<LCPO> VECTOR MECH,STAT+DYNAMICS
<LCPO> VECTOR MECH,STAT+DYNAMICS
12th Edition
ISBN: 9781265566296
Author: BEER
Publisher: MCG
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Chapter 16.2, Problem 16.136P

(a)

To determine

Find the couple M.

(a)

Expert Solution
Check Mark

Answer to Problem 16.136P

The couple M is 78.3Nm_.

Explanation of Solution

Given information:

The mass of the rod BC is mBC=6kg.

The mass of the disk is md=10kg.

The mass of the rod CD is mCD=5kg.

The angular velocity is ωAB=36rad/s.

The angular acceleration is α=150rad/s2.

Calculation:

Consider the acceleration due to gravity as g=9.81m/s2.

Calculate the velocity of disk AB (vB) as shown below.

vB=AB¯ωAB

Substitute 200mm for AB¯ and 36rad/s for ωAB.

vB=200mm×1m1,000mm×36rad/s=7.2m/s

Calculate the velocity of rod BC (vC) as shown below.

The velocity of disk AB is equal to the velocity of rod BC.

vB=vC

Substitute 7.2m/s for vB.

vC=7.2m/s

Calculate the angular velocity of rod CD (ωCD) as shown below.

ωCD=vClCD

Substitute 7.2m/s for vC and 250mm for lCD.

ωCD=7.2m/s(250mm×1m1,000mm)=28.8rad/s

Apply the acceleration analysis as shown below.

Calculate the acceleration for disk AB (aB) as shown below.

aB=αAB×rB/AωAB2rB/A

Substitute 150rad/s2 for αAB, 0.2m for rB/A, and 36m/s2 for ωAB.

aB=150×0.2362×0.2=[30+259.2]m/s2

Calculate the acceleration for rod BC (aC) as shown below.

aC=aB+(aC/B)tωBC2rC/B

Substitute [30+259.2]m/s2 for aB, 0.4αBC for (aC/B)t, and 0 for ωBC.

aC=30+259.2+0.4αBC (1)

Calculate the acceleration for rod CD (aC) as shown below.

aC=(aC/D)tωCD2rC/D

Substitute 0.25αCD for (aC/D)t, 0.25m for rC/D, and 28.8rad/s for ωCD.

aC=0.25αCD28.82×0.25=0.25αCD207.36 (2)

Equating the components of Equations (1) and (2) as shown below.

Along x component.

30cos30°259.2cos60°=0.25αCDcos30°207.36cos60°0.2165αCD=51.9αCD=239.72rad/s2

Along y component.

30cos30°259.2sin60°+0.4αBC=0.25αCDsin30°207.36sin60°0.4αBC=0.125αCD+18.914

Substitute 239.72rad/s2 for αCD.

0.4αBC=0.125×239.72+18.914αBC=122.198rad/s2

Calculate the acceleration of the mass centers as shown below.

Calculate the acceleration of mass center for disk AB (aA) as shown below.

aAB=aA=0

Calculate the acceleration of the mass center at P for rod BC (aP) as shown below.

aP=aB+αBC×rP/BωBC2rP/B

Substitute [30+259.2]m/s2 for aB, 0.2m for rP/B, and 0 for ωBC.

aP=30+259.2+0.2αBC+0=30+259.2+0.2αBC

Substitute 122.198rad/s2 for αBC.

aP=30+259.2+0.2×122.198=30+259.2+24.44

Calculate the acceleration of the mass center at Q for rod CD (aQ) as shown below.

aQ=αCD×rQ/DωCD2rQ/D

Substitute 239.72rad/s2 for αCD, 0.125m for rQ/D, and 28.8rad/s for ωCD.

aQ=0.125×239.7228.82×0.125=29.965103.68

Calculate the inertial terms at mass centers as shown below.

The inertia terms at centers are ma.

For disk AB.

mABaA=0

For rod BC.

mBCaP

Substitute 6kg for mBC and 30+259.2+24.44 for aP.

mBCaP=6×(30+259.2+24.44)=180+1,555.2+179.5656

For rod CD.

mCDaQ

Substitute 5kg for mBC and 29.965103.68 for aP.

mCDaQ=5×(29.965103.68)=149.825+518.4

Calculate the mass moment of inertia (I¯) as shown below.

For disk AB.

I¯AB=12mABrAB2

Substitute 10kg for mAB and 0.2m for rAB.

I¯AB=12×10×0.22=0.2kgm2

For rod BC.

I¯BC=112mBClBC2

Substitute 6kg for mBC and 0.4m for lBC.

