CEE 271: Applied Mechanics II, Dynamics– Lecture 9: Ch.13, Sec.4-5 –

Prof. Albert S. Kim

Civil and Environmental Engineering, University of Hawaii at Manoa

1 / 40

EQUATIONS OF MOTION:RECTANGULARCOORDINATES

Today’s objectives: Studentswill be able to

1 Apply Newton’s second lawto determine forces andaccelerations for particlesin rectilinear motion.

In-class activities:• Reading Quiz• Applications• Equations of Motion Using

Rectangular (Cartesian)Coordinates

• Concept Quiz• Group Problem Solving• Attention Quiz

2 / 40

READING QUIZ

1 In dynamics, the friction force acting on a moving object isalways(a) in the direction of its motion.(b) a kinetic friction.(c) a static friction.(d) zero.

ANS: (b)

2 If a particle is connected to a spring, the elastic springforce is expressed by F = ks. The “s” in this equation isthe(a) spring constant.(b) un-deformed length of the spring.(c) difference between deformed and un-deformed lengths.(d) deformed length of the spring.

ANS: (c)

3 / 40

APPLICATIONS

• If a man is trying to move a 100 lb crate, how large a forceF must he exert to start moving the crate? What factorsinfluence how large this force must be to start moving thecrate?

• If the crate starts moving, is there acceleration present?• What would you have to know before you could find these

answers?4 / 40

APPLICATIONS(continued)

• Objects that move in air (or other fluid) have a drag forceacting on them. This drag force is a function of velocity.

• If the dragster is traveling with a known velocity and themagnitude of the opposing drag force at any instant isgiven as a function of velocity, can we determine thetime and distance required for dragster to come to a stop ifits engine is shut off? How ?

5 / 40

RECTANGULAR COORDINATES (Section 13.4)

• The equation of motion, F = ma, is best used when theproblem requires finding forces (especially forcesperpendicular to the path), accelerations, velocities, ormass. Remember, unbalanced forces cause acceleration!

• Three scalar equations can be written from this vectorequation. The equation of motion, being a vector equation,may be expressed in terms of its three components in theCartesian (rectangular) coordinate system as ΣF = ma or

(ΣFx) i + (ΣFy) j + (ΣFz)k = m(axi + ayj + azk)

or, as scalar equations,

ΣFx = max, ΣFy = may, ΣFz = maz

6 / 40

PROCEDURE FOR ANALYSIS

Free Body Diagram• Establish your coordinate system and draw the particle’s

free body diagram showing only external forces. Theseexternal forces usually include the weight, normal forces,friction forces, and applied forces. Show the ”ma” vector(sometimes called the inertial force) on a separatediagram.

• Make sure any friction forces act opposite to the direction ofmotion! If the particle is connected to an elastic linearspring, a spring force equal to ‘ks’ should be included onthe FBD.

7 / 40

PROCEDURE FOR ANALYSIS (continued)

Equation of Motion• If the forces can be resolved directly from the free-body

diagram (often the case in 2-D problems), use thescalar form of the equation of motion. In more complexcases (usually 3-D), a Cartesian vector is written for everyforce and a vector analysis is often best.

• A Cartesian vector formulation of the second law isΣF = ma or

ΣFxi + ΣFyj + ΣFzk = m(axi + ayj + azk)

• Three scalar equations can be written from this vectorequation. You may only need two equations if the motion isin 2-D.

8 / 40

PROCEDURE FOR ANALYSIS (continued)

Kinematics

• The second law only provides solutions for forces andaccelerations. If velocity or position have to be found,kinematics equations are used once the acceleration isfound from the equation of motion.

• Any of the kinematics tools learned in Chapter 12 may beneeded to solve a problem.

• Make sure you use consistent positive coordinate directionsas used in the equation of motion part of the problem!

9 / 40

EXAMPLE

• Given: The 200lb mine car is hoistedup the incline. The motor M pulls in thecable with an acceleration of 4 ft/s2.

• Find: The acceleration of the mine carand the tension in the cable.

• Plan:1 Draw the free-body and kinetic diagrams of the car.2 Using a dependent motion equation, determine an

acceleration relationship between cable and mine car.3 Apply the equation of motion to determine the cable

tension.

