Package: Vector Mechanics For Engineers: Dynamics With 1 Semester Connect Access Card
Package: Vector Mechanics For Engineers: Dynamics With 1 Semester Connect Access Card
11th Edition
ISBN: 9781259679407
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell
Publisher: McGraw-Hill Education
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Chapter 16.1, Problem 16.36P
To determine

(a)

The angular acceleration of the gear A.

Expert Solution
Check Mark

Answer to Problem 16.36P

The angular acceleration of the gear A is 6.057rad/s2.

Explanation of Solution

Given Information:

The mass of the gear A is 9kg, the radius of gyration of the gear A is 200mm, the mass of the gear B is 9kg, the radius of gyration of the gear B is 200mm, the mass of the gear C is 3kg, the radius of gyration of the gear C is 75mm, the magnitude of the couple M is 5Nm, the radius of gear A is 250mm, the radius of gear B is 250mm and the radius of gear C is 100mm.

Write the expression for the tangential acceleration of the gear teeth.

αt=αr ...... (I)

Here, the angular acceleration is α and the radius of the gear is r.

Write the expression for the tangential acceleration of the gear A.

αtA=rAαA ...... (II)

Here, the angular acceleration of gear A is αA and the radius of the gear A is rA.

Write the expression for the tangential acceleration of the gear B.

αtB=rBαB ...... (III)

Here, the angular acceleration of gear B is αB and the radius of the gear B is rB.

Write the expression for the tangential acceleration of the gear C.

αtC=rCαC ...... (IV)

Here, the angular acceleration of gear C is αC and the radius of the gear C is rC.

Draw the free body diagram for the gear B.

Package: Vector Mechanics For Engineers: Dynamics With 1 Semester Connect Access Card, Chapter 16.1, Problem 16.36P , additional homework tip  1

Figure-(1)

Draw the kinetic diagram of the gear B.

Package: Vector Mechanics For Engineers: Dynamics With 1 Semester Connect Access Card, Chapter 16.1, Problem 16.36P , additional homework tip  2

Figure-(2)

Write the expression for the moment of inertia gear B.

IB=mBkB2 ...... (V)

Here, the mass of the gear B is mB and the radius of gyration of the gear B is kB.

Write the expression for the external moment at B using the Figure-(1).

MB=rBFBC ...... (VI)

Here, the tangential force exerted by gear B on gear A is FBC.

Write the expression for the effective forces using the Figure-(2).

(MB)eff=IBαB ...... (VII)

Since, the system of external forces is equivalent to system of effective forces hence (MB)eff=(MB).

Substitute IBαB for MB in Equation (VI).

IBαB=rBFBCFBC=IBαBrB ...... (VIII)

Draw the free body diagram for the gear C.

Package: Vector Mechanics For Engineers: Dynamics With 1 Semester Connect Access Card, Chapter 16.1, Problem 16.36P , additional homework tip  3

Figure-(3)

Draw the kinetic diagram for the gear C.

Package: Vector Mechanics For Engineers: Dynamics With 1 Semester Connect Access Card, Chapter 16.1, Problem 16.36P , additional homework tip  4

Figure-(4)

Write the expression for the moment of inertia gear C.

IC=mCkC2 ...... (IX)

Here, the mass of the gear is C

mC and the radius of gyration of the gear C is kC.

Write the expression for the external moment at C using the Figure-(3).

MC=FBCrC+FACrC ...... (X)

Here, the tangential force exerted by gear A on gear C is FAC.

Write the expression for the effective forces using the Figure-(4).

(MC)eff=ICαC ...... (XI)

Since, the system of external forces is equivalent to system of effective forces hence (MC)eff=(MC).

Substitute ICαC for (MC) in Equation (VI).

ICαC=FBCrC+FACrCFAC=ICαC+FBCrCrC ...... (XII)

Draw the free body diagram for the gear A.

Package: Vector Mechanics For Engineers: Dynamics With 1 Semester Connect Access Card, Chapter 16.1, Problem 16.36P , additional homework tip  5

Figure-(5)

Draw the kinetic diagram of the gear A.

Package: Vector Mechanics For Engineers: Dynamics With 1 Semester Connect Access Card, Chapter 16.1, Problem 16.36P , additional homework tip  6

Figure-(6) Write the expression for the moment of inertia gear A.

IA=mAkA2 ...... (XIII)

Here, the mass of the gear is mA and the radius of gyration of the gear A is kA.

Write the expression for the external moment at A using the Figure-(5).

MA=MrAFAC ...... (XIV)

Here, the couple force applied on the gear A is M.

