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Transcript
CP Physics Chapter 4 Newton’s Laws Force • Force (F) is a push or a pull • Measured in Newtons (N) for SI, pounds (lb) in US. (4.45 N = 1 lb) •It has magnitude & direction, so it’s a vector Force in general • Name some forces • How strong are they? • Contact vs at-adistance (or field) Force in motion • Forces act on things, or “bodies” • Forces can require physical contact or be “at a distance” • Forces are drawn as vectors and added as vectors • To draw forces (vectors) acting on a body, isolate (or “free”) the body from everything—only the forces acting on the body are drawn. Called a Free Body Diagram (FBD)—see samples • Force, acceleration, and mass (F, a, m) can be used to explain motion with Newton’s Laws of Motion Newton’s First Law of Motion • Law of Inertia • Net Force (F) a vector sum • Constant velocity means F = 0, or ..\My Videos\inertia\train-hurricane.mp4 equilibrium, ..\My Videos\inertia\planeland.mp4 ..\..\Videos\inertia\FirstSnowcarcrashfun.mp4 or a = 0 Bed of nails 1st Law: Inertia • Old: A body at rest will stay at rest, A body in motion will stay in motion, Until acted on by an outside force • New: • A body at rest will stay at rest, A body with velocity will maintain its velocity, Until acted on by a Net Force Example Net Force You place your book on a bathroom scale and you find that it weighs 50 N. A) What force does it exert on the scale? B) You now push down on the book with a force of 10 N, what is the force applied to the book by the Earth? What force is exerted on the scale? C) You now lift the book with a force of 20 N, what is the force exerted on the scale? packet examples Newton’s Second Law of Motion • Net Force = mass acceleration or F = ma Basic Example F = ma What is the net force required to accelerate a 1.5 kg box at 2.0 m/sec2? Basic Example 2 What is the net force exerted on a 1500 kg car if it is accelerated from 5 m/s to 10 m/s in 3 sec? nd 2 Law: Acceleration • F = ma • The most used equation in Physics • Both sides of the equation can be a problem to solve • The F can be the difficult part because F is a vector sum • See the more advanced problems after 3rd law A use of nd 2 Law: What is. . . • Mass, m (kg) • Weight, Fg (N), or force of gravity They are NOT the same thing! nd • From 2 Law, when a is = g Fg = m·g calculates weight Examples How much does a 3 kg object weigh? What is the mass of a 900 N teacher? Newton’s Third Law of Motion “Every action (force) has an equal & opposite reaction (force)” • Forces are always in pairs • When 2 objects interact, the forces they exert are an action-reaction pair • Action-reaction forces are on different objects and are equal but opposite in direction Newton’s Third Law • Action-Reaction Pairs • The hammer exerts a force on the nail to the right. • The nail exerts an equal but opposite force on the hammer to the left. Action “hammer hits nail” Reaction “nail hits hammer” Newton’s Third Law • Action-Reaction Pairs • The rocket exerts a downward force on the exhaust gases. • The gases exert an equal but opposite upward force on the rocket. Ffuel FRocket One more on Newton’s Third Law • Action-Reaction Pairs • The earth pulls down on you (weight)… • You pull up on the earth with an equal and opposite force (same as your weight) Common forces in Problems • Weight (Fg), the force of gravity (= mg) • Support Force; called Normal Force where “Normal” is perpendicular. This force is perpendicular to the supporting surface (FN) • Tension (FT), when a rope, cable, spring, etc is pulled or stretched apart • Friction (Ff), but not yet! Normal Force, FN -Supporting force -Perpendicular to supporting surface -NOT always the same as weight FN Problem solving-• Identify Forces, Knows, and Unknowns • Sketch the problem – Choose your coordinate system (x & y) – Isolate the object – Draw external forces (free body diagram) • Apply Newton’s Laws in x & y (Fx=max & Fy=may ). Know if ax or ay = 0 !! • Solve for unknowns Example A 52 N sled is pulled across a level cement sidewalk. A horizontal force of 36 N is