applying newton’s laws - fcps 5 [compatibility mode].pdfgoals for chapter 5 • to use and apply...
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PowerPoint® Lectures forUniversity Physics, Twelfth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by James Pazun
Chapter 5
Applying Newton’s Laws
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Goals for Chapter 5
• To use and apply Newton’s First Law
• To use and apply Newton’s Second Law
• To study friction and fluid resistance
• To consider forces in circular motion
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Homework
5.61, 5.81, 5.83, 5.93, 5.95 Bonus: 5.122
Problems are like AP free responses
Use the example problems to help.
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Overview
Newton’s three laws of motion (the basis for classical mechanics) can be stated very simply.
Applying them to situations, however, can be very tricky indeed.
We will begin with objects in equilibrium, then move onto objects not in equilibrium.
We will study friction and its effect on an object’s motion.
We will study object in uniform circular motion.
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Particles in EquilibriumEquilibrium when an object is at rest or at a constant velocity in an inertial reference frame
Vector Form
We will most often use the component form:
Some equilibrium problems may seem complicated, but it is important to remember that all problems dealing with particles in equilibrium are done the same way.
0=∑ F
∑ = 0xF ∑ = 0yF 5.2
5.1
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Using Newton’s First Law when forces are in equilibrium
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1-D equilibriumA gymnast with mass mG=50.0kg suspends herself from the lower end of a hanging rope. The upper end of the rope is attached to the gymnasium ceiling.
What is the gymnast’s weight?
What force (magnitude and direction) does the rope exert on her?
What is the tension at the top of the massless rope?
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Draw a sketch of the situation
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Draw a free body diagram for each body in the situation
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We can now answer the questions:What is the gymnast’s weight?
What force (magnitude and direction) does the rope exert on her?
What is the tension at the top of the massless rope?
If the rope has a weight of 120N, find the tension at each end of the rope.
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Two-dimensional equilibriumIn the figure below a car engine with weight w hangs from a chain that is linked at ring O to two other chains, one fastened to the ceiling and the other to the wall. Find the tensions in each of these three chains, assuming that w is given and the weights of the ring and chains are negligible.
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Free body diagrams of each body:
Engine Ring
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Sum your forces for each dimension for each body
Solve:
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A car on an inclined planeA car rests on the slanted tracks of a ramp leading to a car-transporter trailer. Only a cable attached to the car and to the frame of the trailer prevents the car from rolling backward off the trailer. If the weight of the car is w find the tension (T) in the cable and the force with which the tracks push on the car’s tires (n).
Solve on the board
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Consider a mass on a plane tied to a cord over a pulleyBlocks of granite are being hauled u pa 15o slope out of a quarry. For environmental reasons, dirt is also being dumped into the quarry to fill up old holes. You have been asked to find a way to use this dirt to move the granite out more easily. You design a system in which a granite block on a cart with steel wheels (w1) is pulled uphill on steel rails by a dirt-filled bucket (w2) dropping vertically into the quarry. Ignoring friction in the pulley and wheels and the weight of the cable, determine how the weights w1 and w2 must be related in order for the system to move with constant speed.
Solve for w2.
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Problem Sets
1: 1,4,7,10,13
2: 2,5,8,11,14
3: 3,6,9,12,15
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Dynamics of ParticlesWe’ve covered the most common types of equilibrium problems, so now we are ready to explore what happens if the velocity is not constant, dynamics.In dynamics we apply Newton’s second law to an accelerating body We will use the component form:
The problem solving strategy is very similar to that for particles in equilibrium.Be sure to understand that ma is not a force, it is equal to the magnitude of a force.
xx maF =∑ yy maF =∑
amF =∑ 5.3
5.4
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Beware incorrect free-body diagrams—Figure 5.6
• Only gravity acts on the apple.
• Refer to Figure 5.6 and Problem-Solving Strategy 5.2
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Straight-line motion with constant forceAn iceboat is at rest on a perfectly frictionless horizontal surface. A steady wind is blowing so that 4.0s after the iceboat is released, it attains a velocity of 6.0 m/s. The iceboat and rider have a combined mass of 200 kg.What constant horizontal force FW does the wind exert on the iceboat?
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Straight line motion with a time varying forceWith the same iceboat, lets consider that once the iceboat starts to move, its position as a function of time is
Find the force FW exerted by the wind as a function of time in this case.
What is the force at time t = 3.0 s?
For what times is the force zero?
Positive?
Negative?