10 / 40

EXAMPLE (solution)

Solution

1. Draw the free-body and kinetic diagrams of the mine car:• Since the motion is up the incline, rotate the x− y axes.• Motion occurs only in the x-direction. We are also

neglecting any friction in the wheel bearings, etc., on thecart.

11 / 40

2. The cable equation results in sp + 2sc = ltTaking the derivative twice yields

ap + 2ac = 0 (1)

The relative acceleration is ap = ac + ap/c.As the motor is mounted on the car,ap/c = 4 ft/s2. So,

ap = ac + 4 ft/s2 (2)

Solving equations (1) and (2), yields

aC = 1.333 ft/s2

12 / 40

3. Apply the equation of motion in the x-direction:

m = 200/32.2 = 6.211 slug

(+↗) Fx = max

3T −mg(sin 30◦) = max

3T − (200)(sin 30◦) = 6.211(1.333)

T = 36.1 lb

13 / 40

CHECK YOUR UNDERSTANDING QUIZ

1. If the cable has a tension of 3 N,determine the acceleration of block B.(a) 4.26 m/s

2 ↑(b) 4.26 m/s

2 ↓(c) 8.31 m/s

2 ↑(d) 8.31 m/s

2 ↓ANS: (d)

2. Determine the acceleration of the block.

(a) 2.20 m/s2 ↑

(b) 3.17 m/s2 ↑

(c) 11.0 m/s2 ↑

(d) 4.26 m/s2 ↑

ANS: (a)

14 / 40

GROUP PROBLEM SOLVING

• Given: WA = 10 lb, WB = 20 lb,voA = 2 ft/s(→), and µk = 0.2

• Find: vA when A has moved 4 feet tothe right.

• Plan: Since both forces and velocityare involved, this problem requires boththe equation of motion and kinematics.

• First, draw free body diagrams of A and B. Apply theequation of motion to each.

• Using dependent motion equations, derive a relationshipbetween aA and aB and use with the equation of motionformulas.

15 / 40

GROUP PROBLEM SOLVING (Solution)

• Free-body and kinetic diagrams of B:Apply the equation of motion to B:

+ ↓ ΣFy = may

WB − 2T = mBaB

20− 2T =20

32.2aB (3)

• Free-body and kinetic diagrams of A:Apply the equations of motion to A:

(+ ↑) ΣFy = may = 0

N = WA = 10 lb

f = µkN = 2 lb

(+←) ΣFx = max

f − T = mAaA

2− T =10

32.2aA (4)

16 / 40

• Now consider the kinematics.Constraint equation:

sA + 2sB = constant

vA + 2vB = 0

Therefore

aA + 2aB = 0

aA = −2aB (5)

(Notice aA is considered positive tothe left and aB is positive downward.)

17 / 40

• Now combine equations (3), (4), and (5)Constraint equation:

T =22

3= 7.33 lb

aA = −17.16 ft/s2 = 17.16 ft/s2(→)

• Now use the kinematic equation:

(vA)2 = (v0A)2 + 2aA(sA − s0A)

(vA)2 = (2)2 + 2(−17.16)(−4)

vA = 11.9 ft/s(→)

18 / 40

ATTENTION QUIZ

1. Determine the tension in thecable when the 400 kg box ismoving upward with a 4 m/s2

acceleration.(a) 2265 N(b) 3365 N(c) 5524 N(d) 6543 N

������������������������

������������������������

60◦ a = 4m/s2

T

ANS: (c)

19 / 40

ATTENTION QUIZ

2. A 10 lb particle has forces of F1 = (3i + 5j) lb andF2 = (−7i + 9j) lb acting on it. Determine the accelerationof the particle.(a) (−0.4i + 1.4j) ft/s

2

(b) (−0.4i + 14j) ft/s2

(c) (−12.9i + 45j) ft/s2

(d) (13i + 4j) ft/s2

ANS: (c)

20 / 40

EQUATIONS OF MOTION: NORMAL ANDTANGENTIAL COORDINATES

Today’s objectives: Studentswill be able to

1 Apply the equation ofmotion using normal andtangential coordinates.

In-class activities:• Reading Quiz• Applications• Equation of Motion in n− t

Coordinates• Concept Quiz• Group Problem Solving• Attention Quiz

21 / 40

READING QUIZ

1 The ‘normal’ component of the equation of motion iswritten as ΣFn = man, where ΣFn is referred to as the?(a) impulse(b) centripetal force(c) tangential force(d) inertia force

ANS: (b)

2 The positive n direction of the normal and tangentialcoordinates is .(a) normal to the tangential component(b) always directed toward the center of curvature(c) normal to the bi-normal component(d) All of the above.

