Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics
Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics
11th Edition
ISBN: 9781259639272
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 19.4, Problem 19.110P

(a)

To determine

Find the amplitude of the motion (xm) of the bob.

(a)

Expert Solution
Check Mark

Answer to Problem 19.110P

The amplitude of the motion (xm) of the bob is 1.034in._.

Explanation of Solution

Given information:

The weight of the bob (WB) is 2.75lb.

The weight of the collar (WC) is 3lb.

The length of the simple pendulum (l) is 24in. or 2ft.

The amplitude of the collar (xC) is δmsinωft.

The magnitude of static deflection (δm) 0.4in. or 0.03333ft.

The frequency (ff) is 0.5Hz.

The acceleration due to gravity (g) is 32.2ft/s2.

Calculation:

Calculate the mass of the bob (mB) using the formula:

WB=mgmB=WBg

Substitute 2.75lb for WB and 32.2ft/s2 for g.

m=2.75lb32.2ft/s2=0.0854lbs2/ft

Calculate the mass of the collar (mC) using the formula:

mC=WCg

Substitute 2.75lb for WC and 32.2ft/s2 for g.

mC=2.75lb32.2ft/s2=0.0854lbs2/ft

Calculate the frequency of the period (ωf) force using the formula:

ff=ωf2π

Substitute 0.5Hz for ff.

0.5Hz=ωf2πωf=2π(0.5)ωf=πrad/s

Show the system at before and after moving the collar horizontally as in Figure 1.

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 19.4, Problem 19.110P , additional homework tip  1

Refer Figure (1), θ is the angular displacement of the pendulum and mg is the gravitational force due to the mass of the bob.

Before giving horizontal movement, the force mg is equal to the tension in the pendulum.

The expression for the force balance equation for initial condition as follows:

T=mg

Calculate the value of sinθ by referring Figure (1).

sinθ=xxcl

The expression for the force balance equation in x-direction in the displaced condition as follows:

Fx=max

Here, Fx is the sum of forces in x-direction and ax is the acceleration in the x-direction.

The only force in the x-direction is the tension component Tsinθ and it is opposite in direction to that of the acceleration. The acceleration in the x-direction due to the linear displacement is x¨.

Substitute Tsinθ for Fx and x¨ for ax.

Tsinθ=mx¨ (1)

Calculate the differential equation of motion using the formula:

Substitute mg for T and xxcl for sinθ in equation (1).

Tsinθ=mx¨mg(xxcl)=mx¨g(xxcl)=x¨gxl+gxcl=x¨x¨+glx=glxc

Substitute δmsinωft for xc.

x¨+glx=glδmsinωft (2)

Here, δm is the amplitude of deflection and t is the time.

Compare the differential equation (2) with the general differential equation of motion for forced vibration (x¨+ωn2x=ωn2δmsinωft).

Calculate the natural circular frequency of vibration (ωn) as follows:

ωn2=gl

Substitute 32.2ft/s2 for g and 2ft for l.

ωn2=32.2ft/s22ft=16.1rad/s2

Calculate the amplitude of the forced vibration (xm) using the formula:

xm=δm1ωf2ωn2

Substitute 0.03333ft for δm, π for ωf, and 16.1rad/s2 for ωn2.

xm=0.03333ft1π216.1rad/s2=0.033330.38698=0.08613ft×12in.1ft=1.034in.

Therefore, the amplitude of the motion (xm) of the bob is 1.034in._.

(b)

To determine

Find the force that must be applied to collar C (F) to maintain the motion.

(b)

Expert Solution
Check Mark

Answer to Problem 19.110P

The force that must be applied to collar C (F) to maintain the motion is 0.103sinπtlb_.

Explanation of Solution

Given information:

The weight of the bob (WB) is 2.75lb.

The weight of the collar (WC) is 3lb.

The length of the simple pendulum (l) is 24in. or 2ft.

The amplitude of the collar (xC) is δmsinωft.

The magnitude of static deflection (δm) 0.4in. or 0.03333ft.

The frequency (ff) is 0.5Hz.

The acceleration due to gravity (g) is 32.2ft/s2.

Calculation:

Consider the collar as a free body and show the free body diagram equation as in Figure (2).

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics, Chapter 19.4, Problem 19.110P , additional homework tip  2

Refer Figure (2), F is the force applied to the collar to maintain the motion is F and x¨c is the acceleration of the collar.

The expression for the force balance equation from Figure 2 as follows:

F+Tsinθ=mcx¨c

Substitute mg for T and xxcl for sinθ.

F+(mg)(xxcl)=mcx¨cF=mgl(xxc)+mcx¨c

Substitute ωn2 for gl.

F=mgl(xxc)+mcx¨c=m(ωn2)(xxc)+mcx¨c=mωn2(xxc)+mcx¨c (5)

The expression for the relation for (xc) as follows:

xc=δmsinωft

Differentiate the relation for xc with respect to t.

xc=δmsinωftx˙c=δm(ωfcosωft)x¨c=δm((ωf)2(sinωft))x¨c=δmωf2sinωft

Substitute xmsinωft for x, δmsinωft for xc, and δmωf2sinωft for x¨c in equation (5).

F=mωn2(xxc)+mcx¨c=mωn2((xmsinωft)(δmsinωft))+mc(δmωf2sinωft)=mωn2xmsinωft+mωn2δmsinωftmcδmωf2sinωft=(mωn2xm+mωn2δmmcδmωf2)sinωft (6)

Substitute 0.0854lbs2/ft for m, 16.1rad/s2 for ωn2, 0.08613 ft for xm, 0.03333 ft for δm, 0.0854lbs2/ft for mc, and πrad/s for ωf in equation (6).

F=(mωn2xm+mωn2δmmcδmωf2)sinωft={(0.0854lbs2/ft)(16.1rad/s2)(0.08613ft)+(0.0854lbs2/ft)(16.1rad/s2)(0.03333ft)(0.0854lbs2/ft)(0.03333ft)(πrad/s)2}sin(πrad/s)t=(0.1184+0.04580.03)sinπt=0.103sinπtlb

Therefore, the force that must be applied to collar C (F) to maintain the motion is 0.103sinπtlb_.

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Chapter 19 Solutions

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

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