in scenarios involving elevators or similar contexts. According to Newton's second law of motion, the net force acting on an object is directly proportional to the acceleration of that object. This relationship is expressed by the equation:
  • \( F = ma \),
where \( F \) is the net force, \( m \) is the mass of the object, and \( a \) is the acceleration.
In our elevator problem, two scenarios are considered: one where the elevator is decelerating while descending, and another where it is accelerating upwards. In both cases, Newton’s second law helps us understand how forces are balanced by accounting for the different accelerations.
When dealing with a descending elevator, deceleration implies an upward net force as it slows down. Conversely, when the elevator is ascending with an acceleration, tension in the cord must overcome gravity and the added force from the upward acceleration. Recognizing how force and acceleration interact offers insights into why tension changes under varying accelerations.
Tension in Physics
Tension is a force exerted by a rope, cable, or cord when it is pulled tight by forces acting on each end. In this context, we focus on the tension in the cord from which the lamp hangs in an elevator. Understanding tension involves analyzing how forces work in harmony to either maintain an object’s position or to control its movement.
When the elevator decelerates, the tension must counteract gravity and contribute to slowing down the lamp's descent. This establishes a system where tension is reduced since some of the gravitational force is used in deceleration. The governing equation is:
  • \( T = mg - ma \),
where \( mg \) is the gravitational force and \( ma \) is the force due to deceleration.
In the ascending scenario, tension must exert additional force to lift the lamp against its weight and to accelerate it upwards, necessitating greater tension. Thus, the equation becomes:
  • \( T = ma + mg \).
Through these equations, we glimpse into how tension plays a pivotal role in maintaining stability and movement in elevator systems.
Gravitational Force
Gravitational force is one of the fundamental forces acting on objects, pulling them toward the center of the Earth. It's a consistent force represented by the equation:
  • \( F_g = mg \),
where \( m \) is mass and \( g \) is the gravitational acceleration, typically approximated as \( 9.8 \, \text{m/s}^2 \).
In the physics behind an elevator moving up or down, gravitational force is a constant factor acting downward on the lamp. It's pivotal in calculating both the tension in the cord and the necessary counterforces in various elevator motions.
When the elevator decelerates while descending, gravity’s pull is slightly opposed, influencing the tension calculation. On the flip side, in ascent, gravitational force combines with additional forces to dictate how much tension is required to support movement upwards. Recognizing gravity’s role in these calculations aids in grasping the broader implications of motion dynamics.

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Most popular questions from this chapter

Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward force (thrust) of \(3260 \mathrm{~N}\), the craft descends at constant speed; if the engine provides only \(2200 \mathrm{~N}\), the craft accelerates downward at \(0.39\) \(\mathrm{m} / \mathrm{s}^{2}\). (a) What is the weight of the landing craft in the vicinity of Callisto's surface? (b) What is the mass of the craft? (c) What is the magnitude of the free-fall acceleration near the surface of Callisto?

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 \(\vec{F}_{1}=(2 \mathrm{~N}) \hat{\mathrm{i}}+\) \((3 N) \hat{j}+(-2 N) \hat{k}\) and \(\vec{F}_{2}=(-5 N) \hat{i}+(8 N) \hat{j}+(-2 N) \hat{k} .\) What is the third force?

A \(10 \mathrm{~kg}\) monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a \(15 \mathrm{~kg}\) package on the ground (Fig. 5-54). (a) What is the magnitude of the least acceleration the monkey must have if it is to lift the package off the ground? If, after the package has been lifted, the monkey stops its climb and holds onto the rope, what are the (b) mag. nitude and (c) direction of the monkey's acceleration and (d) the tension in the rope?

A \(40 \mathrm{~kg}\) skier skis directly down a frictionless slope angled at \(10^{\circ}\) to the horizontal. Assume the skier moves in the negative direction of an \(x\) axis along the slope. A wind force with component \(F_{x}\) acts on the skier. What is \(F_{R}\) if the magnitude of the skier's velocity is (a) constant, (b) increasing at a rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\), and (c) increasing at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

Holding on to a towrope moving parallel to a frictionless ski slope, a \(50 \mathrm{~kg}\) skier is pulled up the slope, which is at an angle of \(8.0^{\circ}\) with the horizontal. What is the magnitude \(F_{\text {ropo }}\) of the force on the skier from the rope when (a) the magnitude \(v\) of the skier's velocity is constant at \(2.0 \mathrm{~m} / \mathrm{s}\) and (b) \(v=2.0 \mathrm{~m} / \mathrm{s}\) as \(v\) increases at a rate of \(0.10 \mathrm{~m} / \mathrm{s}^{2} ?\)

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