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6. Chapter Dynamics I: Motion Along a Line
Chapter Goal: To learn how to solve problems about motion in a straight line.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Student Learning Objectives Ch. 6
To draw and make effective use of free-body diagrams. To recognize and solve simple equilibrium problems. To distinguish mass, weight, and apparent weight. To learn and use simple models of friction. To apply the full strategy for force and motion problems to problems in single-particle dynamics.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Equilibrium
An object on which the net force is zero is said to be in equilibrium. Static equilibrium: object is at rest. Dynamic equilibrium: moving along a straight line with constant velocity. Both are identical from a Newtonian perspective because the net force and the acceleration are zero.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Problem-Solving Strategy: Equilibrium Problems
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Problem-Solving Strategy: Equilibrium Problems
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example Equilibrium Problem (#31)
A 500 kg piano is being lowered into position by a crane while 2 people steady it with ropes pulling to the side. Bobs rope pulls left, 150 below horizontal, with 500 N of tension. Ellens rope pulls right, 250 below horizontal. a. What tension must Ellen maintain in her rope to keep the piano descending at a steady speed? b. What is the tension in the main cable?
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example Problem (#31) Freebody Diagram
T3
T1 is Bobs side, T2 is Ellens side, T3 is the main cable. Same system for the angles
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Problem #31 - Apply Newtons 1st Law
T3
F y = 0 , F x = 0
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Problem #31 Solve and assess
T2 = 533 N, T3 = 5.25 x 103 N. The cable must supports the weight of the piano (4900 N) plus the added downward components of the tension in the supporting ropes.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Using Newtons 2nd Law: Workbook exercises 5-12
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Using Newtons 2nd Law: Workbook exercises Answers 5-6
-F3 sin
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Using Newtons 2nd Law: Workbook exercises Answers: 7-12
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Problem-Solving Strategy for Newtons 2nd Law Problems
1. Use the problem-solving strategy outlined for Newtons 1st Law problems to draw the free body diagram and determine known quantities. 1. Use Newtons Law in component form to find the values for any individual forces and/or the acceleration. 2. If necessary, the objects trajectory (time, velocity, position, acceleration) can be determined by using the equations of kinematics. 3. Reverse # 2 and 3 if necessary.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example Dynamics Problem
A 75-kg snowboarder starts down a 50-m high, 100 slope on a frictionless board. What is his speed at the bottom?
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example Dynamics Problem Visualize Freebody Diagram
A 75-kg snowboarder starts down a 50-m high, 100 slope on a frictionless board. What is his speed at the bottom?
n
FG
Find v1 a
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example Dynamics Problem Newtons 2nd Law in component form to solve for acceleration
A 75-kg skier starts down a 50m high, 100 slope on a frictionless board. What is his speed at the bottom?
n
Fy = may = 0 Fx = max a = 1.7 m/s2
Supports earlier statement that a = g sin
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
FG
Example Dynamics Problem Use kinematics to find speed. Is time important?
A 75-kg snowboarder starts down a 50-m high, 100 slope on a frictionless board. What is his speed at the bottom? Note (the slope is 50 m high, not long!) v1 = 31.3 m/s. Thats about 60 mph!
Find v1
a = 1.7 m/s2, from previous
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example Problem #9 and more
The figure shows a force acting on a 2.0-kg object moving along the x-axis. The object is at rest at the origin at t=0. What is the velocity and acceleration of the object at t = 2, 4, and 6s?
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example Problem #9 and more
The figure shows a force acting on a 2.0-kg object moving along the xaxis. The object is at rest at the origin at t=0. What is the velocity and acceleration of the object at t = 2, 4, and 6s? a(2) = 2 m/s2, v(2) = 4m/s a(4) = -1 m/s2 , v(4) = 5 m/s a (6) = 0 m/s2, v(6) = 4m/s
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Mass and Gravity
Mass is a scalar quantity that describes the amount of matter in an object. Mass is an intrinsic property of an object. The force of gravity is an attractive, long-range inverse square force between any two objects.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The Earth and the Moon
The figure shows the moon (m1) and the earth (m2). The earth is approximately 80 times as massive as the moon. The red arrow shown is the force that the earth exerts on the moon (F2on1 ). The moon also exerts a force on the earth, F1on2, shown in blue (not to scale!). The magnitude of this force is: a. b. c. d. about 80 smaller F2on1 somewhat smaller than F2on1 Equal to F2on1 Not related to F2on1.
moon
?
F1on 2
earth
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The Earth and the Moon
The figure shows the moon (m1) and the earth (m2). The earth is approximately 80 times as massive as the moon. The red arrow shown is the force that the earth exerts on the moon (F2on1 ). The moon also exerts a force on the earth, F1on2, shown in blue (not to scale!). The magnitude of this force is: a.Equal to F2on1
earth moon
F1on 2
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Consider an object of mass m, on or near the surface of a planet. We can write the gravitational force even more simply as:
gravitational (weight )force
where the quantity g is defined to be M, R represent the mass and radius of the planet. The weight force is an intrinsic property of an object and does not have a unique value. The direction of the gravity vector defines true vertical.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Apparent Weight
Apparent weight w, is a contact force (e.g. T, n, or Fsp), which can be thought of as what the scale says, although there is not always a scale. If object and scale are in vertical static or dynamic equilibrium w = FG = mg. If object and scale accelerate vertically, w mg. It must be calculated using Newtons 2nd Law. The use of w is optional, as long as you know which force is the apparent weight.
Copyright 2008 Pearson Education, Inc., publishing as Pearson elevator Addison-Wesley.
An suspended by a cable is moving upward and slowing to a stop. As it does, your apparent weight is:
A. less than your true weight. B. equal to your true weight. C. more than your true weight. D. zero. Hint: Draw the freebody diagram, and determine direction of acceleration to establish net force.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
