Textbook Question
An 85 kg cheerleader stands on a scale that reads in kg. What does the scale read if the 85 kg cheerleader lifts the 50 kg cheerleader upward with an acceleration of 2.0 m/s²?
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Ch 07: Newton's Third Law
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 7, Problem 28
A rope of length L and mass m is suspended from the ceiling. Find an expression for the tension in the rope at position y, measured upward from the free end of the rope.
Verified step by step guidance1
Start by understanding that the tension in the rope at a position y is due to the weight of the rope below that point. The weight of the rope below y depends on the mass of the rope segment below y.
The mass per unit length of the rope, also called the linear mass density, is \( \lambda = \frac{m}{L} \), where \( m \) is the total mass of the rope and \( L \) is its total length.
The length of the rope below the position y is \( L - y \). Therefore, the mass of the rope below y is \( \lambda (L - y) = \frac{m}{L}(L - y) \).
The weight of the rope below y is the force due to gravity acting on this mass, which is \( F = \text{mass} \times g = \frac{m}{L}(L - y)g \), where \( g \) is the acceleration due to gravity.
The tension at position y is equal to the weight of the rope below that point. Thus, the tension is given by \( T(y) = \frac{m}{L}(L - y)g \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Tension in a Rope
Tension is the force exerted along the length of a rope or string when it is pulled tight by forces acting from opposite ends. In a hanging rope, the tension varies along its length due to the weight of the rope itself. The tension at any point in the rope must balance the weight of the rope below that point.
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Calculating Tension in a Pendulum with Energy Conservation
Weight of the Rope
The weight of the rope is the force due to gravity acting on its mass, calculated as W = mg, where m is the mass of the rope and g is the acceleration due to gravity. This weight contributes to the tension experienced at different points along the rope, as the tension must support not only the segment of the rope above a given point but also the weight of the rope below that point.
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Static Equilibrium
Static equilibrium refers to a state where an object is at rest and the sum of forces acting on it is zero. In the context of the rope, this means that at any point along the rope, the upward tension force must equal the downward gravitational force acting on the segment of the rope below that point. This principle allows us to derive expressions for tension based on the position along the rope.
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Related Practice
Textbook Question
A 75 kg archer on ice skates is standing at rest on very smooth ice. He shoots a 450 g arrow horizontally. When released, the arrow reaches a speed of 110 m/s in 0.25 s. Assume that the force of the bow string on the arrow is constant. What is the archer's recoil speed?
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Textbook Question
The 100 kg block in FIGURE EX7.24 takes 6.0 s to reach the floor after being released from rest. What is the mass of the block on the left? The pulley is massless and frictionless.
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Textbook Question
Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of the blocks, with a mass of 6.0 kg, accelerates downward at (3/4)g. What is the mass of the other block?
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Textbook Question
Your forehead can withstand a force of about 6.0 kN before fracturing, while your cheekbone can withstand only about 1.3 kN. Suppose a 140 g baseball traveling at 30 m/s strikes your head and stops in 1.5 ms. What is the magnitude of the force that stops the baseball?
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Textbook Question
The 1.0 kg block in FIGURE EX7.23 is tied to the wall with a rope. It sits on top of the 2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. The coefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block is μk = 0.40. What is the tension in the rope attached to the wall?
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