Announcements - Walla Walla Universitycurt.nelson/engr222/spring...Is the tension force in the cable greater than the weight of the elevator and its load? Applications - continued

Engr222 Spring 2004 Chapter 13 1

Announcements

• No class Friday• HW# 14 and 15 due Wed, May 12

Newton’s Law of Motion (Equation of Motion) Sections 13.1-3

Today’s Objectives:Students will be able to:a) Write the equation of motion

for an accelerating body.b) Draw the free-body and

kinetic diagrams for an accelerating body.

In-Class Activities:• Reading quiz• Applications• Newton’s laws of motion• Newton’s law of

gravitational attraction• Equation of motion for a

particle or system of particles

• Concept quiz• Group problem solving• Attention quiz


Reading Quiz1. Newton’s second law can be written in mathematical form

as ΣF = ma. Within the summation of forces ΣF, ________ are(is) not included.

A) external forces B) weight

C) internal forces D) All of the above

2. The equation of motion for a system of n-particles can be written as ΣFi = Σ miai = maG, where aG indicates _______.

A) summation of each particle’s acceleration

B) acceleration of the center of mass of the systemC) acceleration of the largest particleD) None of the above

Applications

The motion of an object depends on the forces acting on it.

Knowing the drag force, how can we determine the acceleration or velocity of the parachutist at any point in time?

A parachutist relies on the atmospheric drag resistance force to limit his velocity.


A freight elevator is lifted using a motor attached to a cable and pulley system as shown.

How can we determine the tension force in the cable required to lift the elevator at a given acceleration?

Is the tension force in the cable greater than the weight of the elevator and its load?

Applications - continued

Newton’s Laws of Motion

The motion of a particle is governed by Newton’s three laws of motion.

First Law: A particle originally at rest, or moving in a straight line at constant velocity, will remain in this state if the resultant force acting on the particle is zero.

Second Law: If the resultant force on the particle is not zero, the particle experiences an acceleration in the same direction as the resultant force. This acceleration has a magnitude proportional to the resultant force.

Third Law: Mutual forces of action and reaction between two particles are equal, opposite, and collinear.


The first and third laws were used in developing the concepts of statics. Newton’s second law forms the basis of the study of dynamics.

Mathematically, Newton’s second law of motion can be written

F = mawhere F is the resultant unbalanced force acting on the particle, and a is the acceleration of the particle. The positive scalar m is called the mass of the particle.

Newton’s second law cannot be used when the particle’s speed approaches the speed of light, or if the size of the particle is extremely small (~ size of an atom).

Newton’s Laws of Motion - continued

Newton’s Law of Gravitational Attraction

F = G(m1m2/r2)

where F = force of attraction between the two bodies,G = universal constant of gravitation ,m1, m2 = mass of each body, andr = distance between centers of the two bodies.

When near the surface of the earth, the only gravitational force having any sizable magnitude is that between the earth and the body. This force is called the weight of the body.

Any two particles or bodies have a mutually attractivegravitational force acting between them. Newton postulated the law governing this gravitational force as


Mass and Weight

It is important to understand the difference between the mass and weight of a body!

Mass is an absolute property of a body. It is independent of the gravitational field in which it is measured. The mass provides a measure of the resistance of a body to a change in velocity, as defined by Newton’s second law of motion (m = F/a).

The weight of a body is not absolute, since it depends on the gravitational field in which it is measured. Weight is defined as

W = mgwhere g is the acceleration due to gravity.

Units: SI System vs. FPS SystemSI system: In the SI system of units, mass is a base unit and weight is a derived unit. Typically, mass is specified in kilograms (kg), and weight is calculated from W = mg. If the gravitational acceleration (g) is specified in units of m/s2, then the weight is expressed in newtons (N). On the earth’s surface, g can be taken as g = 9.81 m/s2.

W (N) = m (kg) g (m/s2) => N = kg·m/s2

FPS System: In the FPS system of units, weight is a base unitand mass is a derived unit. Weight is typically specified in pounds (lb), and mass is calculated from m = W/g. If g is specified in units of ft/s2, then the mass is expressed in slugs. On the earth’s surface, g is approximately 32.2 ft/s2.

m (slugs) = W (lb)/g (ft/s2) => slug = lb·s2/ft


Equation of MotionThe motion of a particle is governed by Newton’s second law, relating the unbalanced forces on a particle to its acceleration. If more than one force acts on the particle, the equation of motion can be written

�F = FR = mawhere FR is the resultant force, which is a vector summation of all the forces.

