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 23a

Chapter 7, Problem 23a

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 μ_k = 0.40. What is the tension in the rope attached to the wall?

Verified step by step guidance

Step 1: Analyze the forces acting on the system. The 1.0 kg block is tied to the wall with a rope, so there is a tension force in the rope preventing it from moving. The 2.0 kg block is being pulled to the right with a tension force of 20 N. Both blocks experience friction due to the coefficient of kinetic friction (μk = 0.40).

Step 2: Calculate the frictional force between the 1.0 kg block and the 2.0 kg block. The normal force exerted by the 1.0 kg block on the 2.0 kg block is equal to its weight, which is given by F_normal = m * g, where m = 1.0 kg and g = 9.8 m/s². The frictional force is then F_friction = μk * F_normal.

Step 3: Calculate the frictional force between the 2.0 kg block and the surface it is on. The normal force exerted by the 2.0 kg block on the surface is equal to the combined weight of both blocks, which is F_normal = (m1 + m2) * g, where m1 = 1.0 kg and m2 = 2.0 kg. The frictional force is F_friction = μk * F_normal.

Step 4: Determine the net force acting on the 2.0 kg block. The net force is given by F_net = Tension_force - Friction_force (from both surfaces). Use the calculated frictional forces from Steps 2 and 3 to find the net force.

Step 5: Use Newton's second law to analyze the forces acting on the 1.0 kg block. Since the block is stationary, the tension in the rope attached to the wall must balance the frictional force between the 1.0 kg block and the 2.0 kg block. Therefore, the tension in the rope is equal to the frictional force calculated in Step 2.

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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 equation F = ma, where F is the net force, m is the mass, and a is the acceleration. In this scenario, understanding how the forces acting on the blocks relate to their motion is crucial for determining the tension in the rope.

Frictional Force

Frictional force is the resistance that one surface or object encounters when moving over another. It is calculated using the equation F_friction = μN, where μ is the coefficient of friction and N is the normal force. In this problem, the coefficient of kinetic friction (μ_k = 0.40) will affect the motion of the blocks and must be considered when calculating the net forces acting on them.

Tension in a Rope

Tension is the force transmitted through a rope or string when it is pulled tight by forces acting from opposite ends. In this scenario, the tension in the rope attached to the wall must balance the forces acting on the 1.0 kg block, including the frictional force opposing the motion and the net force due to the 20 N tension pulling the 2.0 kg block. Analyzing these forces is essential to find the tension in the rope.

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Textbook Question

A mobile at the art museum has a 2.0 kg steel cat and a 4.0 kg steel dog suspended from a lightweight cable, as shown in FIGURE EX7.21. It is found that $theta sub 1$ = 20° when the center rope is adjusted to be perfectly horizontal. What are the tension and the angle of rope 3?

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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 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

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

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

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.

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