Concept explainers
(a)
To find:
the value the couple M applied at disk A.
Answer to Problem 16.135P
At disk A, couple applied magnitude is
Explanation of Solution
Given information:
Rod BC mass, m = 6kg
Disk mass, m = 10kg
Rod CD mass, m = 5kg
Disk AB velocity
Disk radius,
Disk angular velocity,
Since point C velocity is parallel to point B velocity, the point C velocity magnitude and direction is same as point B.
Rod CD angular velocity
Disk B acceleration,
Rod BC acceleration tangential component,
Rod BC acceleration,
Rod CD acceleration tangential component,
Rod CD acceleration,
Equation forces horizontal component from equations A and B,
Equation forces vertical component from equations A and B,
Acceleration of point A is zero since it is pivoted
Rod BC acceleration of mass centre P,
Rod CD acceleration of mass centre Q,
Disk AB effective force at mass centre,
Disk AB moment of inertia,
Rod BC effective force at mass centre,
Rod BC moment of inertia,
Rod CD effective force at mass centre,
Rod CD moment of inertia,
Rod BC free body diagram
Figure A
Moment at point B from above figure,
Rod CD free body diagram
Figure B
Moment at point D from above figure,
Combined disk AB and rod BC free body diagram
Figure C
From above figure, take moment at point A,
M is couple applied at point A
At disk A, couple applied magnitude is
Conclusion:
At disk A, couple applied magnitude is
(b)
To find:
the force components exerted on rod BC
Answer to Problem 16.135P
The force horizontal component exerted at point C is
Explanation of Solution
Given information:
Rod BC mass, m = 6kg
Disk mass, m = 10kg
Rod CD mass, m = 5kg
Rod BC free body diagram
Figure A
Moment at point B from above figure,
Rod CD free body diagram
Figure B
Moment at point D from above figure,
Combined disk AB and rod BC free body diagram
Figure C
From above figure, take moment at point A,
M is couple applied at point A
At disk A, couple applied magnitude is
Conclusion:
The force horizontal component exerted at point C is
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Chapter 16 Solutions
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