To illustrate the equation, consider a particle acted on by two forces.

First, draw the particle’s free-body diagram, showing all forces acting on the particle. Next, draw the kinetic diagram, showing the inertial force maacting in the same direction as the resultant force FR.

Inertial Frame of Reference

This equation of motion is only valid if the acceleration is measured in a Newtonian or inertial frame of reference. What does this mean?

For problems concerned with motions at or near the earth’s surface, we typically assume our “inertial frame” to be fixed to the earth. We neglect any acceleration effects from the earth’s rotation.

For problems involving satellites or rockets, the inertial frame of reference is often fixed to the stars.


System of ParticlesThe equation of motion can be extended to include systems of particles. This includes the motion of solids, liquids, or gas systems.

As in statics, there are internal forces and external forces acting on the system. What is the difference between them?

Using the definitions of m = �mi as the total mass of all particles and aG as the acceleration of the center of mass G of the particles, then maG =�miai .

The text shows the details, but for a system of particles: �F = maG where �F is the sum of the external forces acting on the entire system.

Key Points

1) Newton’s second law is a “Law of Nature”--experimentally proven and not the result of an analytical proof.

2) Mass (property of an object) is a measure of the resistance to a change in velocity of the object.

3) Weight (a force) depends on the local gravitational field. Calculating the weight of an object is an application of F = ma, i.e., W = mg.

4) Unbalanced forces cause the acceleration of objects. This condition is fundamental to all dynamics problems!


Procedure for the Application of the Equation of Motion

1) Select a convenient inertial coordinate system. Rectangular, normal/tangential, or cylindrical coordinates may be used.

2) Draw a free-body diagram showing all external forcesapplied to the particle. Resolve forces into their appropriate components.

3) Draw the kinetic diagram, showing the particle’s inertial force, ma. Resolve this vector into its appropriate components.

4) Apply the equations of motion in their scalar component form and solve these equations for the unknowns.

5) It may be necessary to apply the proper kinematic relationsto generate additional equations.

Example

Given: A crate of mass m is pulled by a cable attached to a truck. The coefficient of kinetic friction between the crate and road is µk.

Find: Draw the free-body and kinetic diagrams of the crate.

Plan: 1) Define an inertial coordinate system.2) Draw the crate’s free-body diagram, showing all

external forces applied to the crate in the proper directions.

3) Draw the crate’s kinetic diagram, showing the inertial force vector ma in the proper direction.


1) An inertial x-y frame can be defined as fixed to the ground.Solution:

3) Draw the kinetic diagram of the crate:

The crate will be pulled to the right. The acceleration vector can be directed to the right if the truck is speeding up or to the left if it is slowing down.

2) Draw the free-body diagram of the crate:

The weight force (W) acts through the crate’s center of mass. T is the tension force in the cable. The normal force (N) is perpendicular to the surface. The friction force (F = uKN) acts in a direction opposite to the motion of the crate.

y

x

W = mgT

30°

NF = uKN

ma

Example - continued

Concept Quiz

1. The block (mass = m) is moving upward with a speed v. Draw the FBD if the kinetic friction coefficient is µk.

A) B)

C) D) None of the above.µµµµkmg

mg

N

mg

N

µµµµkN

N

mg

µµµµkN

v


2. Packaging for oranges is tested using a machine that exerts ay = 20 m/s2 and ax = 3 m/s2, simultaneously. Select the correct FBD and kinetic diagram for this condition.

A) B)

C) D)

= •

may

max

W

Ry

Rx

= • max

W

Ry

Rx

= •

may

max

W

Ry

= •

may

Ry

y

x

Concept Quiz - continued

Group Problem SolvingGiven: Each block has a mass m. The

coefficient of kinetic friction at all surfaces of contact is µ. A horizontal force P is applied to the bottom block.

Find: Draw the free-body and kinetic diagrams of each block.

Plan: 1) Define an inertial coordinate system.2) Draw the free-body diagrams for each block, showing

all external forces.3) Draw the kinetic diagrams for each block, showing the

inertial forces.


The friction forces oppose the motion of each block relative to the surfaces on which they slide.

1) An inertial x-y frame can be defined as fixed to the ground.