An elevator suspended by a cable is moving upward and slowing to a stop. As it does, your apparent weight is:
A. less than your true weight. Net force on you is down, just like the elevator. Therefore weight force is greater than normal.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Workbook Problem # 18
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Workbook Problem # 18 - ans
Explanation: S is the normal force which is the apparent weight. From Ns 2nd Law: S= ma + |mg|
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
speed and direction are not relevant.
Apparent Weight Problem
A 50-kg woman gets in a 1000-kg elevator at rest. The elevator has a scale in it (I hate when that happens). As the elevator begins to move, the scale reads 600 N for the first 3 seconds. a. How far has the elevator moved in those 3 s? b. In which direction?
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Apparent Weight Problem
b. Which direction has the elevator going? Since the scale reads heavy (I hate when that happens), acceleration is up. Shes either going up and speeding up, or going down and slowing down. Only one of these is physically possible, given the problem statement!
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Apparent Weight Problem
How far has the elevator moved in those 3 s? Time is important so: y = a t12 y = 9.9 m
a0 = 2.2 m/s2, determined from Ns 2nd Law y 1, t 1 = 3 s v 1
0m
y0 = t0 = v0 = 0
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Friction
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Kinetic Friction
Experiments show that the kinetic friction force is nearly constant and proportional to the magnitude of the normal force.
where the proportionality constant k is called the coefficient of kinetic friction.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Static Friction
The box is in static equilibrium, so the static friction must exactly balance the pushing force:
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Static friction
An object remains at rest as long as fs < fs max The object slips when fs = fs max A static friction force fs > fs max is not physically possible. fs max >fk for the same surfaces
where the proportionality constant s is called the coefficient of static friction.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Rolling Friction
Rolling friction acts much like kinetic friction, but values for ur are much less than those for uk.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
A model of friction
motion indicates motion relative to the two surfaces the max value static friction, fs max occurs at the very instant the object begins to move (which often means 1 ns before, for problem-solving purposes.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Rank order, from largest tosmallest, the size of the friction forces tofa in fthese five different e situations. The box and the floor are made of the same materials in all situations.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Rank order, from largest to smallest, the size of the friction forces in these five different situations. The box and the floor are made of the same materials in all situations.
fb > fc = fd = fe > fa.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example kinetic friction problem
A 75-kg snowboarder starts down a 50-m high, 100 slope s = 0.12 and k = 0.06. What is his speed at the bottom? This is the same problem as before only the slope is no longer frictionless. Before, the velocity was 31.3 m/s. How does friction change that
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example kinetic friction problem
A 75-kg snowboarder starts down a 50-m high, 100 slope s = 0.12 and k = 0.06 What is his speed at the bottom?
n fk FG
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example kinetic friction problem
A 75-kg snowboarder starts down a 50-m high, 100 slope on a frictionless board. What is his speed at the bottom? v1 = 25.4 m/s. Friction acts to slow him down, although not by much.
Find v1
a, from previous
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example static friction problem
A truck is hauling a crate when it starts up a 10.0 hill. The coefficients of friction are s = 0.35, and k = 0.15, respectively. What is the maximum acceleration the truck can have as he goes up the hill?
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example static friction problem
n fs
Fg
known = 10 us = .35 uk = .15
find amax
When does amax occur? Find n using Newtons 2nd law in the y direction. Find amax using Newtons 2nd law in the y direction.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Example static friction problem
n fs
Fg
known = 10 us = .35 uk = .15
find amax
amax =1.68 m/s/s
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Keep the picture up
A person is trying to judge whether a picture of mass 1.10 kg is properly positioned by pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between picture and wall is 0.660. What is the minimum amount of pressing force required? Draw a freebody diagram. In which direction is the normal force in this problem? Does it have anything to do with the weight of the picture?
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Keep the picture up Freebody diagram
fbd - picture fs n Fpush Knowns m = 1.10 kg s = 0.660 Find Fpush
Fg Forces which are usually x are y in this problem and vice versa (with the exception of gravity). Newtons Laws still work. Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Keep the picture up Freebody diagram
fbd - picture fs n Fpush Knowns m = 1.10 kg s = 0.660 Find Fpush Newtons Law in the x direction tells us that n = Fpush but nothing else about the value of either. Moving right along to the ydirection:
Fg
Fy = may = 0 = fs FG or fs = mg. No matter how hard you press, the picture will not levitate up. Fact. However, if you dont push hard enough.
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Keep the picture up - answer
fbd - picture fs n Fpush Knowns m = 1.10 kg s = 0.660 Find Fpush The minimum value of n must be the value that allows fsmax to be equal to the weight of the picture: Fy = 0 = fsmax FG or s |n| = mg n = Fpush = 16.3 N
FG
Copyright 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

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National Taiwan University - COMPUTER S - 101

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National Taiwan University - COMPUTER S - 101

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National Taiwan University - COMPUTER S - 101

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National Taiwan University - COMPUTER S - 101

8(1) 20x1 1 1 182,000()25,0007,000150,000(2)182,000+110,000(NCI)=292,000292,000-250,000()=42,000() 920x1 1 1 ()20,00020,00020x3 1 1 ()90,00020,00010,00060,00020x3 12 31 3,0003,00020x4 12 31 3,0003,00020x5 1 1 150,000150,0001

National Taiwan University - COMPUTER S - 101

1 1.(1)(2)$210,000180,000240,00090,00050,000(80,000)(270,000)210,000$020,00030,00010,00020,0002. ,$1,000,000$1,000,000$1,500,000500,000$2,000,000$(1,000,000)$(50,000)(200,000)500,000(100,000)(300,000)(850,000)$(1,000,000)$