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

Date: __________________

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ABSOLUTE DEPENDENT MOTION ANALYSIS OFTWO PARTICLES

Today’s objectives: Studentswill be able to

1 Relate the positions,velocities, andaccelerations of particlesundergoing dependentmotion.

In-class activities:• Reading Quiz• Applications• Define Dependent Motion• Develop Position, Velocity,

and AccelerationRelationships

• Concept Quiz• Group Problem Solving• Attention Quiz

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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)

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APPLICATIONS

• The cable and pulley systemshown can be used to modify thespeed of the mine car, A, relativeto the speed of the motor, M .

• It is important to establish therelationships between the variousmotions in order to determine thepower requirements for the motorand the tension in the cable.

• For instance, if the speed of the cable (P ) is known(because we know the motor characteristics), how can wedetermine the speed of the mine car? Will the slope of thetrack have any impact on the answer?

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APPLICATIONS(continued)

• Rope and pulley arrangements areoften used to assist in liftingheavy objects. The totallifting force required from thetruck depends on boththe weight and the acceleration ofthe cabinet.

• How can we determine theacceleration and velocity of thecabinet if the acceleration of thetruck is known?

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DEPENDENT MOTION (Section 12.9)

• In many kinematics problems, themotion of one object will depend onthe motion of another object.

• The blocks in this figure areconnected by an inextensible cordwrapped around a pulley. If block Amoves downward along the inclinedplane, block B will move up the otherincline.

• The motion of each block can be related mathematically bydefining position coordinates, sA and sB. Each coordinateaxis is defined from a fixed point or datum line, measuredpositive along each plane in the direction of motion of eachblock.

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DEPENDENT MOTION (continued)

• In this example, position coordinatessA and sB can be defined fromfixed datum lines extending from thecenter of the pulley along eachincline to blocks A and B.

• If the cord has a fixed length, theposition coordinates sA and sB arerelated mathematically by

sA + lCD + sB = lT

were lT is the total cord length and lCD is the length of cordpassing over the arc CD on the pulley.

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DEPENDENT MOTION (continued)

• The velocities of blocks A and B canbe related by differentiating theposition equation. Note that lCD andlT remain constant, so

dlCD

dt=

dlTdt

= 0

dsAdt

+dsBdt

= 0 ⇒ vB = −vA

• The negative sign indicates that as A moves down theincline (positive sA direction), B moves up the incline(negative sB direction).

• Accelerations can be found by differentiating the velocityexpression. Prove to yourself that aB = −aA .

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DEPENDENT MOTION EXAMPLE

• Consider a more complicatedexample. Position coordinates (sAand sB) are defined from fixeddatum lines, measured along thedirection of motion of each block.

• Note that sB is only defined to thecenter of the pulley above block B,since this block moves with thepulley. Also, h is a constant.

• The red colored segments of the cord remain constant inlength during motion of the blocks.

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DEPENDENT MOTION EXAMPLE (continued)

• The position coordinates are related bythe equation

2sB + h+ sA = lT

where lT is the total cord length minusthe lengths of the red segments.

• Since lT and h remain constant duringthe motion, the velocities andaccelerations can be related by twosuccessive time derivatives:2vB = −vA and 2aB = −aA

• When block B moves downward (+sB), block A moves tothe left (−sA). Remember to be consistent with your signconvention!

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DEPENDENT MOTION EXAMPLE (continued)

• This example can also be workedby defining the positioncoordinate for B (sB) from thebottom pulley instead of the toppulley.

• The position, velocity, andacceleration relations thenbecome2(h− sB) + h+ sA = lTand 2vB = vA , 2aB = aA

• Prove to yourself that the results are the same, even if thesign conventions are different than the previousformulation.

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DEPENDENT MOTION: PROCEDURES

• These procedures can be used to relate the dependentmotion of particles moving along rectilinear paths (only themagnitudes of velocity and acceleration change, not theirline of direction).

1 Define position coordinates from fixed datum lines, along thepath 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 not change 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 written for each cord.

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

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EXAMPLE

• Given: In the figure on the left, the cordat A is pulled down with a speed of2m/s.

• Find: The speed of block B.• Plan:

1 There are two cords involved in themotion in this example.

2 There will be two position equations(one for each cord).

3 Write these two equations, combinethem, and then differentiate them.

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EXAMPLE (continued)

Solution:1 Define the position coordinates from a fixed datum line.

Three coordinates must be defined: one for point A (sA),one for block B (sB), and one for block C (sC).

• Define the datum line through the toppulley D(which has a fixed position).

• sA can be defined to the point A.• sB can be defined to the center of the

pulley above B.• sC is defined to the center of pulley C.• All coordinates are defined as

positive down and along the direction ofmotion of each point/object.

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EXAMPLE (continued)

2. Write position/length equations foreach cord. Define l1 as the length of thefirst cord, minus any segments ofconstant length. Define l2 in a similarmanner for the second cord:Cord 1: sA + 2sC = l1Cord 2: sB + (sB − sC) = l2

3. Eliminating sC between the twoequations, we get

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

that l1 and l2 are constant lengths.vA + 4vB = 0⇒ vB = −0.25vA = −0.25(2) = −0.5m/sThe velocity of block B is 0.5 m/s up (negative sBdirection).

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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)

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GROUP PROBLEM SOLVING

• Given: In this pulley system, block A ismoving downward with a speed of 4 ft/swhile block C is moving up at 2 ft/s.

• Find: The speed of block B.

• Plan:All blocks are connected to a single cable, so only oneposition / length equation will be required. Define positioncoordinates for each block, write out the position relation,and then differentiate it to relate the velocities.

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GROUP PROBLEM SOLVING (continued)

Solution:1. A datum line can be drawn through the

upper, fixed pulleys and positioncoordinates defined from this line toeach block (or the pulley above theblock).

2. Defining sA, sB, and sC as 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/sThe velocity of block B is 1 ft/s up (negative sB direction). 18 / 20

ATTENTION QUIZ

1 Determine the speed of block B whenblock A is moving down at 6 ft/s whileblock C is moving down at 18 ft/s.(a) 24 ft/s(b) 3 ft/s(c) 12 ft/s(d) 9 ft/s

ANS: (c)

2 Determine the velocity vector of block Awhen block B is moving downward witha 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

note

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