Write the expression for the effective forces using the Figure-(6).

(MA)eff=IAαA ...... (XV)

Since, the system of external forces is equivalent to system of effective forces hence (MA)eff=(MA).

Substitute IAαA for (MA) in Equation (VI).

IAαA=MrAFACFAC=MIAαArA ...... (XVI)

Calculation:

Substitute 250mm for rA in Equation (II).

αtA=(250mm)αA=(250mm(1m103mm))αA=(0.25m)αA ...... (XVII)

Substitute 250mm for rB in Equation (III).

αtB=(250mm)αB=(250mm(1m103mm))αB=(0.25m)αB ...... (XVIII)

Substitute 100mm for rB in Equation (IV).

αtC=(100mm)αC=(100mm(1m103mm))αC=(0.1m)αC ...... (XIX)

The tangential acceleration of the gear A and B is same that is αA=αB.

Substitute (0.25m)αB for αtA in Equation (XVII).

(0.25m)αA=(0.25m)αBαA=αB

The tangential acceleration of the gear A and C is same that is αA=αC.

Substitute (0.1m)αC for αtA in Equation (XVII).

(0.25m)αA=(0.1m)αCαC=2.5αA

Substitute 9kg for mB and 200mm for kB in Equation (V).

IB=(9kg)(200mm)2=(9kg)(200mm(1m103mm))2=(9kg)(0.2m)2=0.36kgm2

Substitute αA for αB, 0.36kgm2 for IB, 250mm for rB in Equation (VIII).

FBC=(0.36kgm2)αA(250mm)=(0.36kgm2)αA(250mm(1m103mm))=(1.44kgm)αA

Substitute 9kg for mA and 200mm for kA in Equation (XIII).

IA=(9kg)(200mm)2=(9kg)(200mm(1m103mm))2=(9kg)(0.2m)2=0.36kgm2

Substitute 5Nm for M, 0.36kgm2 for IA, 250mm for rA in Equation (XVI).

FAC=5Nm(0.36kgm2)αA(250mm)=5Nm(0.36kgm2)αA(250mm(1m103mm))=5Nm(1.44kgm)αA

Substitute 3kg for mC and 75mm for kC in Equation (IX).

IC=(3kg)(75mm)2=(3kg)(75mm(1m103mm))2=(3kg)(0.075m)2=0.01687kgm2

Substitute IBαArB for FBC, 2.5αA for αC, 100mm for rC and 250mm for rB in Equation (XII).

FAC=IC(2.5αA)+(IBαA(250mm))(100mm)(100mm)=IC(2.5αA)+(IBαA(250mm(1m103mm)))(100mm(1m103mm))(100mm(1m103mm))=10.1m(2.5IC+0.4IB)αA ....... (XX)

Substitute 10.1m(2.5IC+0.4IB)αA for FAC and 250mm for rA in Equation (XVI).

10.1m(2.5IC+0.4IB)αA=MIAαA(250mm)10.1m(2.5IC+0.4IB)αA=MIAαA(250mm(1m103mm))M=IAαA+(6.25IC+IB)αAαA=M(IA+6.25IC+IB) ...... (XXI)

Substitute 0.01687kgm2 for IC, 5Nm for M, 0.36kgm2 for IA and 0.36kgm2 for IB in Equation (XXI).

αA=5Nm(0.36kgm2+6.25(0.01687kgm2)+0.36kgm2)=5Nm(1kgm/s21N)0.8254kgm2=6.057rad/s2

Conclusion:

The angular acceleration of the gear A is 6.057rad/s2.

To determine

(b)

The tangential force exerted by gear C on gear A.

Expert Solution
Check Mark

Answer to Problem 16.36P

The tangential force exerted by gear C on gear A is 11.27N.

Explanation of Solution

Write the expression for the tangential force exerted by gear C on gear A.

FAC=10.1m(2.5IC+0.4IB)αA ...... (XXII).

Calculation:

Substitute 6.057rad/s2 for αA, 0.36kgm2 for IB, 0.01687kgm2 for IC in Equation (XXII).

FAC=10.1m(2.5(0.01687kgm2)+0.4(0.36kgm2))(6.057rad/s2)=((0.4217kgm)+(1.44kgm))(6.057rad/s2)=1.8617kgm(1N1kgm/s2)(6.057rad/s2)=11.27N

Conclusion:

The tangential force exerted by gear C on gear A is 11.27N.

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Chapter 16 Solutions

Package: Vector Mechanics For Engineers: Dynamics With 1 Semester Connect Access Card

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