2) Draw the free-body diagram of each block:

3) Draw the kinetic diagram of each block:

y

x

Block B:

T

NBFfB = µNB

WB = mg y

x

P

NBWA = mg

FfB = µNB

FfA = µNANA

Block A:

Block B:maB = 0

Block A:maA

Group Problem Solving - continued

Attention Quiz

1. Internal forces are not included in an equation of motion analysis because the internal forces are_____.

A) equal to zeroB) equal and opposite and do not affect the calculations C) negligibly smallD) not important

2. A 10 lb block is initially moving down a ramp with a velocity of v. The force F is applied to bring the block to rest. Select the correct FBD.

A) B) C)µµµµk10

10

N

F

µµµµk10

10

N

F

µµµµkN

10

N

F

F

v


Textbook Problem 13-

Announcements

• HW 14 and 15 due Wed, May 12


Equations of Motion: Rectangular Coordinates Section 13.4

Today’s Objectives:Students will be able to apply Newton’s second law to determine forces and accelerations for particles in rectilinear motion.

In-Class Activities:• Reading quiz• Applications• Equations of motion using

rectangular (Cartesian) Coordinates

• Concept quiz• Group problem solving• Attention Quiz

Reading Quiz

1. In dynamics, the friction force acting on a moving object is always

A) in the direction of its motion. B) a kinetic friction.

C) a static friction. D) zero.

2. If a particle is connected to a spring, the elastic spring force is expressed by F = ks . The “s” in this equation is the

A) spring constant.

B) unstretched length of the spring.

C) difference between stretched and unstretched length.D) stretched length of the spring.


Applications

If a man is pushing a 100 lb crate, how large a force F must he exert to start moving the crate?

What would you have to know before you could calculate the answer?

Objects that move in any fluid have a drag force acting on them. This drag force is a function of velocity.

If the ship has an initial velocity vo and the magnitude of the opposing drag force at any instant is half the velocity, how long it would take for the ship to come to a stop if its enginesstop?



Equation of MotionThe equation of motion, F = ma, is best used when the problem requires finding forces (especially forces perpendicular to the path), accelerations, velocities or mass. Remember, unbalanced forces cause acceleration!

Three scalar equations can be written from this vector equation.The equation of motion, being a vector equation, may be expressed in terms of its three components in the Cartesian (rectangular) coordinate system as

�F = ma or �Fx i + �Fy j + �Fz k = m(ax i + ay j + az k)

or, as scalar equations, �Fx = max ,�Fy = may , and �Fz = maz .

Procedure for Analysis

• Free Body Diagram

Establish your coordinate system and draw the particle’s free body diagram showing only external forces. These external forces usually include the weight, normal forces, friction forces, and applied forces. Show the ‘ma’ vector (sometimes called the inertial force) on a separate diagram.

Make sure any friction forces act opposite to the direction of motion! If the particle is connected to an elastic spring, a spring force equal to ks should be included on the FBD.


• Equations of Motion

If the forces can be resolved directly from the free-body diagram (often the case in 2-D problems), use the scalar form of the equation of motion. In more complex cases (usually 3-D), a Cartesian vector is written for every force and a vector analysis is often best.

A Cartesian vector formulation of the second law is�F = ma or�Fx i + �Fy j + �Fz k = m(ax i + ay j + az k)

Three scalar equations can be written from this vector equation.You may only need two equations if the motion is in 2-D.

Procedure for Analysis - continued

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

• Kinematics

Any of the tools learned in Chapter 12 may be needed to solve a problem. Make sure you use consistent positive coordinate directions as used in the equation of motion part of the problem.

Procedure for Analysis - continued


Plan: Since both forces and velocity are involved, this problem requires both the equation of motion and kinematics. First, draw free body diagrams of A and B. Apply the equation of motion . Using dependent motion equations, derive a relationship between aA and aB and use with the equation of motion formulas.

Example

Given: WA = 10 lbWB = 20 lbvA = 2 ft/sµk = 0.2

Find: vA when A has moved 4 feet.

Free-body and kinetic diagrams of B:

=

2T

WB mBaB

yy ma F =↓+ �

BBB amTW =− 2

BaT 2.3220220 =− (1)

Apply the equation of motion to B:

Example - continued


0==↑+ � yy maF

lbWN A10==

lbNFk 2== µ

xx maF =←+ �

AA amTF =−

AaT 2.32102 =− (2)

Apply the equations of motion to A:

=

WA

T

N

mAaA

F = µkN

Free-body and kinetic diagrams of A:y

x

Example - continued

Constraint equation:sA + 2 sB = constant

orvA + 2 vB = 0

ThereforeaA + 2 aB = 0aA = -2 aB (3)

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

A

B

sA

sB

Datums

Now consider the kinematics.