I¯BC=112×6×0.42=0.08kgm2

For rod CD.

I¯CD=112mCDlCD2

Substitute 5kg for mCD and 0.25m for lCD.

I¯CD=112×5×0.252=0.0260417kgm2

Calculate the effective couples at mass centers as shown below.

The inertia terms at centers are I¯α.

For disk AB.

I¯ABαAB=0

For rod BC.

I¯BCαBC

Substitute 0.08kgm2 for I¯BC and 122.198rad/s2 for αBC.

I¯BCαBC=0.08×122.198=9.776kgm2/s2×1N1kgm/s2=9.776Nm

For rod CD.

I¯CDαCD

Substitute 0.0260417kgm2 for I¯CD and 239.72rad/s2 for αCD.

I¯CDαCD=0.0260417×239.72=6.243kgm2/s2×1N1kgm/s2=6.243Nm

Sketch the effective force and couples on the system as shown in Figure 1.

<LCPO> VECTOR MECH,STAT+DYNAMICS, Chapter 16.2, Problem 16.136P , additional homework tip  1

Sketch the Free Body Diagram of the rod BC as shown in Figure 2.

<LCPO> VECTOR MECH,STAT+DYNAMICS, Chapter 16.2, Problem 16.136P , additional homework tip  2

Refer to Figure 2.

Apply the Equilibrium of moment about B as shown below.

MB=I¯Gα+madlBCCy12lBCmBCg=I¯BCαBC+mad

Substitute 0.4m for lBC, 6kg for mBC, 9.81m/s2 for g, 9.776Nm for I¯BCαBC, 180+1,555.2+179.5656 for ma, and 0.2m for d.

0.4Cy12×0.4×6×9.81=9.776+(180sin30°1,555.2sin60°+179.5656)×0.20.4Cy=193.9Cy=484.75N

Sketch the Free Body Diagram of the rod CD as shown in Figure 3.

<LCPO> VECTOR MECH,STAT+DYNAMICS, Chapter 16.2, Problem 16.136P , additional homework tip  3

Refer to Figure 3.

Apply the Equilibrium of moment about D as shown below.

MD=I¯Gα+madCxlCDcos30°+484.75×0.125mCDg×0.1252=I¯CDαCD+mad0.866CxlCD+60.5940.0625mCDg=I¯CDαCD+mad

Substitute 0.25m for lCD, 5kg for mCD, 9.81m/s2 for g, 6.243Nm for I¯CDαCD, 149.825N for ma, and 0.125m for d.

0.866Cx×0.25+65.5350.0625×5×9.81=6.243+149.825×0.1250.2165Cx=37.49825Cx=173.2N

Sketch the Free Body Diagram of the rod AB and BC as shown in Figure 4.

<LCPO> VECTOR MECH,STAT+DYNAMICS, Chapter 16.2, Problem 16.136P , additional homework tip  4

Refer to Figure 4.

Apply the Equilibrium of moment about A as shown below.

MA=I¯Gα+mad[M484.75×(0.4+0.2sin30°)+173.2×(0.2cos30°)mBCg×(0.2+0.2sin30°)]=I¯BCαBC+madM212.3750.3mBCg=I¯BCαBC+mad

Substitute 6kg for mBC, 9.81m/s2 for g, 9.776Nm for I¯BCαBC, 180+1,555.2+179.5656 for ma, and 0.2m for d.

M212.3750.3×6×9.81=[9.776+180(0.2+0.2sin30°)1,555.2cos30°×0.2+179.5656×(0.2+0.2sin30°)]M230.033=151.722M=78.3NmM=78.3Nm

Therefore, the couple M is 78.3Nm(Counter-clockwise)_.

(b)

To determine

The components of force exerted at C on rod BC.

(b)

Expert Solution
Check Mark

Answer to Problem 16.136P

The components of force exerted at C on rod BC is Cx=173N andCy=485N_.

Explanation of Solution

Given information:

The mass of the rod BC is mBC=6kg.

The mass of the disk is md=10kg.

The mass of the rod CD is mCD=5kg.

The angular velocity is ωAB=36rad/s.

The angular acceleration is α=150rad/s2.

Calculation:

Refer to part (a).

The components of force exerted at C on rod BC along x direction is Cx=173.2N173N

The components of force exerted at C on rod BC along y direction is Cy=484.75N485N

Therefore, the components of force exerted at C on rod BC is Cx=173N andCy=485N_.

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Chapter 16 Solutions

<LCPO> VECTOR MECH,STAT+DYNAMICS

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