CEE 271: Applied Mechanics II, Dynamics– Lecture 6: Ch.12, Sec.10 –

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

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RELATIVE-MOTION ANALYSIS OF TWO PARTICLESUSING TRANSLATING AXES

Today’s objectives: Studentswill be able to

1 Understand translatingframes of reference.

2 Use translating frames ofreference to analyzerelative motion.

In-class activities:• Reading Quiz• Applications• Relative Position, Velocity

and Acceleration• Vector and Graphical

Methods• Concept Quiz• Group Problem Solving• Attention Quiz

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READING QUIZ

1 The velocity of B relative to A is defined as(a) vB − vA(b) vA − vB(c) vB + vA(d) vA + vB

ANS: (a)

2 Since two dimensional vector addition forms a triangle,there can be at most unknowns (either magnitudes and/ordirections of the vectors).(a) one(b) two(c) three(d) four

ANS: (b)

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APPLICATIONS

• When you try to hit a movingobject, the position, velocity, andacceleration of the object all haveto be accounted for by your mind.

• You are smarter than you thought!

• Here, the boy on the ground is at d = 10 ft when the girl inthe window throws the ball to him.

• If the boy on the ground is running at a constant speed of4 ft/s, how fast should the ball be thrown?

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APPLICATIONS(continued)

• When fighter jets take off or landon an aircraft carrier, the velocityof the carrier becomes an issue.

• If the aircraft carrier is underway with a forward velocity of50 km/hr and plane A takes off at a horizontal air speed of200 km/hr (measured by someone on the water), how do wefind the velocity of the plane relative to the carrier?

• How would you find the same thing for airplane B?• How does the wind impact this sort of situation?

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RELATIVE POSITION (Section 12.10)

• The absolute position of twoparticles A and B with respect tothe fixed x, y, z reference frameare given by rA and rB.

• The position of B relative to A isrepresented by

rB/A = rB − rA

• Therefore, if rB = (10i+ 2j) m

and rA = (4i+ 5j) m,then rB/A = (6i− 3j) m.

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RELATIVE VELOCITY

• To determine the relative velocityof B with respect to A, the timederivative of the relative positionequation is taken.

vB/A = vB − vA

orvB = vA + vB/A

• In these equations, vB and vA are called absolute velocitiesand vB/A is the relative velocity of B with respect to A.

• Note that vB/A = −vA/B .

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RELATIVE ACCELERATION

• The time derivative of the relativevelocity equation yields a similarvector relationship between theabsolute and relative accelerationsof particles A and B.

• These derivatives yield:

aB/A = aB − aA

oraB = aA + aB/A

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SOLVING PROBLEMS

• Since the relative motion equations are vector equations,problems involving them may be solved in one of two ways.

• For instance, the velocity vectors in vB = vA + vB/A couldbe written as two dimensional (2-D) Cartesian vectors andthe resulting 2-D scalar component equations solved for upto two unknowns.

• Alternatively, vector problems can be solved ‘graphically ’by use of trigonometry. This approach usually makes useof the law of sines or the law of cosines.

• Could a CAD system be used to solve these types ofproblems?

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LAWS OF SINES AND COSINES

• Since vector addition or subtraction forms atriangle, sine and cosine laws can be applied to solve forrelative or absolute velocities and accelerations. As areview, their formulations are provided below.

• Law of Sines:a

sinA=

b

sinB=

c

sinC

• Law of Cosines:

a2 = b2 + c2 − 2bc cosA

b2 = a2 + c2 − 2ac cosB

c2 = a2 + b2 − 2ab cosC

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EXAMPLE

• Given: vA = 650 km/hvB = 800 km/h

• Find: vB/A

• Plan:1 Vector Method: Write vectors vA and vB in Cartesian form,

then determine vB − vA2 Graphical Method: Draw vectors vA and vB from a common

point. Apply the laws of sines and cosines to determinevB/A.

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EXAMPLE (continued)

Solutiona). Vector Method:

vA = (650i)

vB = −800 cos 60i− 800 sin 60j

vB/A = vB − vA = (−1050i− 692.8j) (1)

vB/A = (−1050)2 + (−692.8)2 = 1258 (2)

θ = tan−1

(−692.8−1050

)= 33.4◦ (3)

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EXAMPLE (continued)

Solutionb). Graphical Method:

Note that the vector thatmeasures the tip of B relative toA is vB/A.

• Law of Cosines:(vB/A)

2 = (800)2 + (650)2 −(800)(650) cos 120◦

vB/A = 1258 km/h

• Law of Sines:vB/A

sin 120◦=

vAsin θ

⇒ θ = 33.4◦

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CONCEPT QUIZ

1. Two particles, A and B, are moving in the directionsshown. What should be the angle θ so that vB/A isminimum?(a) 0◦

(b) 180◦

(c) 90◦

(d) 270◦

ANS: (a)

2. Determine the velocity ofplane A with respect to planeB.(a) (400i+ 520j) km/hr(b) (1220i− 300j) km/hr(c) (−181i− 300j) km/hr(d) (−1220i+ 300j) km/hr

ANS: (b) 14 / 18

GROUP PROBLEM SOLVING

• Given: vA = 30mi/hvB = 20mi/haB = 1200mi/h2

aA = 0mi/h2

• Find: vB/A and aB/A

• Plan: Write the velocity and acceleration vectors for A andB and determine vB/A and aB/A by using vector equations.

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GROUP PROBLEM SOLVING (solution continued)

Solution:• The velocity of B is:vB = −20 sin(30)i+ 20 cos(30)j =(−10i+ 17.32j)mi/h

• The velocity of A is:vA = −30i (mi/h)

• The relative velocity of B with respect to A is (vB/A):vB/A = (−10i+ 17.32j)− (−30i) = (20i+ 17.32j) mi/h orvB/A =

√(20)2 + (17.32)2 = 26.5 mi/h

θ = tan−1(17.3220 ) = 40.9◦

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GROUP PROBLEM SOLVING (solution continued)

• The acceleration of B is:

aB = (at)B + (an)B

= [−1200 sin 30i+ 1200 cos 30j]

+ [202

0.3 cos 30i+ 202

0.3 sin 30j]

= 554.7i+ 1706j(mi/h2)

• The acceleration of A is zero : aA = 0

• The relative acceleration of B with respect to A is:

aB/A = aB − aA = 554.7i+ 1706j(mi/h2) (4)

aB/A =√(554.7)2 + (1706)2 = 1790mi/h2 (5)

β = tan−1(1706

554.7) = 72◦ (6)

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ATTENTION QUIZ

1 Determine the relative velocity of particle B with respect toparticle A.(a) (+48i+ 30j) km/h(b) (−48i+ 30j) km/h(c) (+48i− 30j) km/h(d) (−48i− 30j) km/h

ANS: (c)

2 If theta equals 90◦ and A and B start moving from thesame point, what is the magnitude of rB/A at t = 5 s?(a) 20 ft(b) 15 ft(c) 18 ft(d) 25 ft

ANS: (d)

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