Physics 111. Applying Newton s Laws. Lecture 9 (Walker: 5.4-5) Newton s Third Law Free Body Diagram Solving 2-D Force Problems Weight & Gravity

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1 Phsics 111 Lecture 9 (Walker: 5.4-5) Newton s Third Law ree Bod Diagram Solving -D orce Problems Weight & Gravit Sept. 1, 009 Quiz Wednesda - Chaps. 3 & 4 Lecture 9 1/6 Newton s Third Law of Motion orces alwas come in pairs, acting on different objects: If Object 1 eerts a force on Object, then Object eerts a force on Object 1. These forces are called action-reaction pairs. Alternate Wording: or ever action there is an equal and opposite reaction. r A on B r = B on A Lecture 9 /6 Question Appling Newton s Laws Small car is pushing large truck that has a dead batter. Mass of truck is much larger than mass of car. Which of the following statements is true? a. Car eerts force on truck, but truck doesn t eert force on car. b. Car eerts a larger force on truck than truck eerts on car. c. Car eerts the same force on truck as truck eerts on car. d. Truck eerts a larger force on car than car eerts on truck. e. Truck eerts a force on car, but car doesn t eert a force on truck. Lecture 9 3/6 Assumptions Objects behave as particles can ignore rotational motion (for now) Masses of strings or ropes are negligible Interested onl in the forces acting on the object can neglect reaction forces Start with a ree Bod Diagram Lecture 9 4/6

2 ree Bod Diagram (BD) Diagram that shows all force vectors acting on the object of interest. How to construct: Draw object as a dot (point particle) Replace everthing that touches object with appropriate force vectors Add weight vector (alwas points straight down) Choose a convenient coordinate sstem Resolve force vectors into components ample of a free-bod diagram: Lecture 9 5/6 Lecture 9 6/6 Drawing orce Vectors ample 5-4 A bo of mass m 1 =10.0 kg rests on a floor net to a bo of mass m =5.00 kg. If ou push bo 1 with a horizontal force of magnitude =0.0N in the positive -direction, (a) what is the acceleration of the boes; (b) What is the magnitude of contact force 1? Of? Lecture 9 7/6 Lecture 9 8/6

N N 1 N 1 W W 1 W Moving a Satellite (Deep Outer Space) BD Combined BD Bo 1 BD Bo Boes Onl acceleration in each case ( forces cancel) Combined Boes: a = /(m 1 +m ) = 0N/15kg = 1.

3 N N 1 N 1 W W 1 W Moving a Satellite (Deep Outer Space) BD Combined BD Bo 1 BD Bo Boes Onl acceleration in each case ( forces cancel) Combined Boes: a = /(m 1 +m ) = 0N/15kg = 1.33 m/s Bo 1: a 1 = a = ( - 1 )/m 1 so 1.33 m/s = (0N - 1 )/10kg Thus 1 = 6.7 N Bo : a = a = /5kg so 1.33 m/s = /5kg Thus = 6.7 N (as epected from Third Law!!) Lecture 9 9/6 Lecture 9 10/6 D Newton s nd Law Problems Draw BD and choose good coordinate sstem Get force components: = cosθ; = sinθ Use Newton nd Law = ma in component form (treat & separatel) = ma a or a ma be zero = ma Lecture 9 11/6 Lecture 9 1/6

= 41N @ 5 CCW from -a = cos 5 = 5. N = sin 5 = 3.3 N 1 = 6 N 1 = 0 Net force: = 1 + = 6 N + 5. N = 51. N = 3.3 N = + θ = arctan = 60.5N = 3. Lecture 9 13/6 inding Acceleration a = /m = 51.

