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No Announcements
class Friday HW# 14 and 15 due Wed, May 12
Newtons Law of Motion (Equation of Motion)
Todays 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.
Sections 13.1-3
In-Class Activities: Reading quiz Applications Newtons laws of motion Newtons law of gravitational attraction Equation of motion for a particle or system of particles Concept quiz Group problem solving Attention quiz
Engr222 Spring 2004 Chapter 13
1
Reading Quiz
1. Newtons second law can be written in mathematical form as F = ma. Within the summation of forces F, ________ are(is) not included. A) external forces C) internal forces B) weight D) All of the above
2. The equation of motion for a system of n-particles can be written as Fi = miai = maG, where aG indicates _______. A) summation of each particles acceleration B) acceleration of the center of mass of the system C) acceleration of the largest particle D) None of the above
Applications
The motion of an object depends on the forces acting on it. A parachutist relies on the atmospheric drag resistance force to limit his velocity. Knowing the drag force, how can we determine the acceleration or velocity of the parachutist at any point in time?
Engr222 Spring 2004 Chapter 13
2
Applications - continued
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?
Newtons Laws of Motion
The motion of a particle is governed by Newtons 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.
Engr222 Spring 2004 Chapter 13
3
Newtons Laws of Motion - continued
The first and third laws were used in developing the concepts of statics. Newtons second law forms the basis of the study of dynamics. Mathematically, Newtons second law of motion can be written F = ma where 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. Newtons second law cannot be used when the particles speed approaches the speed of light, or if the size of the particle is extremely small (~ size of an atom).
Newtons Law of Gravitational Attraction
Any two particles or bodies have a mutually attractive gravitational force acting between them. Newton postulated the law governing this gravitational force as F = G(m1m2/r2) where F = force of attraction between the two bodies, G = universal constant of gravitation , m1, m2 = mass of each body, and r = 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.
Engr222 Spring 2004 Chapter 13
4
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 Newtons 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 = mg where g is the acceleration due to gravity.
Units: SI System vs. FPS System
SI 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 earths surface, g can be taken as g = 9.81 m/s2. W (N) = m (kg) g (m/s2) => N = kgm/s2 FPS System: In the FPS system of units, weight is a base unit and 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 earths surface, g is approximately 32.2 ft/s2. m (slugs) = W (lb)/g (ft/s2) => slug = lbs2/ft
Engr222 Spring 2004 Chapter 13
5
Equation of Motion
The motion of a particle is governed by Newtons 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 = ma where 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 particles freebody diagram, showing all forces acting on the particle. Next, draw the kinetic diagram, showing the inertial force ma acting 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 earths surface, we typically assume our inertial frame to be fixed to the earth. We neglect any acceleration effects from the earths rotation. For problems involving satellites or rockets, the inertial frame of reference is often fixed to the stars.
Engr222 Spring 2004 Chapter 13
6
System of Particles
The 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) Newtons 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!
Engr222 Spring 2004 Chapter 13
7