Example - continued


Now combine equations (1), (2), and (3).

lbT 33.7322 ==

→→→→=−= 22 16.1716.17 sft

sftaA

Now use the kinematic equation:

)(222oAAAoAA ssavv −+=

)4)(16.17(22 22 +=Av

→→→→= sftvA 9.11

Example - continued

Concept Quiz

1. If the cable has a tension of 3 N, determine the acceleration of block B.

A) 4.26 m/s2 down B) 4.26 m/s2 up

C) 8.31 m/s2 down D) 8.31 m/s2 up

10 kg

4 kg

µµµµk=0.4

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

5 kg60 N

•30°°°°


Group Problem Solving

Given: The 400 kg mine car is hoisted up the incline. The force in the cable isF = (3200t2) N. The car has an initial velocity ofvi = 2 m/s at t = 0.

Find: The velocity when t = 2 s.

Plan: Draw the free-body diagram of the car and apply the equation of motion to determine the acceleration. Apply kinematics relations to determine the velocity.

1) Draw the free-body and kinetic diagrams of the mine car:

Solution:

Since the motion is up the incline, rotate the x-y axes.

θ = tan-1(8/15) = 28.07°

Motion occurs only in the x-direction.

=

θx

y

W = mg F

N

ma



2) Apply the equation of motion in thex-direction:

3) Use kinematics to determine the velocity:

+ � Fx = max => F – mg(sinθ) = max

=> 3200t2 – (400)(9.81)(sin 28.07°) = 400a

=> a = (8t2 – 4.616) m/s2

a = dv/dt => dv = a dt

� dv = � (8t2 – 4.616) dt, v1 = 2 m/s, t = 2 s

v – 2 = (8/3t3 – 4.616t)� = 12.10 => v = 14.1 m/s

v

v1

t

02

0


Attention Quiz

2. A 10 lb particle has forces of F1= (3i + 5j) lb and F2= (-7i + 9j) lb acting on it. Determine the acceleration of the particle.

A) (-0.4 i + 1.4 j) ft/s2 B) (-4 i + 14 j) ft/s2

C) (-12.9 i + 45 j) ft/s2 D) (13 i + 4 j) ft/s2

1. Determine the tension in the cable when the 400 kg box is moving upward with a 4 m/s2

acceleration.

A) 2265 N B) 3365 N

C) 5524 N D) 6543 N

T

60

a = 4 m/s2


Textbook Problem 13-15

The driver attempts to tow the crate using a rope that has a tensile strength of 200 lb. If the crate is originally at rest and has a weight of 500 lb, determine the greatest acceleration it can have if the coefficient of static friction between the crate and the road is µµµµs=0.4, and the coefficient of kinetic friction is µµµµk=0.3.

Announcements

• New schedule– Mon, May 10 13.1-4– Wed, May 12 13.5 HW #14,15 due– Fri, May 14 13.6 HW #16 due– Mon, May 17 14.1-3– Wed, May 19 Mid-term #2 HW #17 due

• Mid-term #2 covers 12.6-10, 13.1-6

– Fri, May 21 14.5-6 HW #18 due– Mon, May 24 … Back on schedule

• Note: We will not cover section 14.4 (scratch HW#19)


Equations of Motion: Normal and Tangential Components - Section 13.5

Today’s Objectives:Students will be able to apply the equation of motion using normal and tangential coordinates.

In-Class Activities:• Reading quiz• Applications• Equation of motion in n-t

coordinates• Concept quiz• Group problem solving• Attention quiz

Reading Quiz

2. The positive n direction of the normal and tangential coordinates is ____________.

A) normal to the tangential component B) always directed toward the center of curvature C) normal to the bi-normal componentD) All of the above

1. The “normal” component of the equation of motion is written as ΣFn = man, where ΣFn is referred to as the _______.

A) impulse B) centripetal force

C) tangential force D) inertia force


Applications

Race tracks are often banked in the turns to reduce the frictional forces required to keep the cars from sliding at high speeds.

