/*! 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 90 A child of mass \(m\) sits on to... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A child of mass \(m\) sits on top of a rectangular slab of mass \(M=35 \mathrm{kg},\) which in turn rests on the frictionless horizontal floor at a pizza shop. The slab is attached to a horizontal spring with spring constant \(k=430 \mathrm{N} / \mathrm{m}\) (the other end is attached to an immovable wall, Fig. 45. The coefficient of static friction between the child and the top of the slab is \(\mu=0.40 .\) The shop owner's intention is that, when displaced from the equilibrium position and released, the slab and child (with no slippage between the two) execute SHM with amplitude \(A=0.50 \mathrm{m} .\) Should there be a weight restriction for this ride? If so, what is it?

Short Answer

Expert verified
The child's mass must not exceed 19.84 kg to avoid slipping.

Step by step solution

01

Understanding SHM and Friction

Simple Harmonic Motion (SHM) is described by a restoring force proportional to the displacement. The maximum force due to spring is \( F_{spring} = kA \). To maintain SHM without slippage, the static friction \( F_{friction} = \mu (m+M)g \) must be able to counteract the spring force when the child is at the maximum amplitude.
02

Calculate Maximum Spring Force

The maximum force exerted by the spring happens at maximum displacement, calculated as \( F_{spring} = kA = 430 \times 0.5 = 215 \ \mathrm{N} \). This is the force required from static friction to prevent slippage.
03

Calculate Maximum Static Friction Force

The maximum static friction is calculated using \( F_{friction} = \mu (m+M)g \). Here, \( g = 9.81 \ \mathrm{m/s^2} \), and \( M = 35 \ \mathrm{kg} \). Thus, \( F_{friction} = 0.40 (m + 35) \times 9.81 \).
04

Set Static Friction Equal to Spring Force

To prevent slippage, \( F_{friction} \) must equal \( F_{spring} \). Equating them gives: \[ 0.40 (m + 35) \times 9.81 = 215 \]
05

Solve for Maximum Allowable Mass of Child

Rearrange the equation from Step 4 to solve for \( m \):\[ m + 35 = \frac{215}{0.40 \times 9.81} \] Calculate the maximum value of \( m \):\( m + 35 \approx 54.84 \)Thus, \( m \approx 19.84 \ \mathrm{kg} \).
06

Conclusion on Weight Restriction

The child must have a mass \( m \leq 19.84 \ \mathrm{kg} \) to prevent slippage and ensure safe SHM on the slab.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Static Friction
Static friction is the force keeping two surfaces at rest relative to each other. In simpler terms, it prevents things from slipping. When you gently push on a book lying on the table, it doesn't move immediately. This is because static friction keeps it in place.
When considering simple harmonic motion (SHM), where a child and slab move back and forth together on a spring, static friction is crucial. If static friction is strong enough, it will keep the child on the slab even when the system is in motion.
The maximum static friction force can be calculated by the formula \( F_{friction} = \mu (m+M)g \), where \( \mu \) is the coefficient of static friction, \( m \) and \( M \) are the masses, and \( g \) is the acceleration due to gravity.
  • The static friction force ensures that the child does not slip off the slab.
  • A higher coefficient of static friction allows for greater forces before motion begins.
Spring Constant
The spring constant, often denoted as \( k \), is a measure of a spring's stiffness. It tells us how much force is needed to stretch or compress the spring by a certain distance.
In our problem, the spring constant is given as \( 430 \, \mathrm{N/m} \). This means that 430 Newtons of force will stretch the spring by one meter.
  • The larger the spring constant, the stiffer the spring.
  • In simple harmonic motion, the spring constant helps determine the system's frequency and energy.
The force exerted by the spring when it is displaced is given by \( F_{spring} = kA \), where \( A \) is the displacement from the equilibrium position. Thus, the maximum force in our example at a displacement of \( A = 0.5 \, \mathrm{m} \) is \( 215 \, \mathrm{N} \).
Understanding the spring constant helps us predict how much force will be exerted as the system oscillates.
Weight Restriction
Weight restriction refers to the maximum weight that can be safely handled by the system—in this case, the child riding on the slab. To maintain smooth, slip-free motion, we must ensure the weight doesn't exceed the maximum static friction available.
By setting the static friction force equal to the spring force, we calculate the maximum permissible weight of the child.
  • If static friction cannot match the spring force, the child will slip, and the system becomes unsafe.
  • The weight restriction ensures the operation remains within safety parameters.
For this exercise, by equating forces, we determined the maximum mass for the child to be approximately \( 19.84 \, \mathrm{kg} \). This means any child weighing more than this could cause slipping, preventing the system from functioning as intended.
Displacement
Displacement in the context of simple harmonic motion is the distance from the equilibrium position - the point where the spring is neither stretched nor compressed. Displacement is crucial as it determines how much potential energy is stored in the system.
In our scenario, the displacement amplitude is \( 0.50 \, \mathrm{m} \). This is the maximum distance the slab (and the child) move from their resting position.
  • The amplitude of displacement contributes to the calculation of maximum spring force \( F_{spring} = kA \).
  • Greater displacement results in higher energy and force being exerted.
Understanding displacement in SHM helps predict the motion characteristics such as speed and acceleration at various points during the oscillation cycle. This foundational concept guides the safe setup and operation of systems like the child and slab scenario.

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

A physical pendulum consists of an \(85-\mathrm{cm}\) -long, 240 -g-mass, uniform wooden rod hung from a nail near one end (Fig. 14-38). The motion is damped because of friction in The motion is damped because of friction in the pivot; the damping force is approximately proportional to \(d \theta / d t\). The rod is set in oscillation by displacing it \(15^{\circ}\) from its equilibrium position and releasing it. After \(8.0 \mathrm{~s}\) the amplitude of the oscillation has been reduced to \(5.5^{\circ} .\) If the angular displacement can be written as \(\theta=A e^{-\gamma t} \cos \omega^{\prime} t,\) find (a) \(\gamma,\) (b) the approximate period of the motion, and \((c)\) how long it takes for the amplitude to be reduced to \(\frac{1}{2}\) of its original value.

(a) At what displacement of a SHO is the energy half kinetic and half potential? (b) What fraction of the total energy of a SHO is kinetic and what fraction potential when the displacement is one third the amplitude?

Imagine that a 10 -cm-diameter circular hole was drilled all the way through the center of the Earth (Fig, \(44 ) .\) At one end of the hole, you drop an apple into the hole. Show that, if you assume that the Earth has a constant density, the subsquent motion of the apple is simple harmonic. How long will the apple take to return? Assume that we can ignore all frictional effects.

Carbon dioxide is a linear molecule. The carbon-oxygen bonds in this molecule act very much like springs. Figure \(14-43\) shows one possible way the oxygen atoms in this molecule can oscillate: the oxygen atoms oscillate symmetrically in and out, while the central carbon atom remains at rest. Hence each oxygen atom acts like a simple harmonic oscillator with a mass equal to the mass of an oxygen atom. It is observed that this oscillation occurs with a frequency of \(f=2.83 \times 10^{13} \mathrm{~Hz} .\) What is the spring constant of the \(\mathrm{C}-\mathrm{O}\) bond?

(1I) The amplitude of a driven harmonic oscillator reaches a value of 23.7\(F_{0} / k\) at a resonant frequency of 382 Hz. What is the \(Q\) value of this system?

See all solutions

Recommended explanations on Physics Textbooks

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.