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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 \(=195\) 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 upward at ~4.26 m/s², and the tension in the rope is ~1077.39 N.

Step by step solution

01

Analyze Forces Involved

Start by identifying the forces acting on both Robin Hood and the chandelier. For Robin, the forces acting are his weight \(m_Rg\) downward and the tension \(T\) upward. For the chandelier, the forces are its weight \(m_Cg\) downward and the tension \(T\) upward, acting at the point where the rope supports Robin. The forces on Robin and the chandelier will determine the equations of motion for each.
02

Apply Newton's Second Law to Robin

Use Newton’s second law \(\sum F = ma\) for Robin. The net force on Robin is \(T - m_Rg = m_Ra\), where \(m_R = 77.0 \, \text{kg}\) and \(a\) is Robin's upward acceleration.
03

Apply Newton's Second Law to the Chandelier

Apply Newton’s second law for the chandelier: the downward force is \(m_Cg - T = m_Ca\), where \(m_C = 195 \, \text{kg}\) and the chandelier experiences the same magnitude of acceleration \(a\) as Robin since they are connected by the same rope.
04

Solve the System of Equations

We have two equations now: \(T = m_Ra + m_Rg\) (from Robin) and \(T = m_Cg - m_Ca\) (from the chandelier). Set the expressions for \(T\) equal to each other:\[m_Ra + m_Rg = m_Cg - m_Ca\] Simplify and solve for \(a\):\[a(m_R + m_C) = m_Cg - m_Rg\] \[a = \frac{(m_C - m_R)g}{m_R+m_C}\] Substitute the given values: \(a = \frac{(195 - 77)(9.81)}{195+77}\).
05

Calculate Acceleration

Substitute the values into the formula: \[a = \frac{(118)(9.81)}{272} \approx 4.26 \, \text{m/s}^2\]. This is the acceleration with which Robin is pulled upwards.
06

Calculate the Tension in the Rope

Substitute \(a = 4.26 \, \text{m/s}^2\) into either equation for \(T\), let's use the equation \(T = m_Ra + m_Rg\):\[T = 77 \times 4.26 + 77 \times 9.81\]Compute the tension:\[T \approx 322.02 + 755.37 = 1077.39 \, \text{N}\]. This is the tension in the rope.

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Key Concepts

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

Acceleration Calculation
When calculating acceleration, we're focusing on how quickly an object speeds up or slows down. In our scenario with Robin Hood and the chandelier, we're delving into Newton's Second Law. This law tells us that acceleration is produced when a force acts on a mass. To find the acceleration that pulls Robin upwards, we begin by identifying the forces at play. We have Robin's weight pulling him down and the tension from the rope pulling him up.
The equation that brings this all together is: \[ T - m_Rg = m_Ra \]Where:
  • \( T \) represents the tension in the rope.
  • \( m_R \) is Robin's mass \(77.0 \, \text{kg}\).
  • \( g \) is the acceleration due to gravity, approximately \(9.81 \, \text{m/s}^2\).
  • \( a \) is the acceleration we are solving for.
With the chandelier, a similar setup applies, and once we have both equations, we can cleverly combine and solve them. This way, we uncover the secret of how fast Robin rises to safety.
Tension in Rope
Tension is a vital concept in physics, especially in problems involving ropes and pulleys. It's the force carried through a rope, caused by forces acting at both ends. In the problem with Robin Hood, the tension in the rope is what helps pull him upwards when the chandelier falls. This scenario provides a classic example of a pulley problem where tension plays a critical role.
To find the tension, we use the same equations as for acceleration, rearranging them to isolate tension. So we have:\[ T = m_Ra + m_Rg \]We also have another approach using the chandelier:\[ T = m_Cg - m_Ca \]Where:
  • \( m_C \) is the mass of the chandelier \(195 \, \text{kg}\).
By solving these equations, we discover the tension in the rope, essential for ensuring Robin's upward journey is safe and speedy.
Forces and Motion
Forces and motion are the heart of physics problems like this one. When analyzing forces, we look at all the elements that are causing movement or trying to resist it. For Robin Hood, there is a delicate dance between gravitational forces and the tension in the rope.
Newton's Laws lay the foundation here, demonstrating how unbalanced forces (in this case, the differencing weights of Robin and the chandelier) create acceleration. When Robin cuts the rope, the chandelier's weight makes it fall, and this same force pulls Robin upward.
This interaction showcases the beauty of physics: how separate objects can impact each other's motion, all governed by the clear and predictable principles of Newton's Second Law.
Physics Problem Solving
Solving physics problems involves a systematic approach and a good grasp of key principles. When tackling a problem like Robin Hood's dilemma, it's crucial to break it down into manageable parts, each guided by fundamental laws.
First, identify the forces at play. Then, apply Newton's Second Law to form equations that relate these forces to movement and tension.
  • Always start by listing known values, like mass and gravitational acceleration.
  • Form clear equations using Newton's laws, ensuring all forces are considered.
  • Solve algebraically to find unknowns such as acceleration and tension.
Finally, double-check your work for accuracy. This not only solidifies the solution but deepens your understanding, equipping you for future challenges in physics problem solving.

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Most popular questions from this chapter

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?

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?

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?

A car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of \(+11 \mathrm{~m} / \mathrm{s}\) in a time of \(28 \mathrm{~s}\). The combined mass of the boat and trailer is 410 kg. The frictional force acting on the trailer can be ignored. What is the tension in the hitch that connects the trailer to the car?

An arrow, starting from rest, leaves the bow with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). If the average force exerted on the arrow by the bow were doubled, all else remaining the same, with what speed would the arrow leave the bow?

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