exerted. A) What is the acceleration of the sled? B) What is the acceleration of the sled if the sled’s mass is 30 kg? Example A) A 24 N fish hangs from a scale in an elevator. The elevator accelerates upward at 3 m/sec2. What does the scale read? B) Again; this time the car accelerates downward at 1.8 m/sec2. So, when is the elevator cable most likely to break? Example An 800 N student stands in an elevator. The elevator accelerates upward at 2.5 m/sec2. What is the tension in the cables if the elevator car’s mass is 1000 kg? When the car accelerates downward at 2 m/sec2? Example What is the acceleration of an Atwood machine if the two masses hanging on it is 3 kg and 4 kg? What is the tension in the cable? CP Physics Chapter 4 Newton’s Laws with friction Study two types of Friction •Static (s) stopped •Kinetic (k) moving Drag is fluid friction , which works differently What Affects the Force of Friction (Ff ) on a sliding object? • Speed? (assuming v > 0) • Area in contact? • Weight of object? • Materials in contact (surface roughness)? What if . . . ? Fpush < Ff, max Fpush = Ff, max Fpush > Ff, max • Ff is NOT an “applied” force (mu) is the coefficient of friction General: Ff = FN Fs sFN so Fs,max sFN Fk = kFN FYI: s > k is a property of the materials, which depends on their surface roughness and contact force, FN s=static k=kinetic Example #1 What is the acceleration of a 30 kg crate if the coefficient of friction is 0.034 and the object has a 15 N horizontal pulling force applied to it? Try this problem again but the crate weighs 30 N. Example #2A A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefficient of friction between the metal runners and the cement? Example #2B Now the sled runs on packed snow where the coefficient of friction is 0.12. If a 650 N person sits on the sled: A)What is the applied force needed to move it at constant speed? B) How much force is needed to accelerate the sled at 2 m/s2? Example #3A What coefficient of friction is needed to keep the system at equilibrium? m1=4 kg =? m2=3 kg Example #3B What is the acceleration of the system shown below? What is the tension in the cable? m1=4 kg =0.2 m2=3 kg 1) What is the minimum mass of the fish needed to just cause the Spam to move? mspam = 20 kg s = 0.5 mfish = ? 2) What is if the fish was 12.5 kg causing acceleration of 1.2 m/s2? Example #4 A penguin slides across the ice 15 m to a stop. The coefficient of friction between the penguin and the ice is 0.01. What was the acceleration and how fast was the penguin going initially? CP Physics Chapter 4 Newton’s Laws if any Angles SOH CAH TOA ! Vector Addition • Calculate x & y components of all vectors. Add all components for x, and all for y direction to find the total in the x (Fx) and y (Fy). • Knowing Fx and Fy , calculate Resultant Force with Pythagorean Thm: F R Fx F 2 2 y Example • A 80 kg skier starts from rest and accelerates down a hill which is angled 35 above horizontal. The coefficient of friction between the skis and snow is 0.10. • Draw the forces acting on the skier • What direction is the acceleration in? Problem solving-• Identify Forces, Knows, and Unknowns • Sketch the problem – – – – Choose your coordinate system Isolate the object(s) RESOLVE any Forces at angles into x & y Draw external forces • Apply Newton’s Laws in x & y (Fx=max & Fy=may ) • Solve for unknowns Example A sled of mass 30 kg is pulled with a force of 50 N at a 25 degree angle. If the coefficient of friction between the runners and the snow is 0.1, what is the acceleration of the sled? Example A 50 kg boy is inside a 20 kg crate that is pushed with a 200 N force at 20 degrees below the horizontal. a) What coefficient of friction would keep the crate from accelerating? b) What coefficient of friction would keep the crate accelerating at 1.5 m/sec2? F = 200 N Review You are standing on a scale while riding an elevator and your mass is 35 kg. Determine how much the Normal force is when: • The elevator is coming down at a constant speed of 2.3 m/s • The elevator is accelerating down at 1.35 m/s2 • The elevator is accelerating up at 2.25 m/s2