General Physics I Spring Applying Newton s Laws

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1 General Physics I Spring 2011 Applying Newton s Laws 1

2 Equilibrium An object is in equilibrium if the net force acting on it is zero. According to Newton s first law, such an object will remain at rest or moving at constant velocity. Thus, the acceleration of the object is zero (consistent with Newton s second law). Since the net force on the object is zero, we have Fnet = F + F + F +... = For two-dimensional situations, we resolve into components: ( Fnet ) x = F + F + F +... = 0. 1x 2x 3x ( F ) = F + F + F +... = 0. net y 1y 2y 3y A shorthand way of writing the two equations above is to use the summation symbol: F x = 0 and F = 0. (Equilibrium) y 2

3 Example: Equilibrium in One and Two Dimensions Engine Ring 3

4 Equilibrium Example (Continued) For the engine, we have Fy = T w= (i) 1 Thus, T = w. 1 For the ring, we have: Fx = T cos60 T = (ii) 3 2 F = T sin60 T = (iii) y 3 1 Suppose the weight of the engine = 2000 N. From Eq. (i), we have: T = w = 2000 N. 1 T N From Eq. (iii), we have: T3 = = = 2309 N. sin 60 sin 60 From Eq. (ii), we have: T = T cos60 = (2309 N)cos 60 = 1155 N

5 Example: Ramp or Incline 5

6 Workbook: Chapter 5, Question 6 6

7 Dynamics and Newton s Second Law If the net force on an object is not zero, then the object will accelerate, according to Newton s second law. If we know the forces acting, we can find the acceleration. We can then use the kinematics equations to find velocity and position, which give complete information about the motion of the object. This is the subject of dynamics. As we have seen previously, Newton s second law is: F = ma. net Rewriting in component form gives: F = ma and F = ma. x x y y 7

8 Solving Dynamics Problems Identify all the forces acting on the object under investigation. Choose a coordinate system such that the acceleration is along one axis (e.g., x-axis) Draw a free-body diagram (very important!) Use Newton s second law in component form to find the net force along the x and y directions. Solve for acceleration if given all the forces. Solve for unknown force if given acceleration and other forces. 8

9 Workbook: Chapter 5, Questions, 7, 8 9

10 Textbook: Chapter 5, Problem 11 10

11 Mass and Weight An object s mass is a measure of its inertia. The mass can be determined from Newton s 2 nd law: apply a net force, measure the resulting acceleration and determine the mass from their magnitudes: m= F net An object s mass is independent of its location. a. 11

12 Mass and Weight An object s weight is the gravitational force acting on it due to a nearby massive body (e.g., Earth, the Moon, etc.). Thus, if the weight of an object is w and the gravitational acceleration it produces is a g, then a g w = m, or, w= mag. Now, a g = g, so in terms of magnitudes, w= mg. Close to the Earth s surface, the gravitational (free-fall) acceleration has a nearly constant magnitude of 9.8 m/s 2 and points toward the center of the Earth (i.e., locally vertically downward). 12

13 Mass and Weight Note that g is independent of mass. Note further that the weight is proportional to the gravitational acceleration. Thus, weight depends on location since g varies with location. It varies slightly at different locations on Earth. On the Moon, g = 1.6 m/s 2. The mass, however, is the same everywhere. It is easier to lift things on the Moon than on Earth, but just as difficult to push something horizontally. 13

14 You throw a ball straight up. Moments before it reaches its maximum height, what force(s) are acting on the ball after it has been released? 1. Weight (down) 2. Weight (down) and inertia (up) 3. Weight (down) and the force of the throw (up) 4. Inertia (up) 14

15 Apparent Weight When you stand on a scale at rest, it does not directly read your weight. Rather, it measures the normal force that you exert on it due to contact with your feet (which causes springs to be compressed). Since you are in equilibrium, the upward normal force that the scale exerts on you is equal in magnitude to your weight. Thus, the normal force that you exert on the scale (which is the reaction to the force exerted by the scale on you) is equal in magnitude to your weight. n w a = 0; n= w 15

16 Apparent Weight If you stand on a scale that is accelerating, its reading will not be equal to your weight. Its reading, i.e., the magnitude of the normal force, is said to be your apparent weight. Suppose that a person stands on a scale in an elevator, which has an upward acceleration of magnitude a. We can find her apparent weight using Newton s second law: Fy = n w= may = ma. n= w+ ma= m( g+ a). Thus, n> w. 16

17 If the elevator has a downward acceleration, then n w = m( a). So, n= m(g a). Apparent Weight If the elevator cable breaks, then a = g, and so n = 0, Hence, the person (and everything else in the elevator) has an apparent weight of zero. This is apparent weightlessness. This is what astronauts in orbit (or people on the "Vomit Comet") feel. This is because the astronauts and the space station are all in free fall around Earth with exactly the same acceleration, just like the person, scale, and elevator are all in free fall in the runaway elevator. n a y = a w "Zero-g" 17

18 Workbook: Chapter 5, Question 19 18

19 Textbook: Chapter 5, Problem 52 (a) Choose positive y direction upward. Rocket initially at rest so no air drag force. Fy = F w= may. thrust F w F mg F a = = = g thrust thrust thrust y m m m N 9.8 m/s m/s 2. = = 20,000 kg (b) F = F mg = ma. y thrust F = m( g+ a ), i.e., thrust y F m= thrust. Assuming the same thrust, we have ( g + a ) y y 5 m= N = kg. (9.8 m/s m/s 2) Thus, mass of fuel burned = 20,000 kg 19,000 kg= 1000 kg. F thrust w 19

Dynamic equilibrium: object moves with constant velocity in a straight line. = 0, a x = i

Dynamic equilibrium: object moves with constant velocity in a straight line. We note that F net a s are both vector quantities, so in terms of their components, (F net ) x = i (F i ) x = 0, a x = i (a