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Forces (Part the First)
Chapter 5 Forces (Part the First)
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Forces “A push or a pull” Can cause acceleration
Aristotelian (wrong) vs Newtonian (mostly right) mechanics Limitations of Newtonian mechanics Objects moving very fast (special relativity) Very small objects (quantum mechanics) Very small objects moving very fast (quantum field theory) 6/28/04
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Forces Unit: kg m/s2 [Newtons] Force is a vector quantity
Direction and magnitude The sum of the forces on an object is Fnet 6/28/04
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The Four Fundamental Interactions
6/28/04
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The Practical Forces of Nature
Gravity Restoring forces Springs, etc. Constraining forces Tension, normal force Dissipative forces Friction, air resistance 6/28/04
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Newton’s 1st Law If there are no forces acting on an object, the velocity of the object will not change The “law of inertia” An object in motion will remain in motion, and object at rest will remain at rest, until acted upon by an outside force Then why do things seem to slow down when we stop pushing them? 6/28/04
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Newton’s First Law vs Hollywood
Is this possible? 6/28/04
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Inertial Reference Frame
A reference frame in which Newton’s laws hold Non-accelerating The earth is approximately an inertial reference frame Rotation, motion around the sun/galaxy/universe Coriolis effect 6/28/04
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Newton’s 2nd Law The net force on a body is equal to the product of the body’s mass and the acceleration of the body Can break force into components Treat each direction independently 6/28/04
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What is Mass? Intrinsic property
NOT dependent on an objects size, weight or density Proportionality constant between force and acceleration Difference between pushing a tennis ball and a bowling ball 6/28/04
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Newton’s 3rd Law When object A exerts a force on object B, object B exerts a force on object A of equal magnitude and opposite direction “Action and reaction” A B 6/28/04
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Forces: Gravity An object of mass m in freefall will experience a force: Weight is equal to magnitude of Fg A peculiarity in Halliday, Resnick, and Walker Note: Mass is intrinsic, while weight depends on where you are. They are NOT the same thing. 6/28/04
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Demo: Newton’s Carts Two people push off each other
The force is the same in magnitude, opposite in direction FB on A FA on B A B 6/28/04
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Demo: Newton’s Carts Double the mass on one cart. What happens? A B
FB on A FA on B A B 6/28/04
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Forces: Normal “Normal” in the mathematical sense meaning perpendicular When an object pushes against a surface, the surface pushes back perpendicular to its surface with a force N N Fg 6/28/04
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Forces: Friction Force that resists motion when an object slides over a surface Always opposite to the direction of motion We’ll discuss more next chapter v Ffr 6/28/04
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Forces: Tension Force exerted by a rope
Always points in the same direction as the rope Ropes are assumed to be massless unless otherwise specified v T Ffr 6/28/04
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Tension (Continued) T T 1. A rope can only pull (cannot push)
Frope 1. A rope can only pull (cannot push) 2. Objects at the ends of a taut rope have the same magnitude velocity and acceleration T 3. Frictionless and massless pulleys only change direction of rope, not the tension. 6/28/04
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Free Body Diagrams The first step in solving any force problem:
Sketch the object in question Draw an arrow for each force acting on the object Label each force Indicate the direction of acceleration off to the side (acceleration is NOT a force) F2 F1 a F3 6/28/04
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Net Force A free body diagram is used to calculate the net force on one object. Fnet= m a Note: The two equal forces in Newton’s Third Law are on different objects. They don’t appear on the same free body diagram. 6/28/04
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How to Solve a Force Problem
Draw a free body diagram Guess at forces of unknown magnitude or direction Break forces into components Sum of all the forces in one direction = mass * acceleration in that direction i.e. Fnet,x=m ax Repeat step 3 for each direction Solve for unknown quantities 6/28/04
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Example: A Standing Person
