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Part \(a\) of the drawing shows a bucket of water suspended from the pulley of a well; the tension in the rope is \(92.0 \mathrm{~N}\). Part \(b\) shows the same bucket of water being pulled up from the well at a constant velocity. What is the tension in the rope in part \(b\) ?

Short Answer

Expert verified
The tension in the rope in part b is 92.0 N.

Step by step solution

01

Understanding the Situation

In part \(a\), the bucket is stationary, which means the tension in the rope equals the gravitational force acting on the bucket. Therefore, \( T = F_g = 92.0 \, \text{N}\).
02

Analyze Part b: Constant Velocity

When the bucket is being pulled up at a constant velocity, there is no net acceleration (Newton's First Law). Thus, the tension in the rope must still balance the gravitational force.
03

Applying Newton's First Law

Since the bucket is moving at a constant velocity in part \(b\), the forces are balanced. Therefore, the tension in the rope, \( T_b \), is equal to the gravitational force \( F_g \).
04

Conclusion: Tension in Part b

The tension in part \(b\) is the same as in part \(a\): \( T_b = 92.0 \, \text{N} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force
Gravitational force is a fundamental concept in physics and plays a crucial role in this exercise. It's the pulling force that attracts two bodies towards each other, especially noticeable between Earth and objects on it. The formula to calculate gravitational force is:\[ F_g = m \cdot g \]where:
  • \( F_g \) is the gravitational force,
  • \( m \) is the mass of the object, and
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \) on Earth.
In part (a) of the problem, the gravitational force acting on the bucket is equal to the tension in the rope, which is given as \( 92.0 \, \text{N} \). This shows that the force required to hold the bucket stationary (and thus the gravitational force) is \( 92.0 \, \text{N} \). When analyzing real-world scenarios, gravitational force helps determine how much force is needed to lift or support objects. It's essential to consider gravitational force whenever an object is in contact with Earth or near Earth's surface.
Constant Velocity
The term "constant velocity" is pivotal in understanding part (b) of this problem. When an object moves at constant velocity, it means that both the speed and direction are consistent. In simple terms, there's no speeding up or slowing down. At constant velocity, the net force acting on the object is zero.This is because all forces are balanced. In the case of the bucket being pulled up, this balance is between tension in the rope and the gravitational force pulling the bucket down. Using Newton's laws, constant velocity signifies that:
  • The upward tension force is equal to the downward gravitational force.
  • There is no acceleration in the system.
This means the tension in the rope remains \( 92.0 \, \text{N} \), the same as when the bucket is stationary. Any variance in speed or direction would require an additional net force, altering the tension needed.
Newton's First Law
Newton's First Law, also known as the law of inertia, states that an object will remain at rest, or move at a constant velocity, unless acted upon by a net force. It's a fundamental principle that helps explain motion and stability in physics.In the context of this exercise:
  • The bucket remains in motion at constant velocity, meaning no unbalanced force acts on it.
  • This stability is why the tension remains at \( 92.0 \, \text{N} \), as it perfectly counteracts the gravitational pull of the Earth.
When considering real-world applications, Newton's First Law explains why objects in motion stay in motion unless something changes their speed or direction. This concept is central to understanding why, in this exercise, pulling the bucket at constant velocity doesn't alter the tension. The rope's tension perfectly balances the forces, ensuring no acceleration occurs. Hence, equilibrium is maintained, showcasing Newton's First Law in action.

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Most popular questions from this chapter

At a time when mining asteroids has become feasible, astronauts have connected a line between their 3500 -kg space tug and a 6200 -kg asteroid. Using their ship's engine, they pull on the asteroid with a force of \(490 \mathrm{~N}\). Initially the tug and the asteroid are at rest, \(450 \mathrm{~m}\) apart. How much time does it take for the ship and the asteroid to meet?

A 350 -kg sailboat has an acceleration of \(0.62 \mathrm{~m} / \mathrm{s}^{2}\) at an angle of \(64^{\circ}\) north of east. Finc the magnitude and direction of the net force that acts on the sailboat.

Refer to Concept Simulation \(4.4\) at for background relating to this problem. The drawing shows a large cube (mass \(=25 \mathrm{~kg}\) ) being accelerated across a horizontal frictionless surface by a horizontal force \(\vec{P}\). A small cube \((\mathrm{mass}=4.0 \mathrm{~kg})\) is in contact with the front surface of the large cube and will slide downward unless \(\vec{P}\) is sufficiently large. The coefficient of static friction between the cubes is \(0.71\). What is the smallest magnitude that \(\overrightarrow{\mathrm{P}}\) can have in order to keep the small cube from sliding downward?

The central ideas in this problem are reviewed in Multiple-Concept Example \(9 .\) One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and its magnitude is slowly increased. When the force reaches \(47.0 \mathrm{~N}\), the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block?

A person is attempting to push a refrigerator across a room. He exerts a horizontal force on the refrigerator, but it does not move. (a) What other horizontal force must be acting on the refrigerator? (b) How are the magnitude and direction of this force related to the force that the person exerts? (c) Suppose that the person applies the force such that it is the largest possible force before the refrigerator begins to move. What factors determine the magnitude of this force? Problem Consult Multiple-Concept Example 9 to explore a model for solving this problem. A person pushes on a 57 -kg refrigerator with a horizontal force of \(-267 \mathrm{~N}\); the minus sign indicates that the force is directed along the \(-x\) direction. The coefficient of static friction is \(0.65 .\) (a) If the refrigerator does not move, what is the magnitude and direction of the static frictional force that the floor exerts on the refrigerator? (b) What is the magnitude of the largest pushing force that can be applied to the refrigerator before it just begins to move?

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