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Ch. 04 - Dynamics: Newton's Laws of Motion
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 04 - Dynamics: Newton's Laws of MotionProblem 86
Chapter 4, Problem 86

Three mountain climbers who are roped together in a line are ascending an icefield inclined at 29° to the horizontal (Fig. 4–67). The last climber slips, pulling the second climber off his feet. The first climber is able to hold them both. If each climber has a mass of 75 kg, calculate the tension in each of the two sections of rope between the three climbers. Ignore friction between the ice and the fallen climbers.
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Identify the forces acting on each climber. For each climber, the forces include their weight (gravitational force), the tension in the rope, and the normal force from the ice surface. Since the ice is frictionless, there is no frictional force to consider.
Break the forces into components parallel and perpendicular to the incline. The weight of each climber can be expressed as \( F_g = m \cdot g \), where \( m \) is the mass of the climber and \( g \) is the acceleration due to gravity. The component of the weight parallel to the incline is \( F_{g, \text{parallel}} = m \cdot g \cdot \sin(\theta) \), and the component perpendicular to the incline is \( F_{g, \text{perpendicular}} = m \cdot g \cdot \cos(\theta) \).
Analyze the forces acting on the last climber (Climber 3). The tension in the rope connecting Climber 3 to Climber 2 must balance the parallel component of Climber 3's weight. Thus, \( T_2 = m \cdot g \cdot \sin(\theta) \).
Analyze the forces acting on the second climber (Climber 2). Climber 2 is being pulled by Climber 3's weight and is also pulling Climber 1. The tension in the rope connecting Climber 2 to Climber 1 must balance the combined parallel components of the weights of Climbers 2 and 3. Thus, \( T_1 = 2 \cdot m \cdot g \cdot \sin(\theta) \).
Substitute the known values into the equations. Use \( m = 75 \ \text{kg} \), \( g = 9.8 \ \text{m/s}^2 \), and \( \theta = 29^\circ \) to calculate the tensions \( T_1 \) and \( T_2 \). Ensure that the trigonometric functions are applied correctly to find the numerical values of the tensions.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Forces and Tension

In this scenario, the tension in the ropes connecting the climbers is a result of the gravitational force acting on the climbers and the forces exerted by the climbers on each other. Tension is the force transmitted through the rope when it is pulled tight by forces acting from opposite ends. Understanding how to calculate tension involves analyzing the forces acting on each climber and applying Newton's second law.
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Inclined Plane Dynamics

The icefield's incline at 29° introduces a component of gravitational force acting parallel to the slope, which affects the climbers' ability to remain stationary. The gravitational force can be resolved into two components: one acting perpendicular to the incline and the other parallel to it. This concept is crucial for determining how much force is needed to hold the climbers in place and how it influences the tension in the ropes.
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Newton's Second Law of Motion

Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma). In this problem, it is essential to apply this law to each climber to analyze the forces acting on them when one slips. By setting up equations based on the forces and the mass of the climbers, we can solve for the tension in the ropes.
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Related Practice
Textbook Question

Consider the system shown in Fig. 4–68 with mA = 8.2kg and mB = 11.5kg. The angles θA = 59° and θB = 32°. In the absence of friction, what force F→\(\overrightarrow{F}\) would be required to pull the masses at a constant velocity up the fixed inclines?

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Textbook Question

The double Atwood machine shown in Fig. 4–55 has frictionless, massless pulleys and cords. Determine the tensions FTA and FTC in the cords.

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Textbook Question

Consider the system shown in Fig. 4–68 with mA = 8.2kg and mB = 11.5kg. The angles θA = 59° and θB = 32°. In the absence of F→\(\overrightarrow{F}\), what is the tension in the string?

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Textbook Question

Determine a formula for the acceleration of the system shown in Fig. 4–49 (see Problem 55) if the cord has a non-negligible mass mC. Specify in terms of ℓA and ℓB , the lengths of cord from the respective masses to the pulley. (The total cord length is ℓA + ℓB.)

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Textbook Question

The double Atwood machine shown in Fig. 4–55 has frictionless, massless pulleys and cords. Determine the acceleration of masses mA, mB, and mC.

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