( ) ( )22 3
31.2 0.20m mssx t t= −
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Consider motion against frictionSuppose the wind is once again blowing steadily in the +x-direction as in the previous problem, so that the iceboat has a constant acceleration ax = 1.5 m/s2. Now, however, there is a constant horizontal friction force with magnitude 100N that opposes the motion of the iceboat. In this case, what force Fw must the wind exert on the iceboat?
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Tension in an elevator cable
An elevator and its load have a total mass of 800 kg. The elevator is originally moving downward at 10.0 m/s; it slows to a stop with constant acceleration in a distance of 25.0 m.
Find the tension T in the supporting cable while the elevator is being brought to a rest.
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Apparent weight in an accelerating elevator
A 50.0 kg woman stands on a bathroom scale while riding in the previous problems elevator. What is the reading on the scale?
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Apparent Weight
In general:
Apparent Weight (n) = m( g + ay)
Where ay is the acceleration of the object.
When ay is positive, the apparent weight is larger (you would feel pushed against the bottom of the elevator)
When ay is negative, the apparent weight is less. Take this to the extreme, when ay = g, the object is weightless.
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Acceleration down a hill
A toboggan loaded with vacationing students (total weight w) slides down a long, snow-covered slope. The hill slopes at a constant angle α and the toboggan is so well waxed that there is virtually no friction.
What is its acceleration?
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Two bodies with the same acceleration
You are pushing a 1.00 kg food tray through the cafeteria line with a constant 9.0 N force. As the tray moves, it pushes in turn on a 0.50 kg carton of milk. The tray and carton slide on a horizontal surface that is so greasy that friction can be neglected.
Find the acceleration of the system and the horizontal force that the tray exerts on the carton.
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Two bodies with the same magnitude of acceleration
An air-track glider with mass m1 moves on a level, frictionless air track in the physics lab. The glider is connected to a lab weight with mass m2 by a light, flexible, nonstretching string that passes over a small frictionless pulley.
Find the acceleration of each body and the tension in the string.
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Problem Sets
1: 16,19,22,25
2: 17,20,23,26
3: 18,21,24
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Frictional Forces
We have seen several problems that have involved forces such as the normal force or friction force, contact forces.
These forces require direct contact with the body to affect its motion.
In this section we will concern ourselves with the force of friction and air resistance or drag.
Frictional forces always act to oppose an objects motion.
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Frictional forces, kinetic and static
• Friction can keep an object from moving or slow its motion from what we last calculated on an ideal, frictionless surface.
• Microscopic imperfections cause non-ideal motion.
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Moving Mass Demo: Force Before and After Motion
Analyze the applied force before motion.
Analyze the applied force during constant velocity
Where is force maximized?
How does the “constant velocity” force compare to the maximum force before the motion?
These forces are forces required to overcome friction.
Before motion is static friction, after is kinetic friction.
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Kinetic and Static Friction
When a body slides or rests on a surface, we can represent the contact force on the body by the components of force perpendicular and parallel to the surface.
The perpendicular force is the normal force, the parallel is the friction force.
By definition:
fn
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Kinetic Friction ForceActs on a body from the surface it is moving across.
Increases when normal force increases
Therefore it is directly proportional to the objects mass or the acceleration due to gravity:
The k denotes kinetic friction, and the Greek letter mu is the coefficient of kinetic friction.
The coefficient of kinetic friction varies for each object and surface.
A coefficient has no units.
nf kk μ=
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Coefficients of friction
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Underlying cause of Friction
Page 172, below CAUTION, second full paragraph.
Microscopic electrical attraction forces
Kinetic Friction always varies, as number of bonds varies.
Lubrication helps to limit these electrical attractions
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Static Friction Force
Occurs when there is no relative motion.
Like kinetic friction, it is proportional to the normal force and hence, mass or acceleration due to gravity.
Notice that the force is less than or equal to the product of the coefficient of static friction and the normal force.
When would the force of static friction be maximized?
• Think back to the friction demo.
nf ss μ≤
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Applied force is proportional until the object moves--Figure 5.19&20
• Notice the transition between static and kinetic friction.
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Notice the effect of friction in horizontal motionA delivery company has just unloaded a 500 N crate full of home exercise equipment in your driveway. You find that to get it started moving toward your garage, you have to pull with a horizontal force of magnitude 230 N. Once it “breaks loose” and starts to move, you can keep it moving at a constant velocity with only 200N.
What are the coefficients of static and kinetic friction?
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Notice the effect of friction in horizontal motion
What is the friction force if the applied force for the previous problem was 50N?
What type of friction is acting?