ANS: (d)

22 / 40

APPLICATIONS

• Race tracks are often banked in the turns to reduce thefrictional forces required to keep the cars from sliding up tothe outer rail at high speeds.

• If the car’s maximum velocity and a minimum coefficient offriction between the tires and track are specified, how canwe determine the minimum banking angle (θ) required toprevent the car from sliding up the track?

23 / 40

APPLICATIONS(continued)

• The picture shows a ride at the amusement park. Thehydraulically-powered arms turn at a constant rate, whichcreates a centrifugal force on the riders.

• We need to determine the smallest angular velocity of thecars A and B so that the passengers do not loose contactwith the seat. What parameters do we need for thiscalculation?

24 / 40

APPLICATIONS(continued)

• Satellites are held in orbit around the earth by using theearth’s gravitational pull as the centripetal force - the forceacting to change the direction of the satellite’s velocity.

• Knowing the radius of orbit of the satellite, we need todetermine the required speed of the satellite to maintainthis orbit. What equation governs this situation?

25 / 40

NORMAL & TANGENTIAL COORDINATES(Sec. 13.5)

• When a particle moves along acurved path, it may be moreconvenient to write the equation ofmotion in terms of normal andtangential coordinates.

• The normal direction (n) alwayspoints toward the path’s center ofcurvature. In a circle, the center ofcurvature is the center of the circle.

• The tangential direction (t) is tangent to the path, usuallyset as positive in the direction of motion of the particle.

26 / 40

EQUATIONS OF MOTION

• Since the equation of motion is a vectorequation, ΣF = ma, it may be written interms of the n and t coordinates asΣFtut + ΣFnun + ΣFbub = mat +man

•∑Ft and

∑Fn are the sums of the force components

acting in the t and n directions, respectively.• This vector equation will be satisfied provided the

individual components on each side of the equation areequal, resulting in the two scalar equations:∑

Ft = mat and∑

Fn = man

• Since there is no motion in the binormal (b) direction, wecan also write

∑Fb = 0.

27 / 40

NORMAL AND TANGENTIAL ACCERLERATIONS

• The tangential acceleration, at = dv/dt, represents the timerate of change in the magnitude of the velocity. Dependingon the direction of ΣFt, the particle’s speed will either beincreasing or decreasing.

• The normal acceleration, an = v2/ρ, represents the timerate of change in the direction of the velocity vector.Remember, an always acts toward the path’s center ofcurvature. Thus, ΣFn will always be directed toward thecenter of the path.

• Recall, if the path of motion is defined as y = f(x), theradius of curvature at any point can be obtained from

ρ =[1 + ( dydx)2]3/2∣∣∣ d2ydx2

∣∣∣28 / 40

SOLVING PROBLEMS WITH n− t COORDINATES

• Use n− t coordinates when a particle is moving along aknown, curved path.

• Establish the n− t coordinate system on the particle.• Draw free-body and kinetic diagrams of the particle. The

normal acceleration (an) always acts ’inward’ (the positiven-direction). The tangential acceleration (at) may act ineither the positive or negative t direction.

• Apply the equations of motion in scalar form and solve.• It may be necessary to employ the kinematic relations:

at =dv

dt= v

dv

ds(6)

an =v2

ρ(7)

29 / 40

EXAMPLE

• Given: At the instant θ = 45◦, the boywith a mass of 75 kg, moves a speed of6 m/s, which is increasing at 0.5 m/s2.Neglect his size and the mass of theseat and cords. The seat is pinconnected to the frame BC.

• Find: Horizontal and vertical reactionsof the seat on the boy.

• Plan:1 Since the problem involves a curved path and requires

finding the force perpendicular to the path, use n− tcoordinates.

2 Draw the boy’s free-body and kinetic diagrams.3 Apply the equation of motion in the n− t directions.

30 / 40

EXAMPLE (Solution)

1. The n− t coordinate system can beestablished on the boy at angle θ = 45◦.Approximating the boy and seattogether as a particle, the free-bodyand kinetic diagrams can be drawn.

