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Ch 07: Newton's Third Law
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 07: Newton's Third LawProblema 24
Capitolo 7, Problema 24

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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Identify the forces acting on the system: The 100 kg block on the right experiences a downward gravitational force \( F_{g, \text{right}} = m_{\text{right}} g \), where \( m_{\text{right}} = 100 \ \text{kg} \) and \( g = 9.8 \ \text{m/s}^2 \). The block on the left (mass \( m_{\text{left}} \)) experiences a downward gravitational force \( F_{g, \text{left}} = m_{\text{left}} g \). The tension in the rope is the same throughout because the pulley is massless and frictionless.
Write the equations of motion for each block: For the 100 kg block (right block), the net force is \( F_{\text{net, right}} = m_{\text{right}} a = m_{\text{right}} g - T \), where \( T \) is the tension in the rope. For the left block, the net force is \( F_{\text{net, left}} = m_{\text{left}} a = T - m_{\text{left}} g \). Here, \( a \) is the acceleration of the system, which is the same for both blocks.
Relate the acceleration to the motion: The 100 kg block takes 6.0 s to reach the floor. Using the kinematic equation \( d = \frac{1}{2} a t^2 \), where \( d \) is the distance the block falls, \( t = 6.0 \ \text{s} \), and \( a \) is the acceleration, solve for \( a \).
Combine the equations of motion: Substitute \( a \) from the kinematic equation into the equations of motion for both blocks. Add the two equations to eliminate \( T \), resulting in \( m_{\text{right}} g - m_{\text{left}} g = (m_{\text{right}} + m_{\text{left}}) a \). Solve for \( m_{\text{left}} \) in terms of known quantities.
Substitute the known values: Use \( m_{\text{right}} = 100 \ \text{kg} \), \( g = 9.8 \ \text{m/s}^2 \), and the calculated \( a \) from the kinematic equation to find \( m_{\text{left}} \).

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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. This principle is crucial for analyzing the forces acting on the blocks and the resulting motion, allowing us to calculate the acceleration and the forces involved in the system.
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Intro to Forces & Newton's Second Law

Free Fall and Acceleration due to Gravity

Free fall refers to the motion of an object under the influence of gravity alone, with an acceleration of approximately 9.81 m/s² near the Earth's surface. Understanding free fall is essential for determining how long it takes for the block to reach the ground and how this relates to the forces acting on the other block in the pulley system.
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Acceleration Due to Gravity

Pulley Systems and Tension

In a pulley system, tension is the force transmitted through the rope or cable, which affects the motion of the connected masses. Since the pulley is massless and frictionless, the tension is the same throughout the rope, allowing us to set up equations that relate the masses and their accelerations, ultimately leading to the solution for the unknown mass.
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Systems Connected By Pulleys
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