If the car’s maximum velocity and a minimum coefficient of friction between the tires and track are specified, how can we determine the minimum banking angle (θ) required to prevent the car from sliding?

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

Knowing the radius of orbit of the satellite, how can we determine the required speed of the satellite to maintain this orbit?



Normal and Tangential Components

When a particle moves along a curved path, it may be more convenient to write the equation of motion in terms of normal andtangential coordinates.

The normal direction (n) always points toward the path’s center of curvature. In a circle, the center of curvature is the center of the circle.

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

Equations of Motion

This vector equation will be satisfied provided the individual components on each side of the equation are equal, resulting in the two scalar equations: �Ft = mat and �Fn = man .

Here �Ft & �Fn are the sums of the force components acting in the t & n directions, respectively.

Since the equation of motion is a vector equation , �F = ma,it may be written in terms of the n & t coordinates as

�Ftut + �Fnun = mat + man

Since there is no motion in the binormal (b) direction, we can also write �Fb = 0.


Normal and Tangential Accelerations

The tangential acceleration, at = dv/dt, represents the time rate of change in the magnitude of the velocity. Depending on the direction of �Ft, the particle’s speed will either be increasing or decreasing.

The normal acceleration, an = v2/ρ, represents the time rate of change in the direction of the velocity vector. Remember, an always acts toward the path’s center of curvature. Thus, �Fn will always be directed toward the center of the path.

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

[1 + ( )2]3/2dydx

d2ydx2

Solving Problems with n-t Coordinates• Use n-t coordinates when a particle is moving along a

known, 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 positive n-direction). The tangential acceleration (at) may act in either 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 an = v2/ρ


ExampleGiven:At the instant θ = 60°, the boy’s

center of mass G is momentarily at rest. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords.

Find: The boy’s speed and the tension in each of the two supporting cords of the swing when θ = 90°.

1) Since the problem involves a curved path and finding the force perpendicular to the path, use n-t coordinates. Draw the boy’s free-body and kinetic diagrams.

2) Apply the equation of motion in the n-t directions.3) Use kinematics to relate the boy’s acceleration to his speed.

Plan:

Solution:1) The n-t coordinate system can be established on the

boy at some arbitrary angle θ. Approximating the boy and seat together as a particle, the free-body and kinetic diagrams can be drawn.

T = tension in each cordW = weight of the boy

n

t

man

mat

Kinetic diagram

W

n

t

2T θ

Free-body diagram

=

Example - continued


2) Apply the equations of motion in the n-t directions.

Note that there are 2 equations and 3 unknowns (T, v, at). One more equation is needed.

Using an = v2/ρ = v2/10, w = 60 lb, and m = w/g = (60/32.2),

we get: 2T – 60 sin θ = (60/32.2)(v2/10) (1)

(b) �Ft = mat => W cos θ = mat

=> 60 cos θ = (60/32.2) at

Solving for at: at = 32.2 cos θ (2)

(a) �Fn = man => 2T – W sin θ = man

Example - continued

v dv = at ds where ds = ρ dθ = 10 dθ3) Apply kinematics to relate at and v.

This v is the speed of the boy at θ = 90°. This value can be substituted into equation (1) to solve for T.

2T – 60 sin(90°) = (60/32.2)(9.29)2/10T = 38.0 lb (the tension in each cord)

=> v dv = 32.2 cosθ ds = 32.2 cosθ (10 dθ )

=> v dv = 322 cosθ dθ

=> = 322 sinθ => v = 9.29 ft/s

��90

600

v

90

60

v2

2

Example - continued


Concept Quiz1. A 10 kg sack slides down a smooth surface. If the normal

force on the surface at the flat spot, A, is 98.1 N (↑) , the radius of curvature is ____.

A) 0.2 m B) 0.4 m

C) 1.0 m D) None of the above A

v=2m/s

2. A 20 lb block is moving along a smooth surface. If the normal force on the surface at 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

ρρρρ=7 ft

A

Group Problem Solving

Given: A 200 kg snowmobile with rider is traveling down the hill. When it is at point A, it is traveling at 4 m/s and increasing its speed at 2 m/s2.

Find: The resultant normal force and resultant frictional force exerted on the tracks at point A.

Plan: 1) Treat the snowmobile as a particle. Draw the free-body and kinetic diagrams.

2) Apply the equations of motion in the n-t directions.3) Use calculus to determine the slope and radius of

curvature of the path at point A.