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 forces applied to the particle. Resolve forces into their appropriate components. 3) Draw the kinetic diagram, showing the particles 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 relations to 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 k. Find: Draw the free-body and kinetic diagrams of the crate. Plan: 1) Define an inertial coordinate system. 2) Draw the crates free-body diagram, showing all external forces applied to the crate in the proper directions. 3) Draw the crates kinetic diagram, showing the inertial force vector ma in the proper direction.
Engr222 Spring 2004 Chapter 13
8
Solution: 1) An inertial x-y frame can be defined as fixed to the ground. 2) Draw the free-body diagram of the crate:
y
Example - continued
The weight force (W) acts through the crates center of mass. T is the tension x force in the cable. The normal force (N) is perpendicular to the surface. The friction force (F = uKN) acts in a direction F = uKN opposite to the motion of the crate. N 3) Draw the kinetic diagram of the crate:
W = mg
T 30
ma
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.
Concept Quiz
1. The block (mass = m) is moving upward with a speed v. Draw the FBD if the kinetic friction coefficient is k.
mg mg
A)
N
kN
B)
N
v
kN
C)
mg kmg N
D) None of the above.
Engr222 Spring 2004 Chapter 13
9
Concept Quiz - continued
2. Packaging for oranges is tested using a machine that exerts ay = 20 m/s2 and ax = 3 m/s2, simultaneously. Select the y correct FBD and kinetic diagram for this condition. A)
Rx W may
B)
max Rx
W
x
=
Ry
=
Ry
max
C) =
Ry
may
D)
W
may
Ry
=
max
Group Problem Solving
Given: Each block has a mass m. The coefficient of kinetic friction at all surfaces of contact is . 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.
Engr222 Spring 2004 Chapter 13
10
Group Problem Solving - continued
1) An inertial x-y frame can be defined as fixed to the ground. 2) Draw the free-body diagram of each block:
y x
Block B: WB = mg T NB
y x
NB P
Block A: WA = mg FfB = NB FfA = NA
FfB = NB
The friction forces oppose the motion of each block relative to the surfaces on which they slide. 3) Draw the kinetic diagram of each block:
NA
Block B: maB = 0 maA
Block A:
Attention Quiz
1. Internal forces are not included in an equation of motion analysis because the internal forces are_____. A) B) C) D) equal to zero equal and opposite and do not affect the calculations negligibly small not important
F v
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)
10 F k10 N
B)
10
F k10 N
C)
10
F kN N
Engr222 Spring 2004 Chapter 13
11
Textbook Problem 13-
Announcements
HW 14 and 15 due Wed, May 12
Engr222 Spring 2004 Chapter 13
12
Equations of Motion: Rectangular Coordinates
Section 13.4
Todays Objectives: Students will be able to apply Newtons 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. C) a static friction. B) a kinetic 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) B) C) D) spring constant. unstretched length of the spring. difference between stretched and unstretched length. stretched length of the spring.
Engr222 Spring 2004 Chapter 13
13
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?
Applications - continued
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 engines stop?
Engr222 Spring 2004 Chapter 13
14
Equation of Motion
The 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) Fx = max , Fy = may , and Fz = maz .
or, as scalar equations,
Procedure for Analysis
Free Body Diagram Establish your coordinate system and draw the particles 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.
Engr222 Spring 2004 Chapter 13
15
Procedure for Analysis - continued
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
Kinematics 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. 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.
Engr222 Spring 2004 Chapter 13
16
Example
Given: WA = 10 lb WB = 20 lb vA = 2 ft/s k = 0.2 Find: vA when A has moved 4 feet. 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 - continued
Free-body and kinetic diagrams of B: 2T
=
WB Apply the equation of motion to B: + Fy = ma y WB 2 T = mB aB 20 20 2 T = 32. 2 aB
(1)
mBaB
Engr222 Spring 2004 Chapter 13
