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Transcript
1 UNIT 3 DYNAMICS Dynamics : The study of forces which cause motion ( interaction between forces and matter) Kinematics : The study of motion. UNIVERSAL FORCES: ( 4 FORCES) There are thousands of observable forces but they can be categorized into 4 major categories. 1. Gravitatonal: (Weakest) - Forces existing between particles of mass 2. Electromagnetic: ( physical and magnetic) - Forces that exist between charged particles and magnets (ex. Hand on a table is actually repulsion of electrons) 2 3. Nuclear: ( strongest ) (aka: strong force) - Forces that hold the nucleus together. It holds protons , neutrons together despite the electromagnetic forces of the protons. (holds quarks together) 4. Weak interactive: (decay of nucleus) - These forces are responsible for radioactive decay. β-decay ( decay of neutrons = Proton + beta particle released (electron) -Both weak and strong nuclear forces act only over a very short distance ( about the width of an atom) -Gravitational and Electromagnetic are long range forces: However gravitational is the only one that can act over very long distances ( earth-sun). 3 NEWTON’S LAWS OF MOTION: 1665 Isaac Newton stated three laws of Motion. 1. Newton’s first Law: (Law of inertia) An object continues in its state of rest or uniform motion unless it is acted on by some external force. Inertia: is the tendency of an object not to accelerate. 2. Newton’s second Law: When an unbalanced force acts on an object, the object will be accelerated directly w/ the force applied and inversely with the mass of the object. F=ma for weight W=mg Newton: (N) Kg*m/s2 ( SI unit of force) 4 The amount of force needed to accelerate 1kg mass @ 1m/s2. 3. Newtons 3rd Law of Motion: ( law of reciprocal action) All forces occur in pairs, and these two forces are equal in magnitude and opposite in direction. Mass: The amount of matter in an object Weight: The gravitational pull of the earth on an object. W=mg NET FORCE: Fnet = Applied Force - hindering forces Fnet = ma so a = Fnet m 5 Ex. A 1 Kg mass that is pushed w/ a 50N force. How fast will it be accelerated if the floor causes 15N worth of friction? A 1.50kg ball rolls down a hill with a 40 degree incline. What is the acceleration of the ball down the incline? 6 An arrow with a .40kg mass is shot straight up in the air by a force of 120.00Newtons. Find the acceleration of the arrow. 7 Assignment 1 Pg 100 1-10 (old book Pg 124 Section review 1- 4 Assignment 2 Pg 128 Practice B 1-3 Pg 132 Practice C 1-5 Pg 102 old book pr 11-18 Pg 103 old pr 19-27 Kinetic friction: The retarding frictional force on an object in motion Static friction: The resistance force that keeps the an object from moving Force Normal: is the force that is the force that is equal to the gravitational force on an object but opposite in direction. ( not always equal to the weight of the object) Which is bigger Static or kinetic friction? 8 When are the equal? Coefficient of Friction Is defined as the ratio of the force of friction to the normal force between two surfaces. Coefficient of Kinetic friction is the ratio of the force of kinetic friction to the normal µk = Fk Fn Coefficient of static friction is the ratio of the maximum value of force of static friction to the normal force. µs = Fsmax Fn Force of friction then can be calculated by.. Ff = µFn Look at table 2 pg 138 9 Ex. A 24 kg crate at rest on the floor needs a 75N force horizontally to set it in motion. What is the coefficient of static friction for this system? µs = Fsmax Fn What is Fsmax ? What is Fn ? Assignment 3 Pg 139 Practice D 10 OVERCOMING FRICTION: Spencer attaches a rope to a 50Kg weight. He pulls with a force of 250.0N at an angle of 30o with the horizontal. The µk is .500. Find the acceleration of the box. Fn Fg 30o 1. Lets find the weight of the box: W=mg = (50.0kg) (9.8)= 490N 30o 490N 11 2. Resolve component vectors : Cos30 = X/ 250.0N fx =216.5N fy = sin30 = Y/250.0N fy = 125N 3. Find Fn = Downward force on the box ( in this case = weight of object – upward force) Fn = 490N - 125N = 365N 4. Next we must find the force of kinetic friction Fk Fk= µkFn = .500 X 365N = 182.5N F=ma acceleration = Fx - Ff mass = 216.5N – 182.5N = .68m/s2 50Kg 12 Assignment 4 Pg 141 Practice E Pg 150 standardized test prep 1-9 Pg 107 old 1-3