4 = 5 CCW from -a = cos 5 = 5. N = sin 5 = 3.3 N 1 = 6 N 1 = 0 Net force: = 1 + = 6 N + 5. N = 51. N = 3.3 N = + θ = arctan = 60.5N = 3. Lecture 9 13/6 inding Acceleration a = /m = 51.N / 940kg = m/s a = /m = 3.3N / 940kg = m/s a = a + a a θa = arctan a = 0.064m / s = 3. Note, also, a = net /m and the angle of the acceleration is same as net force angle Lecture 9 14/6 Weight The falling bo is pulled toward arth b long-range force of gravit. The gravitational pull on an object on or near the surface of the arth is called weight, for which we use the smbol W. Weight force is the entire arth pulling on an object. Weight acts equall on objects at rest or in motion. The weight vector alwas points verticall downward, and it can be considered to act at the center of mass of the object. W = mg g = 9.80 N/kg = 9.80 m/s down W Newton s Law of Gravit* Newton proposed that ever object in the universe attracts ever other object with a force that has the following properties: 1. The force is inversel proportional to the square of the distance between the objects.. The force is directl proportional to the product of the masses of the two objects. Lecture 9 15/6 Lecture 9 16/6

5 Newton s Law of Gravit* mm 1 1 on = on 1 = G r The arth s Gravitation* What is the gravitational force on an object of mass m at the surface of the arth? The center of the arth is one arth radius (R ) awa, so that is the distance r: 11 G = N m /kg r is distance between mass m 1 and mass m. Direction of force on m 1 is toward m ; direction of force on m is toward m 1 (third law pair! ) Lecture 9 17/6 Therefore, GM g = R where g = 9.80 N/kg = 9.80 m/s Lecture 9 18/6 arth Gravitation vs. Altitude* The acceleration of gravit decreases slowl with altitude: GM gh ( ) = ( R + h) GM g(0) = = 9.83 m/s R Weight & ree all Weight of bo of mass m is W = mg g = 9.80 N/kg = 9.80 m/s down = mg B Newton s nd Law ma = -mg a = -g = m/s m W Lecture 9 19/6 Lecture 9 0/6

Apparent Weight levator not accelerating: W a =mg levator accelerates up: W a > mg Lecture 9 1/6 orce felt from contact with the floor (or a scale, etc.

6 Apparent Weight Your perception of our weight is based on the contact forces between our bod and our surroundings. If our surroundings are accelerating, our apparent weight ma be more or less than our actual weight. Apparent Weight levator not accelerating: W a =mg levator accelerates up: W a > mg Lecture 9 1/6 orce felt from contact with the floor (or a scale, etc.) in an accelerating sstem. levator accelerates down: W a < mg Lecture 9 /6 Apparent Weight (ample 5.7 from Walker) m = 5.0 kg a) W a (at rest)? b) W a? (a =.5m/s up) c) W a? (a =3. m/s down) Weight Components on Inclined Surface Lecture 9 3/6 Lecture 9 4/6

7 orce Component Down Inclined Plane orces acting on the object: The normal (support) force acts perpendicular to the plane The gravitational force acts straight down Choose coordinate sstem with along the incline and perpendicular to incline Replace the force of gravit with its components Lecture 9 5/6 Acceleration Down Inclined Plane Newton s nd law: = mg sinθ = ma = 0 rom 1 st equation: a = g sin θ Check special cases to see that answer makes sense: o θ = 90 o θ = 0 What is acceleration in each of these cases? Lecture 9 6/6 nd of Lecture 9 Before the net lecture, read Walker Chapter 5, 5.7; Chapter 6, Kinetic riction Homework Assignment #5b should be submitted using WebAssign b 11:00 PM on Thursda, Sept. 4. Lecture 9 7/6

Physics 111. Lecture 10 (Walker: 5.5-6) Free Body Diagram Solving 2-D Force Problems Weight & Gravity. February 18, Quiz Monday - Chaps.

Physics 111. Lecture 10 (Walker: 5.5-6) Free Body Diagram Solving 2-D Force Problems Weight & Gravity. February 18, Quiz Monday - Chaps. Phsics 111 Lecture 10 (Walker: 5.5-6) Free Bod Diagram Solving -D Force Problems Weight & Gravit Februar 18, 009 Quiz Monda - Chaps. 4 & 5 Lecture 10 1/6 Third Law Review A small car is pushing a larger