Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 19

A 150 pound boy stands at the edge of the diving board of Problem 2.30, preparing to execute a dive. During this time, he shifts his weight in a leaping motion, moving up and down at the rate of \(1 \mathrm{cps}\). If, at the apogee of each bob his feet just touch the board so that they are nearly losing contact, determine the steady state motion of the boy.

Short Answer

Expert verified
Answer: The steady-state motion of the boy during each cycle can be represented by the sinusoidal function: x(t) = 1.63 * sin(2 * pi * 1 * t), which repeats every second.

Step by step solution

01

Identify the given information

We know the following: - The boy weighs 150 pounds - He moves up and down at a rate of 1 cps
02

Set up sinusoidal motion equation for displacement

For bobbing motion, the displacement of the boy with respect to time can be described by a sinusoidal function. Let x(t) represent the displacement of the boy at time t. The equation for sinusoidal motion is given by: x(t) = A * sin(2 * pi * f * t) Where: - A is the amplitude (maximum displacement from the equilibrium position) - f is the frequency (number of cycles per second) - t is time. Since the boy moves up and down at a rate of 1 cps, f = 1.
03

Determine the condition for losing contact

At the apogee of each bob, the boy's feet just touch the board, meaning the normal force exerted by the board on his feet becomes zero. By Newton's second law, when the normal force is zero, the acceleration experienced by the boy's body is equal to the acceleration due to gravity (g = 9.81 m/s^2 or 32.2 ft/s^2). Now we need to find the accelerations in the sinusoidal motion, which is the second derivative of the displacement function x(t). a(t) = x''(t) First derivative of x(t): x'(t) = A * (2 * pi * f) * cos(2 * pi * f * t) Second derivative of x(t): a(t) = -A * (2 * pi * f)^2 * sin(2 * pi * f * t)
04

Find the amplitude of the displacement function

Since the acceleration at apogee is equal to the gravitational acceleration: -A * (2 * pi * f)^2 = -g Solving for A: A = g / (2 * pi * f)^2 Substitute the values of g and f: A = 32.2 / (2 * pi * 1)^2 = 1.63 ft
05

Determine the steady-state motion

Now we have the amplitude, which corresponds to the largest displacement from the equilibrium position during the bobbing motion. Therefore, the steady-state motion of the boy during each cycle can be represented by the sinusoidal function: x(t) = 1.63 * sin(2 * pi * 1 * t) The motion repeats every second, as indicated by the frequency of 1 cps.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An aircraft mock-up for ground testing is modeled as shown, where a brace represented as a torsional spring of stiffness \(k_{T}\) has been attached at the pivot support to restrain excessive yawing. It is assumed that during a test, both thrusters will exert a constant force of magnitude \(F_{0}\). However, during a simulation, one of the thrusters deviates from the prescribed force by adding a perturbation in the form of a harmonic thrust of small amplitude. Thus, the force applied by this thruster takes the form \(F(t)=F_{0}(1+\varepsilon \sin \Omega t)\), as indicated, where \(\varepsilon \ll 1\). Determine the steady state yawing motion of the mock-up. What is the amplitude of the response at resonance?

A vehicle of mass \(m\), with suspension of stiffness \(k\) and negligible damping, is traveling at constant speed \(v_{0}\) when it encounters the series of equally spaced semi-circular speed bumps shown. Assuming that no slip occurs between the wheel and the road, determine (a) the motion of the engine block while it is traveling over this part of the roadway and \((b)\) the force transmitted to the block.

Several bolts on the propeller of a fanboat detach, resulting in an offset moment of \(5 \mathrm{lb}-\mathrm{ft}\). Determine the amplitude of bobbing of the boat when the fan rotates at \(200 \mathrm{rpm}\), if the total weight of the boat and passengers is \(1000 \mathrm{lb}\) and the wet area projection is approximately \(30 \mathrm{sq} \mathrm{ft}\). What is the amplitude at $1000 \mathrm{rpm}$ ?

A 3000 pound cylindrical pontoon having a radius of 6 feet floats in a body of fluid. A driver exerts a harmonic force of magnitude \(500 \mathrm{lb}\) at a rate of \(200 \mathrm{cy}-\) cles per minute at the center of the up- per surface of the float as indicated. \((a)\) Determine the density of the fluid if the pontoon is observed to bob with an amplitude of 1 foot. (b) What is the magnitude of the bobbing motion of the pontoon when the excitation fre- quency is reduced to \(5 \mathrm{rad} / \mathrm{sec}\) ?

The damping factor for a shaft-turbine system is measured to be \(0.15\). When the turbine rotates at a rate equal to \(120 \%\) of the undamped natural frequency of the shaft, the system is observed to whirl with an amplitude equal to the radius of the shaft. What will be the amplitude of whirling when the rotation rate of the turbine is reduced to \(80 \%\) of the undamped natural frequency of the shaft if the radius of the shaft is 1 inch?
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Quantum Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Nuclear Physics

Read Explanation

Linear Momentum

Read Explanation

Electromagnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free