Solution:

W = mg = weight of snowmobile and passengerN = resultant normal force on tracksF = resultant friction force on tracks

1) The n-t coordinate system can be established on the snowmobile at point A. Treat the snowmobile and rider as a particle and draw the free-body and kinetic diagrams:

tn

matman

t

θ

θn

N

FW

=


2) Apply the equations of motion in the n-t directions:

� Ft = mat => W sinθ – F = mat

� Fn = man => W cos θ – N = man

Using W = mg and an = v2/ρ = (4)2/ρ=> (200)(9.81) cos θ – N = (200)(16/ρ)

=> N = 1962 cosθ – 3200/ρ (1)

Using W = mg and at = 2 m/s2 (given)

=> (200)(9.81) sin θ – F = (200)(2)

=> F = 1962 sinθ – 400 (2)



3) Determine ρ by differentiating y = f(x) at x = 10 m:

Determine θ from the slope of the curve at A:

From Eq.(1): N = 1962 cos(56.31) – 3200/19.53 = 924 NFrom Eq.(2): F = 1962 sin(56.31) – 400 = 1232 N

y = -5(10-3)x3 => dy/dx = (-15)(10-3)x2 => d2y/dx2 = -30(10-3)x

tan θ = dy/dx

θ = tan-1 (dy/dx) = tan-1 (-1.5) = 56.31°

x = 10 mθ

dy

dx

ρ = =[1 + ( )2]3/2dy

dxd2ydx2

[1 + (-1.5)2]3/2

-0.3x = 10 m


Attention Quiz

1. The tangential acceleration of an objectA) 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.

2. The ball has a mass of 20 kg and a speed of v = 30 m/s at the instant it is at its lowest point. Determine the tension in the cord at this instant.

A) 1596 N B) 1796 N

C) 1996 N D) 2196 N

θθθθ10 m

v = 30m/s



Announcements


Equations of Motion: Cylindrical Coordinates Section 13.6

Today’s Objectives:Students will be able to analyze the kinetics of a particle using cylindrical coordinates.

In-Class Activities:• Reading quiz• Applications• Equations of motion using

cylindrical coordinates• Angle between radial and

tangential directions• Concept quiz• Group problem solving• Attention quiz

Reading Quiz

1. The normal force which the path exerts on a particle is always perpendicular to the _________.

A) radial line B) transverse direction

C) tangent to the path D) None of the above

2. Friction forces always act in the __________ direction.

A) radial B) tangential

C) transverse D) None of the above


Applications

The forces acting on the 100-lb boy can be analyzed using the cylindrical coordinate system.

If the boy slides down at a constant speed of 2 m/s, can we find the frictional force acting on him?

When an airplane executes the vertical loop shown above, the centrifugal force causes the normal force (apparent weight) on the pilot to be smaller than her actual weight.

If the pilot experiences weightlessness at A, what is the airplane’s velocity at A?



Equations of Motion: Cylindrical Coordinates

This approach to solving problems has some external similarity to the normal & tangential method just studied. However, the path may be more complex or the problem may have other attributes that make it desirable to use cylindrical coordinates.

Equilibrium equations or “Equations of Motion” in cylindrical coordinates (using r, θ , and z coordinates) may be expressed in scalar form as:

� Fr = mar = m(r – rθ2)� Fθ = maθ = m(rθ – 2rθ)� Fz = maz = mz

.

. ...

....

Equations of Motion - continued

Note that a fixed coordinate system is used, not a “body-centered” system as used in the n – t approach.

If the particle is constrained to move only in the r – θ plane (i.e., the z coordinate is constant), then only the first two equations are used (as shown below). The coordinate system in such a case becomes a polar coordinate system. In this case, the path is only a function of θ.

� Fr = mar = m(r – rθ2)� Fθ = maθ = m(rθ – 2rθ)

.. .

....


Normal and Tangential ForcesIf a force P causes the particle to move along a path defined by r = f (θ ), the normal force N exerted by the path on the particle is always perpendicular to the path’s tangent. The frictional force F always acts along the tangent in the opposite direction of motion. The directions of N and F can be specified relative to the radial coordinate by using angle ψ .

Determination of the angle ψ

The angle ψ, defined as the angle between the extended radial line and the tangent to the curve, can be required to solve some problems. It can be determined from the following relationship.