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The angle at which tension is applied mattersSuppose you try to move the 500 N crate by tying a rope around it and pulling upward on the rope at an angle of 30o above the horizontal. Assume the coefficient of kinetic friction is 0.40.
How hard do you have to pull to keep the crate moving with constant velocity?
Is this easier or harder than puling horizontally?
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Consider the toboggan ride two more timesConsider a toboggan sliding down a slope with a constant velocity. Derive an expression for the slope angle in terms of w and μs.
What if the same toboggan with the same coefficient of friction is sliding downhill, but on a steeper slope? This time the toboggan accelerates. Derive an expression for the acceleration in terms of g, w, μs, and the angle of the slope, α.
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Motion with Rolling Friction
A typical car weighs about 12,000 N. If the coefficient of rolling friction is μr = 0.015, what horizontal force is needed to make the car move with constant speed on a level road?
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Fluid Resistance and Terminal SpeedThe moving body exerts a force on the fluid to push it out if its way, by Newton’s 3rd Law the fluid exerts an equal and opposite force back on the body, Fluid Resistance.
The force of resistance can be classified into what happens at low speeds and high speeds.
Low Speed High Speed
k and D are proportionality constants that vary with the size and shape of the object and the density of the fluid/air.
kvf = 2Dvf =5.7 5.8
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Effects of Fluid ResistanceBecause this force varies with velocity, the acceleration from this force is not constant.Rock and pond example on page 179.As the rock falls it accelerates. This increase in velocity also increases the fluid resistance force, decreasing the net force causing the acceleration.Figure 5.6 shows the acceleration, velocity and position vs. time graphs for the motion of the rock, both with and without fluid resistance.
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Effects of Fluid Resistance: Analysis
Lets consider down to be a positive velocity, since there are no x-components to the rocks motion, Newton’s 2nd gives:
ΣFy = mg + (-kvy) = may
When the rock 1st starts to move, v = 0 and ay = g. As the speed increases, so does the resisting force until:
mg – kvy = 0 or vy = mg/k
This is the terminal velocity of the rock
The terminal velocity of an object is the velocity when the resistance force equals the applied force, or when ΣF = 0 (equilibrium)
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Effects of Fluid Resistance
For an object falling through the air with a high velocity:
Dmgv t =
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For the sport of skydiving, terminal speed is vitalFor a human body falling through air in a spread eagle position, the numerical value of the constant D is about 0.25 kg/m.
Find the terminal speed of an 80 kg skydiver.
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Problem Sets
1: 27,30,33,36,39,42,45,48
2: 28,31,34,37,40,43,46
3: 29,32,35,38,41,44,47
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The dynamics of uniform circular motionIn uniform circular motion the object’s acceleration and hence net force are both centripetal:
The observed motion of the object is in a circle, so the observed acceleration is centripetal.
2 2
2
2 2
2
4
4
rad
rad
v RaR T
v RF ma m mR T
π
π
= =
= = =∑
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Force in uniform circular motionA sled with a mass of 25.0 kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 5.00 m rope to a post set in the ice. Once given a push, the sled revolves uniformly in a circle around the post. If the sled makes five complete revolutions every minute, find the force F exerted on it by the rope.
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The conical pendulumAn inventor who dares to be different proposes to make a pendulum clock using a pendulum bob with mass m at the end of a thin wire of length L. Instead of swinging back and forth, the bo moves in a horizontal circle with constant speed v, with the wire making a constant angle β with the vertical direction. Assuming that the angle β is known, find the tension F in the wire and the period T.
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Rounding a flat curveThe BMW Z4 roadster is rounding a flat, unbanked curve with radius R. If the coefficient of friction between tires and road is μs, what is the maximum speed vmax at which the driver can take the curve without sliding?
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Rounding a banked CurveFor a car traveling at a certain speed, it is possible to bank a curve at just the right angle so that no friction at all is needed to maintain the car’s turning radius. Then a car can round the curve even on wet ice with Teflon tires. Bobsled racing depends on this same idea. An engineer proposes to rebuild the curve in the previous example so that a car moviig at speed v can safely make the turn even if there is no friction. At what angle β should the curve be banked?
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Uniform circular motion in a vertical circleA passenger on a carnival Ferris wheel moves in a vertical circle of radius R with constant speed v. Assuming that the seat remains upright during the motion, derive expressions for the force the seat exerts on the passenger at the top of the circle and at the bottom.
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Problem Sets
1: 49,52,55,58
2: 50,53,56,59
3: 51,54,57,60
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Homework
5.61, 5.81, 5.83, 5.93, 5.95 Bonus: 5.122
Problems are like AP free responses
Use the example problems to help.