17
Example - continued
y x
Free-body and kinetic diagrams of A: WA T N F = kN
=
m A aA
+
Apply the equations of motion to A: + Fx = ma x Fy = ma y = 0 N = W = 10 lb A F T = mA aA 10 2 T = 32 2 a . A
(2)
F = N = 2 lb k
Example - continued
Now consider the kinematics. Constraint equation: sA + 2 sB = constant or vA + 2 vB = 0 Therefore aA + 2 aB = 0 aA = -2 aB
(3)
sA A
Datums
sB
B
(Notice aA is considered positive to the left and aB is positive downward.)
Engr222 Spring 2004 Chapter 13
18
Example - continued
Now combine equations (1), (2), and (3). T = 22 = 7. 33 lb 3 ft a = 17. 16 fts 2 = 17.16 s2 A Now use the kinematic equation: v A
2 2 = v + 2 aA ( sA soA ) oA
v 2 = 2 2 + 2 (17.16 )(4 ) A ft v = 11. 9 s A
Concept Quiz
1. If the cable has a tension of 3 N, determine the acceleration of block B. A) 4.26 m/s2 down C) 8.31 m/s2 down B) 4.26 m/s2 up D) 8.31 m/s2 up
10 kg k=0.4 4 kg
2. Determine the acceleration of the block. A) 2.20 m/s2 B) 3.17 m/s2
5 kg
30
C) 11.0 m/s2
D) 4.26 m/s2
60 N
Engr222 Spring 2004 Chapter 13
19
Group Problem Solving
Given: The 400 kg mine car is hoisted up the incline. The force in the cable is F = (3200t2) N. The car has an initial velocity of vi = 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.
Group Solving Problem - continued
Solution: 1) Draw the free-body and kinetic diagrams of the mine car: W = mg y x N F ma
=
Since the motion is up the incline, rotate the x-y axes. = tan-1(8/15) = 28.07 Motion occurs only in the x-direction.
Engr222 Spring 2004 Chapter 13
20
Group Problem Solving - continued
2) Apply the equation of motion in the x-direction: + Fx = max => F mg(sin) = max => 3200t2 (400)(9.81)(sin 28.07) = 400a => a = (8t2 4.616) m/s2 3) Use kinematics to determine the velocity: a = dv/dt => dv = a dt
v v1 t
dv =
0
(8t2 4.616) dt, v1 = 2 m/s, t = 2 s
2 0
v 2 = (8/3t3 4.616t)
= 12.10 => v = 14.1 m/s
Attention Quiz
1. Determine the tension in the cable when the 400 kg box is moving upward with a 4 m/s2 acceleration. A) 2265 N C) 5524 N B) 3365 N D) 6543 N
60
a = 4 m/s2
T
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 C) (-12.9 i + 45 j) ft/s2 B) (-4 i + 14 j) ft/s2 D) (13 i + 4 j) ft/s2
Engr222 Spring 2004 Chapter 13
21
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 s=0.4, and the coefficient of kinetic friction is k=0.3.
Announcements
New schedule
Mon, May 10 Wed, May 12 Fri, May 14 Mon, May 17 Wed, May 19 13.1-4 13.5 13.6 14.1-3 Mid-term #2 HW #14,15 due HW #16 due HW #17 due
Mid-term #2 covers 12.6-10, 13.1-6
Fri, May 21 Mon, May 24
14.5-6 HW #18 due Back on schedule
Note: We will not cover section 14.4 (scratch HW#19)
Engr222 Spring 2004 Chapter 13
22
Equations of Motion: Normal and Tangential Components - Section 13.5
Todays 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
1. The normal component of the equation of motion is written as Fn = man, where Fn is referred to as the _______. A) impulse C) tangential force B) centripetal force D) inertia force
2. The positive n direction of the normal and tangential coordinates is ____________. A) B) C) D) normal to the tangential component always directed toward the center of curvature normal to the bi-normal component All of the above
Engr222 Spring 2004 Chapter 13
23
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 cars maximum velocity and a minimum coefficient of friction between the tires and track are specified, how can we determine the minimum banking angle () required to prevent the car from sliding?
Applications - continued
Satellites are held in orbit around the earth by using the earths gravitational pull as the centripetal force the force acting to change the direction of the satellites velocity. Knowing the radius of orbit of the satellite, how can we determine the required speed of the satellite to maintain this orbit?
Engr222 Spring 2004 Chapter 13
24
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 and tangential coordinates. The normal direction (n) always points toward the paths 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