Say a man has a mass of 60 kg. Find the normal force exerted on him by the ground. Draw a diagram N Fg=mg a = 0 2) Sum the forces in the y direction Note: We know the direction of the normal force from the diagram 6/28/04
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Example: A Standing Person
Now say our 60 kg man is jumping with a = 2 m/s2 Draw a diagram a=2 m/s2 N Fg=mg 2) Sum the forces in the y direction Notice: The normal force is variable 6/28/04
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Example: Ring on a Table
Ring centered on round table and strings with weights are attached 0.5 kg How much force is needed to balance forces so the ring stays centered (zero acceleration)? 30° 1 kg 6/28/04
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Solution: Draw a Free-Body Diagram
A free body diagram shows all the forces operating on a single object 9.8 N 4.9 N F1 F2 F3 F3,x F3,y 30° q We had to make an educated guess for where to put F3. If we guessed wrong, that’s OK. We will just pick up a minus sign 6/28/04
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Solution: Break Forces Into Components
F3,x F3,y 30° F1 : F2 : 6/28/04
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Solution: Summing Forces
In the x-direction: In the y-direction: 4.9 N F1=9.8 N F2 F3 F3,x F3,y 30° 6/28/04
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Solution: Does it work? The magnitude of F3 is: The angle of F3 is: q
F1=9.8 N F2 F3 F3,x F3,y 30° q The angle of F3 is: Does it work? 6/28/04
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Pig on Frictionless Inclined Plane
q a Fg Assume the pig has a mass of m and the angle of the ramp is q. Find N and a. Need to consider forces || and to plane… 6/28/04
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Pig Forces: Break Into Components
Acceleration: θ N Fg a y x Fg,y Fg,x Normal Force: Gravity: 6/28/04
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Pig Forces: Sum Forces y-direction: x-direction: y N x a θ Fg,y Fg
Fg,x x-direction: 6/28/04
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Pig on a Plane: Sanity Check
q a Fg N = mg cos θ and a = g sin θ θ = a = 0 and N = mg θ = 90° a = g and N = 0 6/28/04
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Cart on a Plane Now let’s add friction. A cart of mass m is stationary on an inclined plane of angle q. Find Ffr and N. q Ffr Fg N a = 0 6/28/04
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Cart on a Plane: Components
Normal Force: N Ffr Fg,y Fg Gravity: q Fg,x x y Friction: Once again, use a “tilted” coordinate system 6/28/04
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Cart: Summing the Forces
y-direction: q Ffr Fg N Fg,y Fg,x x-direction: 6/28/04
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Does it Work? If we replace the normal force and friction with another force, the cart should stay stationary after the ramp is removed. The mass of the cart is 1 kg and the angle of the ramp is 30° 6/28/04
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It is now attached to another bowling ball. What does it read?
Tension A 16 lb. bowling ball is hung from a rope attached to a scale which is attached to a wall. The scale reads 16 lbs. It is now attached to another bowling ball. What does it read? A) 0 B) 16 lbs C) 32 lbs 6/28/04
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Tension in Two Dimensions
Consider the mass: m mg T1 Summing forces: 6/28/04
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Tension in 2-D (Continued)
Consider the knot: 60o 30o T1 T2 T3 m 60o 30o T1 T3 T2 y-direction: x-direction: 6/28/04
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Tension in 2-D (Continued)
What we know: Jumping through the algebraic hoops (left as an exercise for the ambitious student): 6/28/04
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What happens when we accelerate?
Example: Fuzzy Dice With no acceleration Frope Fg Fnet,y = may = 0 Frope- Fg = 0 Frope= -Fg = mg What happens when we accelerate? 6/28/04
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Fuzzy Dice… Find q: e.g. a=g θ=45° x direction: q Frope y direction:
Fg Frope q x direction: y direction: e.g. a=g θ=45° 6/28/04
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Another Example: m1 m2 m1 m2 Find the acceleration of this system:
Start with free body diagrams of both blocks m1 m2 m1 FG N T a1 x y m2 T FG a2 x y q Recall: a1 = a2 = a 6/28/04
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Solution: m1 Summing forces in the x direction: a1 T N
FG N T a1 x y Summing forces in the y direction: 6/28/04
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Solution m2 T FG a2 x y Summing forces in the y direction: 6/28/04
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Solution What we have: The tensions must be equal to each other m1 m2
6/28/04
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Fan Cart With a Sail: Will It Move?
A) Yes, in the same direction as without the sail B) Yes, in the opposite direction C) No D) No Clue 6/28/04
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