31 / 40

EXAMPLE (continued)

!

"#

$%#

$&#

'#

()*#

!

"#

+,!#

+,"#

-.//01234#35,6.,+# 75"/89#35,6.,+#

2. Apply the equations of motion in the n− t directions.a. man = −Rx cos θ −Ry sin θ +W sin θ

Using an = v2/ρ = 62/10, W = 75(9.81) N, and m = 75 kg,we get: −Rx cos 45◦ −Ry sin 45◦ + 520.3 = (75)(62/10)

b. mat = −Rx sin θ +Ry cos θ −W cos θwe get: −Rx sin 45◦ +Ry cos 45◦ − 520.3 = (75)(0.5)Using above two equations, solve for Rx, Ry.Rx = −217 N and Ry = 572 N.

32 / 40

CONCEPT QUIZ

!"#"$"%"

&"

1. A 10 kg sack slides down a smoothsurface. If the normal force on the surfaceat the flat spot, A, is 98.1 N (↑) , the radiusof curvature is(a) 0.2 m(b) 0.4 m(c) 1.0 m(d) None of the above

ANS: (d)

2. A 20 lb block is moving along a smoothsurface. If the normal force on the surfaceat A is 10 lb, the velocity is .(a) 7.6 ft/s(b) 9.6 ft/s(c) 10.6 ft/s(d) 12.6 ft/s

ANS: (c) 33 / 40

GROUP PROBLEM SOLVING

• Given: A 800 kg car is traveling overthe hill having the shape of aparabola. When it is at point A, it istraveling at 9 m/s and increasing itsspeed at 3 m/s2.

• Find: The resultant normal force and resultant frictionalforce exerted on the road at point A.

• Plan:1 Treat the car as a particle.2 Draw the free-body and kinetic diagrams.3 Apply the equations of motion in the n− t directions.4 Use calculus to determine the slope and radius of curvature

of the path at point A.34 / 40

1. The n− t coordinate system can beestablished on the car at point A. Treatthe car as a particle and draw thefree-body and kinetic diagrams:W = mg = weight of carN = resultant normal force on roadF = resultant friction force on road

35 / 40

2. Apply the equations of motion in the n− t directions:

W cos θ −N = man

Using W = mg and an = v2/ρ = (9)2/ρ

(800)(9.81) cos θ −N = (800)(81/ρ)

N = 7848 cos θ − 64800/ρ (8)

W sin θ − F = mat

Using W = mg and at = 3m/s2 (given)

(800)(9.81) sin θ − F = (800)(3)

F = 7848 sin θ − 2400 (9)

36 / 40

3. Determine ρ by differentiating y = f(x) at x = 80 m:

y = 20(1− x2/6400)

dy/dx = (−40)x/6400

d2y/dx2 = (−40)/640 (10)

ρ|x=80 =[1 + ( dydx)2]3/2∣∣∣ d2ydx2

∣∣∣ =[1 + (−0.5)2]3/2

0.00625= 223.6 meter

Determine θ from the slope of the curve at A:

tan θ = dy/dx|x=80=−40x

6400

∣∣∣∣x=80

= −0.5

θ = tan−1(dy

dx) = tan−1(−0.5) = 26.6◦

37 / 40

• From Eq. (8):

N = 7848 cos θ − 64800/ρ

= 7848 cos(26.6◦)− 64800/223.6

= 6728N (11)

• From Eq.(9) :

F = 7848 sin θ − 2400

= 7848 sin(26.6o)− 2400

= 1114N (12)

38 / 40

ATTENTION QUIZ

1 The tangential acceleration of an object(a) represents the rate of change of the velocity vector’s

direction.(b) represents the rate of change in the magnitude of the

velocity.(c) is a function of the radius of curvature.(d) Both (b) and (c).

ANS: (b)

39 / 40

ATTENTION QUIZ

1 The block has a mass of 20 kg and a speed of v = 30 m/sat the instant it is at its lowest point. Determine the tensionin the cord at this instant.

(a) 1596 N

(b) 1796 N

(c) 1996 N

(d) 2196 N

!"

#$"%"

&"'"($"%)*"

ANS: (c)

40 / 40