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 Here Ft & Fn are the sums of the force components acting in the t & n directions, respectively. 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 . Since there is no motion in the binormal (b) direction, we can also write Fb = 0.
Engr222 Spring 2004 Chapter 13
25
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 particles speed will either be increasing or decreasing. The normal acceleration, an = v2/, represents the time rate of change in the direction of the velocity vector. Remember, an always acts toward the paths 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
=
dy [1 + ( )2]3/2 dx d2 y dx2
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 ndirection). 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/
Engr222 Spring 2004 Chapter 13
26
Given:At the instant = 60, the boys 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 boys speed and the tension in each of the two supporting cords of the swing when = 90. Plan: 1) Since the problem involves a curved path and finding the force perpendicular to the path, use n-t coordinates. Draw the boys free-body and kinetic diagrams. 2) Apply the equation of motion in the n-t directions. 3) Use kinematics to relate the boys acceleration to his speed.
Example
Solution: 1) The n-t coordinate system can be established on the boy at some arbitrary angle . Approximating the boy and seat together as a particle, the free-body and kinetic diagrams can be drawn. Free-body diagram Kinetic diagram n n 2T
Example - continued
=
man mat
W
t T = tension in each cord W = weight of the boy
t
Engr222 Spring 2004 Chapter 13
27
2) Apply the equations of motion in the n-t directions. (a) Fn = man => 2T W sin = man Using an = v2/ = v2/10, w = 60 lb, and m = w/g = (60/32.2), we get: 2T 60 sin = (60/32.2)(v2/10) (1) (b) Ft = mat => W cos = mat => 60 cos = (60/32.2) at Solving for at: at = 32.2 cos (2) Note that there are 2 equations and 3 unknowns (T, v, at). One more equation is needed.
Example - continued
Example - continued
3) Apply kinematics to relate at and v. v dv = at ds where ds = d = 10 d => v dv = 32.2 cos ds = 32.2 cos (10 d )
v 90
=> v dv = 322 cos d
2 => v = 322 sin 2
0 60
90
=> v = 9.29 ft/s
60
This v is the speed of the boy at = 90. This value can be substituted into equation (1) to solve for T. 2T 60 sin(90) = (60/32.2)(9.29)2/10 T = 38.0 lb (the tension in each cord)
Engr222 Spring 2004 Chapter 13
28
Concept Quiz
1. 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 C) 1.0 m B) 0.4 m
v=2m/s
D) None of the above
A
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 C) 10.6 ft/s B) 9.6 ft/s D) 12.6 ft/s
A =7 ft
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 freebody 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.
Engr222 Spring 2004 Chapter 13
29
Group Problem Solving - continued
Solution: 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: W
F N n
=
t
man n
mat t
W = mg = weight of snowmobile and passenger N = resultant normal force on tracks F = resultant friction force on tracks
Group Problem Solving - continued
2) Apply the equations of motion in the n-t directions: Fn = man => W cos N = man Using W = mg and an = v2/ = (4)2/ => (200)(9.81) cos N = (200)(16/) => N = 1962 cos 3200/ Ft = mat => W sin F = mat Using W = mg and at = 2 m/s2 (given) => (200)(9.81) sin F = (200)(2) => F = 1962 sin 400
(2) (1)
Engr222 Spring 2004 Chapter 13
30
Group Problem Solving - continued
3) Determine by differentiating y = f(x) at x = 10 m: y = -5(10-3)x3 => dy/dx = (-15)(10-3)x2 => d2y/dx2 = -30(10-3)x dy [1 + ( )2]3/2 [1 + (-1.5)2]3/2 dx = = d2 y -0.3 x = 10 m dx2 Determine from the slope of the curve at A: dy tan = dy/dx = tan-1
x = 10 m
dx From Eq.(1): N = 1962 cos(56.31) 3200/19.53 = 924 N From Eq.(2): F = 1962 sin(56.31) 400 = 1232 N
(dy/dx) = tan-1 (-1.5) = 56.31
Attention Quiz
1. The tangential acceleration of an object A) represents the rate of change of the velocity vectors 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 l...

