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 15

Chapter 7, Problem 15

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?

Verified step by step guidance

Step 1: Identify the forces acting on each sled. Sled A has a mass of 100 kg, and sled B has a mass of 80 kg. Both sleds experience frictional forces due to the snow, which can be calculated using the formula: \( F_{\text{friction}} = \mu \cdot m \cdot g \), where \( \mu \) is the coefficient of friction (0.10), \( m \) is the mass of the sled, and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

Step 2: Calculate the frictional force for sled A. Using the formula \( F_{\text{friction}} = \mu \cdot m \cdot g \), substitute \( \mu = 0.10 \), \( m = 100 \, \text{kg} \), and \( g = 9.8 \, \text{m/s}^2 \). This will give the frictional force acting on sled A.

Step 3: Calculate the frictional force for sled B. Similarly, use the formula \( F_{\text{friction}} = \mu \cdot m \cdot g \), substituting \( \mu = 0.10 \), \( m = 80 \, \text{kg} \), and \( g = 9.8 \, \text{m/s}^2 \). This will give the frictional force acting on sled B.

Step 4: Analyze the forces acting on sled B. The tension in rope 1 (150 N) is pulling sled B forward, while the frictional force calculated in Step 3 is opposing the motion. The net force on sled B can be calculated using \( F_{\text{net}} = T_1 - F_{\text{friction, B}} \), where \( T_1 \) is the tension in rope 1.

Step 5: Determine the tension in rope 2. The tension in rope 2 must overcome the net force acting on sled B (calculated in Step 4) and the frictional force acting on sled A (calculated in Step 2). Use the equation \( T_2 = F_{\text{net, B}} + F_{\text{friction, A}} \) to find the tension in rope 2.

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Key Concepts

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

Friction

Friction is the force that opposes the relative motion of two surfaces in contact. It is influenced by the nature of the surfaces and the normal force acting between them. The coefficient of friction, given as 0.10 in this scenario, quantifies the frictional force relative to the normal force, affecting how much force is needed to move the sleds across the snow.

Tension in a Rope

Tension is the force transmitted through a rope or cable when it is pulled tight by forces acting from opposite ends. In this problem, the tension in rope 1 (150 N) affects the sleds' movement and the tension in rope 2, which must be calculated. Understanding how tension distributes through connected objects is crucial for solving problems involving multiple forces.

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). This principle is essential for analyzing the forces acting on the sleds and the dog, allowing us to determine the relationship between the tensions in the ropes and the resulting motion of the sleds.

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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

A 2.0-m-long, 500 g rope pulls a 10 kg block of ice across a horizontal, frictionless surface. The block accelerates at 2.0 m/s². How much force pulls forward on he rope? Assume that the rope is perfectly horizontal.

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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

A 500 kg air conditioner sits on the flat roof of a building. The coefficient of static friction between the roof and the air conditioner is 0.90. A massless rope attached to the air conditioner passes over a massless, frictionless pulley at the edge of the roof. In an effort to drag the air conditioner to the edge of the roof, four 100 kg students hang from the free end of the rope, but the air conditioner refuses to budge. What is the magnitude of the rope tension at the point where it is attached to the air conditioner?

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Textbook Question

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?

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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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