Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977305
Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.
Publisher: Mcgraw-hill Education,
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Textbook Question
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Chapter 15.4, Problem 15.128P

The elliptical exercise machine has fixed axes of rotation at points A and E. Knowing that at the instant shown the flywheel AB has a constant angular velocity of 6 rad/s clockwise, determine (a) the angular acceleration of bar DEF, (b) the acceleration of point F.

Fig. P15.127 and P15.128

Chapter 15.4, Problem 15.128P, The elliptical exercise machine has fixed axes of rotation at points A and E. Knowing that at the

Expert Solution
Check Mark
To determine

(a)

Angular acceleration of bar DEF.

Answer to Problem 15.128P

The angular acceleration of bar DEF is 1.466rad/s2 counter clockwise.

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 15.4, Problem 15.128P , additional homework tip  1

Constant angular velocity of AB is 6rad/s clockwise.

Point A and E are fixed.

The absolute value of point A:

vA=vB+vA/B

The relative velocity of A with respect to B is defined as:

vA/B=ωk×rA/B

rA/B - Position vector of A with respect to B

ω - Angular velocity

The absolute acceleration of point B is defined as:

aB=aA+aB/A

aB=aA+(aB/A)t+(aB/A)n

The tangential acceleration is defined as:

(aB/A)t=αk×rB/A

The normal acceleration is defined as:

(aB/A)n=rB/AωAB2

In above equations

α - Angular acceleration

ω - Angular velocity

r - Distance from A to B

Calculation:

Position vector of B relative to A:

rB/A=0.2j

Position vector of D relative to B:

rD/B=1.2i0.2j

Position vector of D relative to E:

rD/E=0.12i0.8j

Absolute velocity of point B:

vB=vA+ωABk×rB/A(1)

Absolute velocity of point D:

vD=vB+ωBDk×rD/B(2)vD=vE+ωDEFk×rD/E(3)

We know that:

vA=0vE=0

Substitute equation 1 into 2 and equate with 3:

ωABk×rB/A+ωBDk×rD/B=ωDEFk×rD/E

Substitute:

6k×0.2j+ωBDk×(1.2i0.2j)=ωDEFk×(0.12i0.8j)1.2i+1.2ωBDj+0.2ωBDi=0.12ωDEFj+0.8ωDEFi

Equate components:

i:1.2+0.2ωBD=0.8ωDEFj:1.2ωBD=0.12ωDEF

Therefore:

ωBD=0.1463rad/sωDEF=1.463rad/s

Similarly:

aB=aA+αABk×rB/AωAB2rB/A(4)aD=aB+αBDk×rD/BωBD2rD/B(5)aD=aE+αDEFk×rD/EωDEF2rD/E(6)

But we know that:

aA=0aE=0αAB=0

Therefore, substitute equation 4 into 5 and equate with 6:

ωAB2rB/A+αBDk×rD/BωBD2rD/B=αDEFk×rD/EωDEF2rD/E

Substitute:

(6)2(0.2j)+αBDk×(1.2i0.2j)(0.1463)2(1.2i0.2j)=αDEFk×(0.12i0.8j)(1.463)2(0.12i0.8j)

Solve further:

7.2j+1.2αBDj+0.2αBDi0.0257i+0.00428j=0.12αDEFj+0.8αDEFi+0.257i+1.7123j

Equate components:

i:0.2αBD0.0257=0.8αDEF+0.257j:7.2+1.2αBD+0.00428=0.12αDEF+1.7123

Solve above equations:

αBD=7.278rad/s2αDEF=1.466rad/s2

Conclusion:

The angular acceleration of bar DEF is 1.466rad/s2 counter clockwise.

Expert Solution
Check Mark
To determine

(b)

Acceleration of point F

Answer to Problem 15.128P

The acceleration of point F is 1.574in/s2 at an angle 47.06° with horizontal downwards.

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 15.4, Problem 15.128P , additional homework tip  2

Constant angular velocity of AB is 6rad/s clockwise.

Point A and E are fixed.

The absolute acceleration of point B is defined as

aB=aA+aB/A

aB=aA+(aB/A)t+(aB/A)n

The tangential acceleration is defined as

(aB/A)t=αk×rB/A

The normal acceleration is defined as

(aB/A)n=rB/AωAB2

In above equations

α - Angular acceleration

ω - Angular velocity

r - Distance from A to B

Calculation:

According to sub part a

αDEF=1.466rad/s2ωDEF=1.463rad/s

Absolute acceleration of point F

aF=aE+αDEFk×rF/EωDEF2rF/E

We know that

aE=0

Therefore

aF=αDEFk×rF/EωDEF2rF/E

Substitute

aF=1.466k×(0.09i+0.6j)(1.463)2(0.09i+0.6j)aF=0.13194j0.8796i0.1926i1.2842j

Then

aF=1.0722i1.1522j

Find the magnitude and the angle

aF=1.574in/s247.06°

Conclusion:

The acceleration of point F is 1.574in/s2 at an angle 47.06° with horizontal downwards.

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Chapter 15 Solutions

Vector Mechanics For Engineers

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