8/10/2019 scee-271-12-sec-9

1/20

CEE 271: Applied Mechanics II, Dynamics

Lecture 5: Ch.12, Sec.9

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

1/20

http://goforward/http://find/http://goback/

8/10/2019 scee-271-12-sec-9

2/20

ABSOLUTE DEPENDENT MOTION ANALYSIS OF

TWO PARTICLES

Todays objectives: Students

will be able to

1 Relate the positions,

velocities, and

accelerations of particlesundergoingdependent

motion.

In-class activities:

Reading Quiz

Applications

Define Dependent Motion

Develop Position, Velocity,

and Acceleration

Relationships

Concept Quiz

Group Problem Solving

Attention Quiz

2/20

http://goforward/http://find/http://goback/

8/10/2019 scee-271-12-sec-9

3/20

READING QUIZ

1 When particles are interconnected by a cable, the motionsof the particles are

(a) always independent.(b) always dependent.(c) not always dependent.

(d) None of the above.ANS:(b)

2 If the motion of one particle is dependent on that of anotherparticle, each coordinate axis system for the particles

(a) should be directed along the path of motion.

(b) can be directed anywhere.(c) should have the same origin.(d) None of the above.

ANS:(a)

3/20

http://find/

8/10/2019 scee-271-12-sec-9

4/20

APPLICATIONS

The cable and pulley system

shown can be used to modify the

speed of the mine car,A, relativeto the speed of the motor,M.

It is important to establish therelationships between the various

motions in order to determine the

power requirementsfor the motor

and thetensionin the cable.

For instance, if the speed of the cable (P) is known(because we know the motor characteristics), how can we

determine the speed of the mine car? Will theslopeof the

track have any impact on the answer?

4/20

http://find/

8/10/2019 scee-271-12-sec-9

5/20

APPLICATIONS(continued)

Rope and pulleyarrangements are

often used to assist in lifting

heavy objects. The total

lifting forcerequired from thetruck depends on both

the weight and the accelerationof

the cabinet.

How can we determine the

acceleration and velocity of thecabinet if the acceleration of the

truck isknown?

5/20

http://find/http://goback/

8/10/2019 scee-271-12-sec-9

6/20

DEPENDENT MOTION (Section 12.9)

In many kinematics problems, the

motion of one object willdependon

the motion of another object.

The blocks in this figure are

connected by aninextensiblecordwrapped around a pulley. If block Amoves downwardalong the inclined

plane, blockB will moveupthe otherincline.

The motion of each block can be related mathematically bydefining position coordinates,sA andsB. Each coordinateaxis is defined froma fixed pointordatum line, measured

positivealong each plane in thedirection of motionof each

block.

6/20

http://find/

8/10/2019 scee-271-12-sec-9

7/20

DEPENDENT MOTION (continued)

In this example, position coordinates

sA andsB can be defined fromfixed datum linesextending from the

centerof the pulley along eachincline to blocksAandB .

If the cord has a fixed length, the

position coordinatessA andsB arerelated mathematically by

sA+lCD+sB =lT

werelTis the total cord length andlCD is the length of cordpassing over the arcCDon the pulley.

7/20

http://find/

8/10/2019 scee-271-12-sec-9

8/20

DEPENDENT MOTION (continued)

The velocities of blocksA andB canbe related by differentiating the

position equation. Note thatlCD andlT remainconstant, so

dlCDdt

=dlTdt

= 0

dsAdt

+dsBdt

= 0 vB = vA

The negative sign indicates that asA moves down theincline (positivesAdirection),B moves up the incline(negativesB direction).

Accelerations can be found by differentiating the velocity

expression. Prove to yourself thataB = aA.

8/20

http://find/

8/10/2019 scee-271-12-sec-9

9/20

DEPENDENT MOTION EXAMPLE

Consider a more complicated

example. Position coordinates (sAandsB) are defined from fixed

datum lines, measured along thedirection of motionof each block.

Note thatsB is only definedtothecenter of the pulley aboveblockB ,since this block moves with the

pulley. Also,h is aconstant.

The red colored segments of the cord remain constantin

length during motion of the blocks.

9/20

http://find/

8/10/2019 scee-271-12-sec-9

10/20

DEPENDENT MOTION EXAMPLE (continued)

The position coordinates are related bythe equation

2sB+h+sA=lT

wherelTis the total cord length minusthe lengths of the red segments.

SincelT andh remainconstantduringthe motion, the velocities and

accelerations can be related by twosuccessive time derivatives:

2vB = vA and 2aB = aA When blockB movesdownward(+sB), blockAmoves to

theleft(sA). Remember to be consistent with your sign

convention!10/20

http://find/

8/10/2019 scee-271-12-sec-9

11/20

DEPENDENT MOTION EXAMPLE (continued)

This example can also be worked

by defining the position

coordinate forB (sB) from thebottompulley instead of the top

pulley. The position, velocity, and

acceleration relations then

become

2(h sB) +h+sA=lTand 2vB =vA , 2aB =aA

Prove to yourself that the results are the same, even if the

sign conventions are different than the previous

formulation.

