Ch 07: Newton's Third Law

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 07: Newton's Third Law

Problem 9b

Chapter 7, Problem 9b

Blocks with masses of 1 kg, 2 kg, and 3 kg are lined up in a row on a frictionless table. All three are pushed forward by a 12 N force applied to the 1 kg block. How much force does the 2 kg block exert on the 1 kg block?

Verified step by step guidance

Step 1: Begin by analyzing the system as a whole. The total mass of the system is the sum of the masses of all three blocks: \( m_{total} = m_1 + m_2 + m_3 \). Substitute \( m_1 = 1 \, \text{kg} \), \( m_2 = 2 \, \text{kg} \), and \( m_3 = 3 \, \text{kg} \) into the equation.

Step 2: Calculate the acceleration of the system using Newton's second law, \( F = m \cdot a \). The net force acting on the system is \( F_{net} = 12 \, \text{N} \). Rearrange the formula to solve for acceleration: \( a = \frac{F_{net}}{m_{total}} \). Substitute the values for \( F_{net} \) and \( m_{total} \).

Step 3: Focus on the interaction between the 1 kg and 2 kg blocks. The force exerted by the 2 kg block on the 1 kg block is equal to the force required to accelerate the 1 kg block at the same rate as the system's acceleration. Use \( F = m \cdot a \) for the 1 kg block, where \( m = 1 \, \text{kg} \) and \( a \) is the system's acceleration calculated in Step 2.

Step 4: Recognize that the force exerted by the 2 kg block on the 1 kg block is part of the internal forces within the system. This force is equal in magnitude but opposite in direction to the force exerted by the 1 kg block on the 2 kg block, according to Newton's third law.

Step 5: Substitute the calculated acceleration into the formula \( F = m \cdot a \) for the 1 kg block to determine the magnitude of the force exerted by the 2 kg block on the 1 kg block. Ensure the units are consistent throughout the calculation.

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Key Concepts

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

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 relationship is expressed by the formula F = ma, where F is the net force, m is the mass, and a is the acceleration. In this scenario, understanding how the total force affects the acceleration of the blocks is crucial for determining the forces they exert on each other.

System of Connected Masses

When multiple masses are connected and subjected to a force, they can be treated as a single system to analyze their motion. The total mass of the system affects how the applied force is distributed among the individual masses. In this case, the 1 kg, 2 kg, and 3 kg blocks move together, and the force exerted by one block on another can be calculated by considering the acceleration of the entire system.

Action and Reaction Forces

According to Newton's Third Law, for every action, there is an equal and opposite reaction. This means that when the 2 kg block exerts a force on the 1 kg block, the 1 kg block exerts an equal and opposite force on the 2 kg block. Understanding this principle is essential for analyzing the interactions between the blocks and determining the force exerted by the 2 kg block on the 1 kg block.

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Newton's Third Law & Action-Reaction Pairs

Related Practice

Textbook Question

A 1000 kg car is pushing an out-of-gear 2000 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push horizontally against the ground with a force of 4500 N. Rolling friction can be neglected. What is the magnitude of the force of the car on the truck?

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Textbook Question

How much force does the astronaut exert on his chair while accelerating straight up at 10 m/s²?

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Textbook Question

The foot of a 55 kg sprinter is on the ground for 0.25 s while her body accelerates from rest to 2.0 m/s. What is the magnitude of the friction force?

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Textbook Question

The sled dog in FIGURE EX7.15 drags sleds A and B across the snow. The coefficient of friction between the sleds and the snow is 0.10. If the tension in rope 1 is 150 N, what is the tension in rope 2?

Textbook Question

Block A in FIGURE EX7.4 is sliding down the incline. The rope is massless, and the massless pulley turns on frictionless bearings, but the surface is not frictionless. The rope and the pulley are among the interacting objects, but you'll have to decide if they're part of the system. Draw a free-body diagram for each object in the system. Use dashed lines to connect members of an action/reaction pair.

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Textbook Question

FIGURE EX7.17 shows two 1.0 kg blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at 3.0 m/s² by force F. What is the tension at the top end of rope 1?

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