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

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1 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 truck that has a dead batter. The mass of the truck is much larger than the mass of the car. Which of the following statements is true? a. Car eerts a force on the truck, but truck doesn t eert a force on the car. b. Car eerts a larger force on truck than truck eerts on the car. c. Car eerts the same force on truck as truck eerts on the 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 10 /6

2 Appling Newton s Laws 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 Free Bod Diagram Lecture 10 3/6 Free Bod Diagram (FBD) 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 Lecture 10 4/6

3 Eample of a free-bod diagram: Lecture 10 5/6 Drawing Force Vectors Lecture 10 6/6

4 Moving a Satellite (Deep Outer Space) Lecture 10 7/6 Lecture 10 8/6

5 D Newton s nd Law Problems Draw FBD and choose good coordinate sstem Get force components: F = Fcosθ; F = Fsinθ Use Newton nd Law F = ma in component form (treat & separatel) F = ma a or a ma be zero F = ma Lecture 10 9/6 F = 5 CCW from -a F = F cos 5 = 5. N F = F sin 5 = 3.3 N F 1 = 6 N F 1 = 0 Net force: F = F 1 + F = 6 N + 5. N = 51. N F = 3.3 N F = F + F F θ = arctan F = 60.5N = 3. Lecture 10 10/6

6 Finding Acceleration a = F /m = 51.N / 940kg = m/s a = F /m = 3.3N / 940kg = m/s a = a + a a θa = arctan a = 0.064m / s = 3. Note, also, a = F net /m and the angle of the acceleration is same as net force angle Lecture 10 11/6 A Catalog of Forces: 1. Weight The falling bo is pulled toward the Earth b the long-range force of gravit. The gravitational pull on a object on or near the surface of the Earth is called weight, for which we use the smbol W. Weight force is the entire Earth pulling on the 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 Lecture 10 1/6

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.

7 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 10 13/6 Newton s Law of Gravit* mm F = F = G r 1 1 on on 1 G = N m /kg 11 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 10 14/6

8 The Earth s Gravitation* What is the gravitational force on an object of mass m at the surface of the Earth? The center of the Earth is one Earth radius (R E ) awa, so that is the distance r: Therefore, g GM = R where E E g = 9.80 N/kg = 9.80 m/s Lecture 10 15/6 Earth Gravitation vs. Altitude* The acceleration of gravit decreases slowl with altitude: GM E gh ( ) = ( R + h) E GM g(0) = = 9.83 m/s R E E Lecture 10 16/6

80 m/s m W Lecture 10 17/6 Eample 5-4 A bo of mass m 1 =10.0 kg rests on a floor net to a bo of mass m =5.

9 Weight & Free Fall Weight of bo of mass m is W = mg g = 9.80 N/kg = 9.80 m/s down F = mg B Newton s nd Law ma = -mg a = -g = m/s m W Lecture 10 17/6 Eample 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 F=0.0N in the positive -direction, (a) what is the acceleration of the boes; (b) What is the magnitude of contact force F 1? Of F? Lecture 10 18/6

10 N N 1 N F F 1 F F W W 1 W FBD Combined FBD Bo 1 FBD Bo Boes Onl acceleration in each case ( forces cancel) Combined Boes: a = F/(m 1 +m ) = 0N/15kg = 1.33 m/s Bo 1: a 1 = a = (F - F 1 )/m 1 so 1.33 m/s = (0N - F 1 )/10kg Thus F 1 = 6.7 N Bo : a = a = F /5kg so 1.33 m/s = F /5kg Thus F = 6.7 N (as epected from Third Law!!) Lecture 10 19/6 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. Lecture 10 0/6

Elevator accelerates down: -W+W a = ma W a =mg + ma W a < mg Lecture 10 1/6 Apparent Weight (Eample 5.

11 Apparent Weight Elevator not accelerating: W+W a =0 Elevator accelerates up: W+W a =ma -W+W a =ma W a =ma + mg W a > mg Force felt from contact with the floor (or a scale, etc.) in an accelerating sstem. Elevator accelerates down: -W+W a = ma W a =mg + ma W a < mg Lecture 10 1/6 Apparent Weight (Eample 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) Lecture 10 /6

12 Weight Components on Inclined Surface Lecture 10 3/6 Force Component Down Inclined Plane Force Component Down Inclined Plane Forces 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 10 4/6

13 Acceleration Down Inclined Plane Newton s nd law: F F = mg sinθ = ma = 0 From 1 st equation: a Check special cases to see that answer makes sense: o θ = 90 o θ = 0 = g sin θ What is acceleration in each of these cases? Lecture 10 5/6 End of Lecture 10 Before the net lecture, read Walker Chapter 5, 5.7; Chapter 6, Kinetic Friction Homework Assignment #5b should be submitted using WebAssign b 11:00 PM on Frida, Feb. 0. Lecture 10 6/6

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

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 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