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Chapter 4: Problem 77

The drawing shows Robin Hood (mass \(=77.0 \mathrm{~kg}\) ) about to escape from a dangerous situation. With one hand, he is gripping the rope that holds up a chandelier (mass \(=195\) \(\mathrm{kg}\) ). When he cuts the rope where it is tied to the floor, the chandelier will fall, and he will be pulled up toward a balcony above. Ignore the friction between the rope and the beams over which it slides, and find (a) the acceleration with which Robin is pulled upward and (b) the tension in the rope while Robin escapes.

Short Answer

Expert verified
Robin is pulled up with an acceleration of \(4.18\,\mathrm{m/s}^2\), and the tension in the rope is \(1077.86\,\mathrm{N}\).

Step by step solution

01

Setting Up the Free-Body Diagrams

To solve the problem, first create free-body diagrams for both Robin Hood and the chandelier. For Robin, the forces are his weight \( m_R g \) acting downward and the tension \( T \) in the rope acting upward. For the chandelier, the forces are its weight \( m_C g \) acting downward and the tension in the rope \( T \) acting upward.
02

Using Newton's Second Law for Robin

Apply Newton's second law to Robin, where the net force is the tension minus Robin's weight. Thus, \(T - m_R g = m_R a\), where \( a \) is the acceleration upwards.
03

Using Newton's Second Law for the Chandelier

Apply Newton's second law to the chandelier, where the net force is its weight minus the tension in the rope. Thus, \( m_C g - T = m_C a \).
04

Solving the System of Equations

Now, you have two equations:1. \( T - m_R g = m_R a \)2. \( m_C g - T = m_C a \)Add these equations to eliminate \( T \): - \( T - m_R g + m_C g - T = m_R a + m_C a \)- \( m_C g - m_R g = (m_R + m_C) a \)- \( a = \frac{(m_C - m_R)g}{m_R + m_C} \).
05

Calculating the Acceleration

Substitute \( m_R = 77.0 \) kg, \( m_C = 195 \) kg, and \( g = 9.8 \) m/s虏 into the equation for acceleration:- \( a = \frac{(195 - 77) imes 9.8}{195 + 77} \)- \( a \approx 4.18 \) m/s虏.
06

Calculating the Tension in the Rope

Substitute the value of \( a \) back into one of the original equations to solve for \( T \). Use \( T - m_R g = m_R a \):- \( T = m_R g + m_R a = 77 imes 9.8 + 77 imes 4.18 \)- \( T \approx 1077.86 \) N.

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Key Concepts

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

Free-Body Diagram
A free-body diagram is a simple way to visualize the forces acting on an object. When solving problems using Newton's Second Law, it's a crucial step. It breaks down complicated motion scenarios into clear, understandable pictures of forces which makes problem-solving easier.
For Robin Hood and the chandelier, we need two separate diagrams.
  • **Robin's Free-Body Diagram**: Here, we have two forces acting on him: his weight (\( m_R g \)), pulling downward, and the tension (\( T \)) from the rope, pulling upward.
  • **Chandelier's Free-Body Diagram**: Here, the forces include the chandelier's weight (\( m_C g \)), which pulls downward, and the same tension in the rope (\( T \)) pulling upward.
By picturing these forces, you can clearly see how Newton's laws apply, allowing you to systematically work out the equations needed to solve the problem.
Tension in the Rope
The tension in a rope is essentially the pulling force that acts along its length. It is crucial in calculations involving pulleys and lifting systems. Understanding how tension works can help us solve for unknown forces in a system using Newton's Second Law.

In the Robin Hood scenario, tension plays a vital role in the transition of forces. Here's how it breaks down:
  • The tension (\( T \)) acts upwards on Robin, offsetting his weight and creating the acceleration that moves him upward.
  • Simultaneously, the tension also acts upwards on the chandelier, countering part of its weight. This results in the chandelier's downward acceleration.
By establishing individual force equations for both Robin and the chandelier, you see that tension can be calculated based on the net force and acceleration. Solving these equations gives you the exact tension in the rope.
Acceleration Calculation
Calculating acceleration involves applying Newton's Second Law which states that Force equals mass times acceleration (\( F = ma \)). Robin's dilemma presents a classic example of simultaneous motion in two different objects tied by a rope.

Here鈥檚 the step-by-step of calculating Robin's acceleration:
  • Write down two equations from Robin's and chandelier's free-body diagrams: \( T - m_R g = m_R a \) for Robin and \( m_C g - T = m_C a \) for the chandelier.
  • Add these equations to cancel the tension (\( T \)) and solve for the acceleration (\( a \)).
  • Substitute known values: \( m_R = 77 \text{ kg}, \) \( m_C = 195 \text{ kg}, \) \( g = 9.8 \text{ m/s}^2 \).
  • Arrive at \( a = \frac{(m_C - m_R)g}{m_R + m_C} \).
  • With the numbers plugged in, the acceleration comes out to be approximately \( 4.18 \text{ m/s}^2. \)
This calculation not only tells us how fast Robin will ascend but also reflects the combined dynamics between the two interconnected objects.

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Most popular questions from this chapter

A \(5.00-\mathrm{kg}\) block is placed on top of a \(12.0-\mathrm{kg}\) block that rests on a frictionless table. The coefficient of static friction between the two blocks is \(0.600 .\) What is the maximum horizontal force that can be applied before the \(5.00-\mathrm{kg}\) block begins to slip relative to the \(12.0-\mathrm{kg}\) block, if the force is applied to (a) the more massive block and (b) the less massive block?

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?

Refer to Multiple-Concept Example 10 for help in solving problems like this one. An ice skater is gliding horizontally across the ice with an initial velocity of \(+6.3 \mathrm{~m} / \mathrm{s}\). The coefficient of kinetic friction between the ice and the skate blades is \(0.081,\) and air resistance is negligible. How much time elapses before her velocity is reduced to \(+2.8 \mathrm{~m} /\) s?

A bicyclist is coasting straight down a hill at a constant speed. The mass of the rider and bicycle is \(80.0 \mathrm{~kg},\) and the hill is inclined at \(15.0^{\circ}\) with respect to the horizontal. Air resistance opposes the motion of the cyclist. Later, the bicyclist climbs the same hill at the same constant speed. How much force (directed parallel to the hill) must be applied to the bicycle in order for the bicyclist to climb the hill?

A fisherman is fishing from a bridge and is using a "45-N test line." In other words, the line will sustain a maximum force of \(45 \mathrm{~N}\) without breaking. (a) What is the weight of the heaviest fish that can be pulled up vertically when the line is reeled in (a) at a constant speed and (b) with an acceleration whose magnitude is \(2.0 \mathrm{~m} / \mathrm{s}^{2} ?\)
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