θθψ

ddrr

drr d ==tan

If ψ is positive, it is measured counterclockwise from the radial line to the tangent. If it is negative, it is measured clockwise.


Example

Plan: Draw a FBD. Then develop the kinematic equations and finally solve the kinetics problem using cylindrical coordinates.

Solution: Notice that r = 2rc cosθ, therefore:r = -2rc sinθ θr = -2rc cosθ θ2 – 2rc sinθ θ

.

.. ..

.

.

Given: The ball (P) is guided along the vertical circular path.W = 0.5 lb, θ = 0.4 rad/s,θ = 0.8 rad/s2, rc = 0.4 ft

Find: Force of the arm OA on the ball when θ = 30°.

...

Free Body Diagram: Establish the inertial coordinate system and draw the particle’s free body diagram. Notice that the radial acceleration is negative.

°=30θ

t r

θ

n

=

θ

ψθ

θ

θ

θ

mg

Ns NOA

θma

rma

Example - continued


Kinematics: at θ = 30°r = 2(0.4) cos(30°) = 0.693 ft

r = -2(0.4) sin(30°)(0.4) = -0.16 ft/s

r = -2(0.4) cos(30°)(0.4)2 – 2(0.4) sin(30°)(0.8) = -0.431 ft/s2..

.

Acceleration components are

ar = r – rθ2 = -0.431 – (0.693)(0.4)2 = -0.542 ft/s2

aθ = rθ + 2rθ = (0.693)(0.8) + 2(-0.16)(0.4) = 0.426 ft/s2

.. .

....

tan ψ = r/(dr/dθ) where dr/dθ = -2rc sinθ

tan ψ = (2rc cosθ)/(-2rc sinθ) = -1/tanθ ∴ ψ = 120°

Example - continued

Kinetics:

� Fr = mar

Ns cos(30°) – 0.5 sin(30°) = (-0.542)

Ns = 0.279 lb

0.532.2

� Fθ = maθ

NOA + 0.279 sin(30°) – 0.5 cos(30°) = (0.426)

NOA = 0.3 lb

0.532.2

Example - continued


Concept Quiz

2. If needing to solve a problem involving the pilot’s weight at Point C, select the approach that would be best.

A) Equations of Motion: Cylindrical Coordinates B) Equations of Motion: Normal & Tangential Coordinates C) Equations of Motion: Polar Coordinates D) No real difference – all are badE) Toss up between B and C

1. When a pilot flies an airplane in a vertical loop of constant radius r at constant speed v, his apparent weight is maximum at

A) Point A B) Point B (top of the loop)C) Point C D) Point D (bottom of the loop)

r A

B

C

D

Group Problem SolvingGiven: A plane flies in a vertical loop

as shown.vA = 80 ft/s (constant)W = 130 lb

Find: Normal force on the pilot at A.

)2cos(600 θ−=rKinematics:r θθ ·· )2sin(1200=r · θθθθ ···· )2sin(1200)2cos(2400 2 +=

At A )90( °=θ 0=r·

Solution:

Plan: Determine θ and θ from the velocity at A and by differentiating r. Solve for the accelerations, and apply the equation of motion to find the force.

. ..


· ··2

22 67.42)133.0(2400)180sin(1200)180cos(2400s

ftr −=−=°+°= θθ··

222 33.53)133.0(60067.42

sftrra r −=−−=−= θ·· ·

Therefore r θθ··· rrv A =+= 22 )()(

Since ftr 600= at A,s

rad133.060080 ==θ

·

Since vA is constant, aθ = rθ + 2rθ = 0 => θ = 0· ·.. ..


Notice that the pilot would experience weightlessness when his radial acceleration is equal to g.

Kinetics: � Fr = mar => -mg – N = mar

N = -130 – (53.3) => N = 85.2 lb13032.2

Free Body Diagram & Kinetic Diagram

θ

rN

mg

rma

=



Attention Quiz

1. For the path defined by r = θ2 , the angle ψ at θ = .5 rad is

A) 10 º B) 14 º

C) 26 º D) 75 º

2. If r = θ2 and θ = 2t, find the magnitude of r and θ when t = 2 seconds.

A) 4 cm/sec, 2 rad/sec2 B) 4 cm/sec, 0 rad/sec2

C) 8 cm/sec, 16 rad/sec2 D) 16 cm/sec, 0 rad/sec2

···


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