/*! 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 97 The spring-held follower \(A B\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The spring-held follower \(A B\) has a mass of \(0.5 \mathrm{kg}\) and moves back and forth as its end rolls on the contoured surface of the cam, where \(r=0.15 \mathrm{m}\) and \(z=(0.02 \cos 2 \theta) \mathrm{m}\). If the cam is rotating at a constant rate of 30 rad/s, determine the maximum and minimum force components \(F_{z}\) the follower exerts on the cam if the spring is uncompressed when \(\theta=90^{\circ}\)

Short Answer

Expert verified
The maximum and minimum force components \(F_{z}\) the follower exerts on the cam are \(1.8+0.02k\) and \(-1.8-0.02k\) respectively assuming k is the spring constant in \(N/m\).

Step by step solution

01

Calculate the vertical acceleration

Determine the vertical acceleration of the cam using the relationship \(z=0.02 \cos 2 \theta \), This will give \( z = 0.02 \cos (2 * 30t)\). Now, differentiate \(z\) twice with respect to \(t\) to get \( \ddot{z} = 4*0.02*30^2 \sin(60t) = 3.6 \sin(60t) \, m/s^2\).
02

Determine the spring force

Substitute \(\theta = 90^{\circ}\) to find that \(z = 0.02 \cos(2*90^\circ) = 0\) when the spring is uncompressed. During any other time, the spring force \( F_{s} = k \Delta z \) where \(k\) is the spring constant and \( \Delta z = z - z_{uncompressed}\). Using \( z = 0.02 \cos(2*30t)\), the spring force becomes \( F_{s} = k * (0.02 \cos(60t))\). This implies that \( F_{s} \) is always >= 0.
03

Calculate the maximum and minimum \(F_{z}\)

The force components \(F_{z}\) can be determined using the follower's vertical forces \(m \cdot \ddot{z} + F_{s}\). That gives \( F_{z} = 0.5 \ddot{z} + k * z_{uncompressed}\). Because \(\ddot{z}\) and \(F_{s}\) are both sinusoidal functions and out of phase by 90 degrees, to obtain the maximum force, when \( F_{z} = 0.5*3.6 + k * 0.02\), and the minimum force, when \( F_{z} = 0.5*-3.6 + k * -0.02\).

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.

Spring Force
In the context of dynamics and mechanical engineering, spring force is an essential concept. It refers to the force exerted by a compressed or stretched spring upon any object that is attached to it. The spring force can be described by Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium, or uncompressed length. This can be mathematically expressed as - \( F_s = k \Delta z \) where:- \( F_s \) is the spring force- \( k \) is the spring constant - \( \Delta z \) is the displacement of the spring from its uncompressed length.
A positive displacement indicates stretching, while a negative displacement indicates compression. It is worth noting that the spring force is always directed opposite to the direction of displacement, acting to restore the spring to its equilibrium position.
Cam-Follower System
The cam-follower system is a crucial mechanism in converting rotational motion into linear motion. It is widely used in engines and machinery for timely opening and closing of valves or any other linear motion requirements.
A cam is a rotating or sliding piece in a mechanical linkage system, used especially in transforming rotary motion into linear motion. The follower rests on the cam surface and traces its contour to produce required motion.
In our exercise, the follower experienced oscillatory motion as it rolls over the contoured surface of the cam. This movement is influenced by the cam rotating at a constant rate, often resulting in a periodic motion that can be mathematically modeled through trigonometric functions.
Vertical Acceleration
To analyze the dynamics of the cam-follower system, determining the vertical acceleration is crucial. Vertical acceleration is the rate of change of the vertical velocity of an object.
In our exercise, it is determined by the second derivative of the position function \( z \), expressed as - \( z = 0.02 \cos(2\theta) \).
When differentiated twice with respect to time \( t \), it leads to the vertical acceleration formula:- \( \ddot{z} = 3.6 \sin(60t) \).
This describes how quickly the vertical speed is changing with time due to the oscillatory motion described by the trigonometric function. Factors like the cam's rotational speed and the profile will influence this acceleration.
Rotational Motion
Rotational motion is the motion of a body about a fixed axis. For the cam in our scenario, rotational motion involves the constant rotational speed of 30 rad/s.
This type of motion is characterized by angular displacement, angular velocity, and angular acceleration. For a cam system, understanding the rotation allows predicting how it will interact with a follower. The rotational motion of the cam leads to linear vertical motion of the follower as it rolls over the surface.
In dynamical systems like the cam-follower, the constant rotational speed must be precisely controlled to ensure that the desired motion is achieved effectively and efficiently. This rotational speed influences the entire system's dynamics, affecting the forces, acceleration, and ultimately the timing and behavior of the mechanical system.

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

Rod \(O A\) rotates counterclockwise with a constant angular velocity of \(\dot{\theta}=5 \mathrm{rad} / \mathrm{s} .\) The double collar \(B\) is pinconnected together such that one collar slides over the rotating rod and the other slides over the horizontal curved rod, of which the shape is described by the equation \(r=1.5(2-\cos \theta)\) ft. If both collars weigh 0.75 lb, determine the normal force which the curved rod exerts on one collar at the instant \(\theta=120^{\circ} .\) Neglect friction.

A freight elevator, including its load, has a mass of 1 Mg. It is prevented from rotating due to the track and wheels mounted along its sides. If the motor \(M\) develops a constant tension \(T=4 \mathrm{kN}\) in its attached cable, determine the velocity of the elevator when it has moved upward \(6 \mathrm{m}\) starting from rest. Neglect the mass of the pulleys and cables.

Cartons having a mass of 5 kg are required to move along the assembly line at a constant speed of \(8 \mathrm{m} / \mathrm{s}\) Determine the smallest radius of curvature, \(\rho,\) for the conveyor so the cartons do not slip. The coefficients of static and kinetic friction between a carton and the conveyor are \(\mu_{s}=0.7\) and \(\mu_{k}=0.5,\) respectively.

The 150-lb man lies against the cushion for which the coefficient of static friction is \(\mu_{s}=0.5 .\) Determine the resultant normal and frictional forces the cushion exerts on him if, it due to rotation about the \(z\) axis, he has a constant speed \(v=20 \mathrm{ft} / \mathrm{s} .\) Neglect the size of the man. Take \(\theta=60^{\circ}\).

The tractor is used to lift the 150 -kg load \(B\) with the 24 -m-long rope, boom, and pulley system. If the tractor travels to the right with an acceleration of \(3 \mathrm{m} / \mathrm{s}^{2}\) and has a velocity of \(4 \mathrm{m} / \mathrm{s}\) at the instant \(s_{A}=5 \mathrm{m},\) determine the tension in the rope at this instant. When \(s_{A}=0, s_{B}=0\)

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.