11/20

http://find/http://goback/

8/10/2019 scee-271-12-sec-9

12/20

DEPENDENT MOTION: PROCEDURES

These procedures can be used to relate the dependentmotion of particles moving along rectilinearpaths (only themagnitudesof velocity and acceleration change, not their

line of direction).1 Defineposition coordinatesfromfixed datum lines, along the

path of each particle. Different datum lines can be used foreach particle.

2 Relate the position coordinates to the cord length.Segments of cord that do notchange in length during themotion may be left out.

3 If a system contains more than one cord, relate the positionof a point on one cord to a point on another cord. Separateequations are writtenfor each cord.

4 Differentiatethe position coordinate equation(s) to relatevelocities and accelerations. Keep track of signs!

12/20

http://find/

8/10/2019 scee-271-12-sec-9

13/20

EXAMPLE

Given: In the figure on the left, the cord

atAis pulled down with a speed of2 m/s.

Find: The speed of blockB .

Plan:1 There aretwocords involved in the

motion in this example.2 There will betwoposition equations

(one for each cord).3 Writethese two equations,combine

them, and thendifferentiatethem.

13/20

http://find/

8/10/2019 scee-271-12-sec-9

14/20

EXAMPLE (continued)

Solution:

1 Define the position coordinates from a fixed datum line.

Three coordinates must be defined: one for pointA(sA),one for blockB (sB), and one for blockC(sC).

Define the datum line through the toppulleyD(which has afixedposition).

sAcan be definedtothe pointA.

sB can be defined tothe center of the

pulley aboveB. sC is definedtothe center of pulleyC.

All coordinates are defined as

positive downand along the direction of

motion of each point/object.

14/20

http://find/

8/10/2019 scee-271-12-sec-9

15/20

EXAMPLE (continued)

2. Write position/length equations foreach cord. Definel1as the length of thefirst cord,minusany segments of

constant length. Definel2 in a similarmanner for the second cord:

Cord 1: sA+ 2sC=l1Cord 2: sB+ (sB sC) =l2

3. EliminatingsCbetween the twoequations, we get

sA+ 4sB =l1+ 2l24. Relate velocities by differentiating this expression. Note

thatl1 andl2 are constant lengths.vA+ 4vB = 0 vB = 0.25vA= 0.25(2) = 0.5 m/sThe velocity of blockB is 0.5 m/s up(negativesBdirection).

15/20

http://goforward/http://find/http://goback/

8/10/2019 scee-271-12-sec-9

16/20

CONCEPT QUIZ

1. Determine the speed of block B .(a) 1 m/s(b) 2 m/s(c) 4 m/s(d) None of the above.

ANS:(a)

2. Two blocks are interconnected by a cable.

Which of the following is correct ?(a) vA = vB(b) (vx)A = (vx)B(c) (vy)A = (vy)B(d) All of the above.

ANS:(d)

16/20

http://find/

8/10/2019 scee-271-12-sec-9

17/20

GROUP PROBLEM SOLVING

Given: In this pulley system, block A ismoving downwardwith a speed of4 ft/swhile blockCis moving up at2 ft/s.

Find: The speed of blockB .

Plan:All blocks are connected to a single cable, so only one

position / length equationwill be required. Define position

coordinates for each block, write out the position relation,

and then differentiate it to relate the velocities.17/20

http://find/

8/10/2019 scee-271-12-sec-9

18/20

GROUP PROBLEM SOLVING (continued)

Solution:

1. A datum line can be drawn through theupper, fixed pulleysand position

coordinates defined from this line to

each block (or the pulley above the

block).

2. DefiningsA,sB, andsCas shown, theposition relation can be written:

sA+ 2sB+sC=l

3. Differentiate to relate velocities:

vA+ 2vB+vC = 0

4 + 2vB+ (2) = 0

vB = 1 ft/s

The velocity of blockB is 1 ft/s up(negativesB direction). 18/20

http://find/

8/10/2019 scee-271-12-sec-9

19/20

ATTENTION QUIZ

1 Determine the speed of blockB whenblockAis moving down at 6 ft/swhileblockCis moving down at 18 ft/s.

(a) 24ft/s(b) 3 ft/s

(c) 12ft/s(d) 9 ft/s

ANS:(c)

2 Determine the velocity vector of blockAwhen blockB is moving downward with

a speed of 10 m/s.(a) (8i+ 6j) m/s(b) (4i+ 3j) m/s(c) (8i 6j) m/s(d) (3i+ 4j) m/s

ANS:(b)19/20

http://find/

8/10/2019 scee-271-12-sec-9

20/20

note

20/20

http://find/