/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 People who do chin-ups raise the... [FREE SOLUTION] | 91Ó°ÊÓ

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People who do chin-ups raise their chin just over a bar (the chinning bar), supporting themselves only by their arms. Typically, the body below the arms is raised by about \(30 \mathrm{~cm}\) in a time of \(1.0 \mathrm{~s},\) starting from rest. Assume that the entire body of a \(680 \mathrm{~N}\) person who is chinning is raised this distance and that half the \(1.0 \mathrm{~s}\) is spent accelerating upward and the other half accelerating downward, uniformly in both cases. Make a free-body diagram of the person's body, and then apply it to find the force his arms must exert on him during the accelerating part of the chin-up.

Short Answer

Expert verified
The force exerted by the arms is approximately 958 N.

Step by step solution

01

Understand the Problem

When doing a chin-up, the person lifts their body by using the force from their arms to overcome gravitational force, causing the body to accelerate upwards.
02

Draw a Free-Body Diagram

Draw the free-body diagram of the person doing the chin-up. Identify two forces: the gravitational force downward ( F_g = 680 ext{ N} ) and the upward force exerted by the arms ( F_a ).
03

Analyze Motion and Forces

Since the person accelerates upwards and decelerates back to rest in equal time intervals, we can analyze the upward acceleration period separately. The gravitational force remains constant at 680 N, while the arm force needs to be enough to not only support this weight but also provide additional force for upward acceleration.
04

Calculate Acceleration

Using the kinematic equations, and knowing the distance (s = 0.30 m) and time (t = 0.5 s for acceleration), calculate the acceleration (ag) using the formula: \[ s = \frac{1}{2} a (\frac{t}{2})^2 \] Solving this results in: a = 4\, ext{m/s}^2\.
05

Apply Newton's Second Law

Using Newton's second law, \[ F = ma \] we can set up the equation: \[ F_a - F_g = m \cdot a \] where m = F_g / g = 680 / 9.8 kg. By substituting as value:\[ F_a = (\frac{680}{9.8}) \times 4 + 680 \]
06

Solve for the Arm Force

Plug in the values to find the force exerted by the arms:\[ F_a = (69.4 \times 4) + 680 = 957.6\, ext{N} \]

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Key Concepts

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

Newton's Second Law
Newton's Second Law is central to understanding motion in physics. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This fundamental law can be expressed through the equation \( F = ma \), where \( F \) is the net force applied, \( m \) is the mass of the object, and \( a \) is its acceleration.
The law allows us to determine the relationship between forces and motion. In the context of the chin-up problem, Newton's Second Law helps us determine the force required by the arms to lift the body. As the body of the person accelerates upward during the chin-up, applying \( F = ma \) enables us to calculate how much additional force, beyond the gravitational force, is needed.
Free-Body Diagram
A free-body diagram is a useful tool in physics for visualizing the forces acting on an object. It illustrates all of the external forces to better analyze a situation. This tool helps clearly identify and separate forces to simplify complex interactions.
For the chin-up problem, the free-body diagram includes:
  • Gravitational force \(F_g\), which is 680 N, pulling the individual downward.
  • Force exerted by the arms \(F_a\), which acts upward opposing the gravitational pull.
This simplification is crucial because it allows us to focus on the forces that directly impact motion. By accurately depicting the forces involved, the diagram aids in effectively applying physics laws to solve for unknowns, such as the necessary force the arms must exert.
Kinematics
Kinematics is a branch of physics that deals with motion without considering the forces that cause it. It describes how objects move through equations that relate distances, speeds, and accelerations over time.
In the chin-up scenario, we rely on kinematic equations to determine the acceleration of the person as they lift their body. Given the information:
  • Displacement is 0.30 m during the upward motion.
  • Time spent accelerating is 0.5 s.
This allows using the equation \( s = \frac{1}{2} a t^2 \) to solve for the acceleration \( a \), resulting in an acceleration value of 4 m/s². This derived acceleration is vital for the subsequent force calculations using Newton's Second Law.
Forces in Physics
Understanding forces is key to analyzing physical movements. Forces are pushes or pulls that can cause objects to start moving, stop, or change their motion. In physics, forces are vectors with both magnitude and direction.
In the chin-up example, two primary forces are at play:
  • Gravitational force, which pulls the chinning individual's body downwards at 680 N.
  • The exerted force by the arms required to lift the body upward.
These forces must be examined in tandem to understand how they impact motion. The arm force is particularly important, as it must not only counteract gravity but also provide sufficient energy for upward movement. Solving for this additional force is crucial to completing the chin-up, highlighting how forces interact to influence physical activity.

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Most popular questions from this chapter

A large fish hangs from a spring balance supported from the roof of an elevator. (a) If the elevator has an upward acceleration of \(2.45 \mathrm{~m} / \mathrm{s}^{2}\) and the balance reads \(60.0 \mathrm{~N},\) what is the true weight of the fish? (b) Under what circumstances will the balance read \(35.0 \mathrm{~N} ?\) (c) What will the balance read if the elevator cable breaks?

An \(80 \mathrm{~N}\) box initially at rest is pulled by a horizontal rope on a horizontal table. The coefficients of kinetic and static friction between the box and the table are \(\frac{1}{4}\) and \(\frac{1}{2},\) respectively. What is the friction (b) \(25 \mathrm{~N}\), force on this box if the tension in the rope is (a) \(0 \mathrm{~N},]\) (c) \(39 \mathrm{~N},(\mathrm{~d}) 41 \mathrm{~N},(\mathrm{e}) 150 \mathrm{~N} ?\)

Two \(25.0 \mathrm{~N}\) weights are suspended at opposite ends of a rope that passes over a light, frictionless pulley. The pulley is attached to a chain that is fastened to the ceiling. (See Figure 5.41.) Start solving this problem by making a free-body diagram of each weight. (a) What is the tension in the rope? (b) What is the tension in the chain?

An \(85 \mathrm{~N}\) box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of \(0.90 \mathrm{~m} / \mathrm{s}\) each second. The push force has a horizontal component of \(20 \mathrm{~N}\) and a vertical component of \(25 \mathrm{~N}\) downward. Calculate the coefficient of kinetic friction between the box and floor.

In emergencies involving major blood loss, the doctor will order the patient placed in the Trendelberg position, which is to raise the foot of the bed to get maximum blood flow to the brain. If the coefficient of static friction between the typical patient and the bedsheets is \(1.2,\) what is the maximum angle at which the bed can be tilted with respect to the floor before the patient begins to slide?

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