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Chapter 4: Problem 74

A helicopter flies over the arctic ice pack at a constant altitude, towing an airborne 129-kg laser sensor that measures the thickness of the ice (see the drawing). The helicopter and the sensor both move only in the horizontal direction and have a horizontal acceleration of magnitude 2.84 \(\mathrm{m} / \mathrm{s}^{2}\) . Ignoring air resistance, find the tension in the cable towing the sensor.

Short Answer

Expert verified
The tension in the cable is 366.36 N.

Step by step solution

01

Understand the Problem

The problem involves a helicopter towing a sensor horizontally at a constant altitude. The helicopter and the sensor are accelerating horizontally with a given magnitude, 2.84 m/s虏. We need to find the tension in the cable towing the 129 kg sensor, while ignoring air resistance.
02

Identify the Forces

Consider the forces acting on the sensor. Since air resistance is ignored, the only forces are the tension in the cable (T) and the gravitational force acting downward on the sensor (weight = mass * gravity). The acceleration is horizontal, so we will primarily concern ourselves with the tension providing this horizontal acceleration.
03

Apply Newton's Second Law

Newton's Second Law states that the net force acting on an object is equal to the product of its mass and acceleration, i.e., \( F = ma \). Here, the net force is the horizontal tension, and the mass is 129 kg while the acceleration is 2.84 m/s虏.
04

Calculate the Tension in the Cable

Using Newton's Second Law, the tension \( T \) is given by \( T = ma \). Plug in the values: \( m = 129 \) kg and \( a = 2.84 \text{ m/s}^2 \). Calculate \( T = 129 \times 2.84 \).
05

Compute the Result

Now perform the multiplication: \( T = 129 \times 2.84 = 366.36 \). Thus, the tension in the cable is 366.36 N.

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Key Concepts

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

Tension Calculation
Tension refers to the force exerted by a string, cable, or similar object when it is pulled tight by forces acting from opposite ends. In this exercise, the key is to determine the tension in the cable towing the sensor. To find tension, we consider that it must provide the necessary horizontal force to accelerate the sensor, as air resistance is ignored.

Newton's Second Law helps us here: it tells us that the net force acting on an object is equal to the mass of the object multiplied by its acceleration, shown mathematically as \( F = ma \).
For the sensor:
  • Mass \( (m) = 129 \text{ kg} \)
  • Acceleration \( (a) = 2.84 \text{ m/s}^2 \)
Therefore, the tension \( T \), being the only force causing the sensor's acceleration, can be calculated as \( T = ma = 129 \times 2.84 \).

Once you plug and compute this, you get \( T = 366.36 \text{ N} \). This means the cable must exert a force of 366.36 N horizontally to achieve the given acceleration.
Horizontal Acceleration
Horizontal acceleration occurs when a force causes an object to accelerate along a straight path parallel to the ground. In the context of the exercise, both the helicopter and the sensor are accelerating with a specific horizontal acceleration of 2.84 \(\text{m/s}^2 \).

Acceleration decides the rate of change of velocity, and in this case, it is applied in a consistent direction, the horizontal. For the sensor, this uniform horizontal acceleration needs a constant force, provided by the tension in the cable.

Key points about horizontal acceleration:
  • Applies to objects moving along the same level plane.
  • Can be measured in meters per second squared (\(\text{m/s}^2\)).
  • Directly affects the velocity of the object in motion.
This fundamental understanding of horizontal acceleration helps to conceptualize how forces like tension can influence movement without changing vertical position.
Force and Mass Relationship
The relationship between force, mass, and acceleration is central to understanding motion dynamics. Newton's Second Law summarizes it in the formula \( F = ma \), where \( F \) is the force applied to an object, \( m \) is its mass, and \( a \) is the acceleration that force produces.

This law indicates two critical insights:
  • The greater the mass of an object, the larger the force needed to achieve the same acceleration.
  • Simplifying problem-solving, the direction of the force is the same direction as the acceleration.
In our exercise situation, the sensor's mass of 129 kg multiplied by its acceleration of 2.84 \(\text{m/s}^2\) is used to resolve the tension (force) in the cable. Understanding how mass influences the amount of force required helps to predict how changes in mass or desired acceleration would alter the required tension force.

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Most popular questions from this chapter

A \(1.14 \times 10^{4}\) -kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is 1.60 \(\mathrm{m} / \mathrm{s}^{2}\) . At an altitude of 165 \(\mathrm{m}\) the craft's downward velocity is 18.0 \(\mathrm{m} / \mathrm{s}\) . To slow down the craft, a retrorocket is firing to provide an upward thrust. Assuming the descent is vertical, find the magnitude of the thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface.

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The drawing (not to scale) shows one alignment of the sun, earth, and moon. The gravitational force \(\overline{\mathbf{F}}_{\text { SM }}\)that the sun exerts on the moon is perpendicular to the force \(\overrightarrow{\mathbf{F}}_{\mathrm{EM}}\) that the earth exerts on the moon. The masses are: mass of sun \(=1.99 \times 10^{30} \mathrm{kg}\) , mass of earth \(=5.98 \times 10^{24} \mathrm{kg}\) , mass of moon \(=7.35 \times 10^{22} \mathrm{kg}\) . The distances shown in the drawing are \(r_{\mathrm{SM}}=1.50 \times 10^{11} \mathrm{m}\) and \(r_{\mathrm{EM}}=3.85 \times 10^{8} \mathrm{m} .\) Determine the magnitude of the net gravitational force on the moon.

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