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Princeton - CS - 325

Pucker Experiments with Real-Time Generation of WhistlingAvi Flamholz 15 May 2007This paper represents my own work in accordance with University regulations.Avi Flamholz11IntroductionWhistled tones have low entropy, are easy to separate

Princeton - COS - 423

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Princeton - COS - 598

OutlineHistoryHistorically, parallel architectures tied to programming models Divergent architectures, with no predictable pattern of growth.Drivers of Parallel Computing Trends in Supercomputers for Scientific Computing Evolution and Converge

Princeton - COS - 522

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Princeton - COS - 598

Princeton - PH - 106

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Princeton - PH - 106

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Princeton - PH - 106

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Princeton - PH - 106

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Princeton - PH - 106

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Princeton - PH - 106

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Princeton - PH - 106

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Princeton - PH - 106

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Princeton - PH - 106

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Princeton - PH - 106

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Princeton - PH - 106

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Princeton - PH - 106

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Maryland - BSCI - 338

Electroreception Electric eld properties Electroreception evolution Bioelectric eld production Electrolocation ElectrocommunicationElectric eld Generated by separation of positive (electron decit) and negative charges (e.g. a dipole such as

Princeton - CS - 511

COS 511: Foundations of Machine LearningRob Schapire Scribe: Qiang Huang Lecture #17 April 8, 20031Review of Bayes RuleIn the previous lecture, we talked about modeling and estimating probability distributions. In this generative approach, we

Princeton - COS - 217

Princeton University COS 217: Introduction to Programming Systems Fall 2004 Midterm Exam AnswersQuestion 1 (a) B, F, I Explanation: A. B. C. D. E. F. G. H. I. Valid. Prints the address of i as a decimal integer. Compiletime error: i is not a pointer

Princeton - COS - 511

COS 511: Foundations of Machine LearningRob Schapire Scribe: Melissa Carroll Lecture #19 April 20, 20061ReviewWhen we left o last time, we established the following problem: Given: a space X where |X| = N examples x1 , ., xm X xi D features

Walla Walla University - ENGR - 480

4100 SERIESSpecifications: Widths up to 12 (305) Lengths up to 6 (1,829) The 1 (25) diameter drive pulley. Load capacity up to 80 lbs.* (36kg). Belt speeds to 255 ft./min. (78 M/min.).Applications: Metal Forming Metal Stamping Machined Pa

Walla Walla University - ENGR - 480

Chapter 1: Getting StartedQuick StartThis example is not intended to tell you everything you need to know about programming and starting-up a complex control system. It is only intended to give you an opportunity to demonstrate to yourself and oth

Texas A&M - CPSC - 689

Trainable Videorealistic Speech AnimationTony Ezzat Gadi Geiger Center for Biological and Computational Learning Massachusetts Institute of Technology Tomaso PoggioAbstractWe describe how to create with machine learning techniques a generative, v

Texas A&M - CPSC - 206

CPSC 206 Spring 2007 Lab 10 The Assignment Extending Lab 9 aka Part IIApril 19, 2007This is due by use of turnin by midnight Friday, April 27, 2007.Part I aka Lab 9This program and documentation will have the following portions 1. The C sourc

Princeton - COS - 318

COS 318: Operating Systems OS Structures and System CallsJaswinder Pal Singh Computer Science Department Princeton University (http:/www.cs.princeton.edu/courses/cos318/)Outline! ! !Protection mechanisms OS structures System and library calls

Princeton - COS - 217

MIDTERM EXAMCS 217 October 28, 1999Name: Precept: Honor Code:Score:Problem 1 2 3 4 5 6 7 Total Score Max 15 5 10 15 5 10 10 7011. Number Systems (a) Translate the following decimal numbers to binary, and binary numbers to decimal. Assume a

Princeton - COS - 318

Todays TopicsCOS 318: Operating Systems DisksKai Li Computer Science Department Princeton University (http:/www.cs.princeton.edu/courses/cos318/)Disk controller Disk layout Technology trend Disk performance Disk scheduling RAID2A Typical Dis

Texas A&M - CPSC - 206

CPSC 206 Spring 2007 Lab 2 The AssignmentJanuary 26, 2007This is due by use of turnin by midnight Friday, February 2, 2007. In this lab you will be required to write a main() function by editing and extending the le lab-02.c and furnish a docum

Texas A&M - CPSC - 310

Texas A&M - CPSC - 310

Summer2007 CPSC310-300/CPSC603-300, Dr. Salih Yurtta. [hw-01] s1. Get project names that are supplied by every existing supplier in the database. r1 SN (Suppliers) r2 JN,SN (Shipments) r3 r2 r1 r4 P rojects 1 r3 r JN ame (r4) JName12. Ge

Texas A&M - CPSC - 310

Summer2007 CPSC310-300/CPSC603-300, Dr. Salih Yurtta. [hw-02] s1. Get project names that are supplied by every existing supplier in the database. {t | u P rojects(t[JN ame] = u[JN ame]) v Suppliers w Shipments(v[SN ] = w[SN ] w[JN ] = u[JN ])}

Ithaca College - GRAD - 0607

Graduate Catalog 2006-2007DirectorySchool of Business | Roy H. Park School of Communications | School of Health Sciences and Human Performance | School of Humanities and Sciences | School of MusicBoard of Trustees

Texas A&M - ENGR - 482

Engineering 482Environmental Ethics and LawReferences Chapter 9- Engineering Ethics Course class notes Engineering 482 website EPA website www.epa.gov Engineering Society Codes of Ethics Engineering Ethics CD-ROMAdditional Environmental La

Maryland - CMSC - 434

Visual RepresentationEvan Golub / Ben Bederson / Saul GreenbergBeyond Simple Screen DesignWhat are the characteristics of good representations? What are the characteristics of good information visualization? What role do metaphors play? How can

LSU - M - 2065

Math 2065 Section 1 Review Exercises for Exam IThe syllabus for Exam I is Chapter 1 (Sections 1, 2) and Chapter 2. You should bring your own paper for the exam, since space will (in general) not be provided on the exam paper for answers. You will ne

Penn State - ASTRO - 293

ASTRO 293 Observational Astronomy Laboratory Spring 2003, Section 1Instructor: Dr. Christopher Palma Oce: 508 Davey Lab Phone: 863-6091 Email: cpalma@astro.psu.edu Time: T 7:35 9:30 Location: 216 Osmond Oce Hours: T 35/F 23 or by appointment OPTIO

Penn State - ASTRO - 480

Chemerda LecuturesJohn Mather, Nobel Prize WinnerTouching the Universe - Measuring the Cosmic Backgroud Radiation, Past and Future, Mon. Mar. 17, 4pm, 102 Thomas Public Lecture: From the Big Bang to the Nobel Prize and on to the James Webb Space Te

Penn State - ASTRO - 485

Deep Xray Surveys with Chandra and XMMNewtonCDFN1.95 MsCDFS0.94 MsGOODSSHDFN GOODSN~ 448 sq. arcmin ~ 582 sourcesBrandt et al. (02) Alexander et al. (03)~ 390 sq. arcmin ~ 369 sourcesGiacconi et al. (02)Superb multiwavelength suppor

Princeton - MC - 019

. . .Telegram NoDated: A p r i l 6 , 1944.SECSTATE WASIIIBGTOB. .: *IYATPCII..Jackpot 1 4 0 and Carib..Ipollowing- summarizes statement from Rreakers : S i t u a t i o n i n Germany r a p i d l y approaching climax w i t h end o f

Princeton - MC - 019

Woxking In the Darkstory of the Central fntelligence Agency is an extremely readable book Ita contents, however, obviously were garnered outside and not inside our countrys most publicized secret organization. Moreover, mme of its highly questionabl

Princeton - MC - 019

~--February 19, 1963_._ -.. .I. .IJohn S. D, Eisenhower . 300 C a r l i s l e StreetGettysburg , Pennsyl-yania Dear John :!I appreciate your l e t t e r of JanGary 25 w i t h regard to the Guatemalan matter,You may have not

Penn State - MATH - 250

Answers to Exam I, Math 250, Spring 2009 Multiple Choices: Problem Answer Partial Credit problems: Problem 7. (a). This is a separable equation. 1 dy = y2 2t dt, 1 = t2 + C, y 1 . t2 1 y= t2 1 . +C No. 1 C No. 2 A No. 3 B No. 4 C No. 5 C No. 6 A

Penn State - MATH - 231

Princeton - MC - 019

FRIDAY. APRXL14, 19671:454:39Take train to Princeton (Trenton)Arrive Trenton-.--.-_-* -:-Reservations made at Princeton Zn for night .- .of A p r i l 14 for Mr. and W s . . Dulles,,63 15D i n n e r , w i t h informal rec

Princeton - MC - 019

iI**.8TUESDAY, 23 JANUARY 1962( I n New York City)-Met With M s Eleanor Roosevelt at Channel r. A.M. W lWW and had a taping of t h e program f o r "PROSPECTS OF " N 3 D . " Had a preliminary t a l k With Mrs. Roosevelt and Miss Braud

Princeton - MC - 019

II 4.The Executive's Bookshelf 0 .I4e.y-zm%-.7~.-. -. -.*s4 * :ix-: -'I'I._.etb.ck to WWurrte the eoup. L typld of t h e p r u l v i t y 011 t h e put o the peoph who f One of t h e momt t r a g l d y Ill-fatcd m o w Hitlcr. ?