Forces. Isaac Newton was the first to discover that the laws that govern motions on the Earth also applied to celestial bodies.

Transcription

1 Forces Now we will discuss the part of mechanics known as dynamics. We will introduce Newton s three laws of motion which are at the heart of classical mechanics. We must note that Newton s laws describe physical phenomena of a vast range. For example, Newton s laws explain the motion of stars and planets. Isaac Newton was the first to discover that the laws that govern motions on the Earth also applied to celestial bodies. Over the next few chapters we will study how bodies interact with one another. Simply, a force is a push or pull on an object. 1

2 Fundamental Forces The four fundamental forces of nature are: Gravity which is the force between two masses; it is the weakest of the four. Strong Force which helps to bind atomic nuclei together; it is the strongest of the four. Weak Force plays a role in some nuclear reactions. Electromagnetic is the force that acts between charged particles. 2

3 How can a force be measured? One way is with a spring scale. By hanging masses on a spring we find that the spring stretch applied force. The units of force are Newtons (N). A force is an example of a vector quantity! 3

4 Newton s First Law Newton s 1 st Law (The Law of Inertia): If no force acts on an object, then its speed and direction of motion do not change. Inertia is a measure of an object s resistance to changes in its motion. If the object is at rest, it remains at rest (speed = 0). If the object is in motion, it continues to move in a straight line with the same speed. No force is required to keep a body in straight line motion when effects such as friction are negligible. 4

5 5

6 Free Body Diagrams: Must be drawn for problems when forces are involved. Must be large so that they are readable. Draw an idealization of the body in question (a dot, a box, ). You will need one free body diagram for each body in the problem that will provide useful information for you to solve the given problem. Indicate only the forces acting on the body. Label the forces appropriately. Do not include the forces that this body exerts on any other body. 6

7 Free Body Diagrams (continued): A coordinate system is a must. Do not include fictitious forces. Remember that ma is itself not a force! You may indicate the direction of the body s acceleration or direction of motion if you wish, but it must be done well off to the side of the free body diagram. 7

8 Newton s Third Law Newton s 3 rd Law: When 2 bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction. Or, forces come in pairs. Mathematically: F = F

9 Example: Consider a box resting on a table. F 1 (a) If F 1 is the force of the Earth on the box, what is the interaction partner of this force? The force of the box on the Earth. 9

10 Example continued: F 2 (b) If F 2 is the force of the box on the table, what is the interaction partner of this force? The force of the table on the box. 10

11 External forces: Any force on a system from a body outside of the system. F Pulling a box across the floor 11

12 Internal forces: Force between bodies of a system. F ext Pulling 2 boxes across the floor where the two boxes are attached to each other by a rope. 12

13 13

14 Gravity Gravity is the force between two masses. Gravity is a longrange or field force. No contact is needed between the bodies. The force of gravity is always attractive! F = GM1M 2 r 2 r is the distance between the two masses M 1 and M 2 and G = Nm 2 /kg 2. M 2 M 1 F 12 F 21. F = F r 14

15 GM Let M 1 = mass of the Earth. F = E M 2 2 r Here F = the force the Earth exerts on mass M 2. This is the force known as weight, w. w = GM E M 2 2 = gm 2. r E M r E E = = km 24 kg GM E where g = = 9.8 N/kg = 9.8 m/s 2 r E 2 Near the surface of the Earth 15

16 Note that g = F m is the gravitational force per unit mass. This is called the gravitational field strength. It is often referred to as the acceleration due to gravity. What is the direction of g? What is the direction of w? 16

17 Example: What is the weight of a 100 kg astronaut on the surface of the Earth (force of the Earth on the astronaut)? How about in low Earth orbit? This is an orbit about 300 km above the surface of the Earth. On Earth: w = mg = 980 N In low Earth orbit: w = GM E mg( r ) = = ( ) o m RE + h 890 N Their weight is reduced by about 10%. The astronaut is NOT weightless! 17

18 Contact Forces Contact forces: these forces arise because of an interaction between the atoms in the surfaces in contact. 18

19 Normal force: this force acts in the direction perpendicular to the contact surface. N Force of the ground on the box w N Force of the ramp on the box w 19

20 Example: Consider a box on a table. FBD for box N y x Apply Newton s 2 nd law F y w = N w So that N = w = = 0 mg This just says the magnitude of the normal force equals the magnitude of the weight; they are not Newton s third law interaction partners. 20

21 Friction: a contact force parallel to the contact surfaces. Static friction acts to prevent objects from sliding. The force of static friction is modeled as Kinetic friction acts to make sliding objects slow down. f s μ N. where μ k is the coefficient of kinetic friction and N is the normal force. s The force of kinetic friction is modeled as N. where μ s is the coefficient of static friction and N is the normal force. f k = μ k 21

22 Example (text problem 2.91): A box full of books rests on a wooden floor. The normal force the floor exerts on the box is 250 N. (a) You push horizontally on the box with a force of 120 N, but it refuses to budge. What can you say about the coefficient of friction between the box and the floor? FBD for box N y F x f s w Apply Newton s 2 nd Law (1) (2) F y F x = N = F w = 0 f s = 0 22

23 Example continued: F f From (2): = = μ μ = = s This is the minimum value of μ s, so μ s > s N s F N (b) If you must push horizontally on the box with 150 N force to start it sliding, what is the coefficient of static friction? F f Again from (2): = = μ μ = = s s N s F N 23

24 Example continued: (c) Once the box is sliding, you only have to push with a force of 120 N to keep it sliding. What is the coefficient of kinetic friction? FBD for box y N Apply (1) Fy = N w = 0 F Newton s x 2 nd Law (2) Fx = F fk = 0 f k w From 2: F = μ f k F N = μ N 120 N 250 N k = = = k

25 Tension This is the force transmitted through a rope from one end to the other. An ideal cord has zero mass, does not stretch, and the tension is the same throughout the cord. 25

26 Example (text problem 2.73): A pulley is hung from the ceiling by a rope. A block of mass M is suspended by another rope that passes over the pulley and is attached to the wall. The rope fastened to the wall makes a right angle with the wall. Neglect the masses of the rope and the pulley. Find the tension in the rope from which the pulley hangs and the angle θ. y FDB for the mass M T x Apply Newton s 2 nd Law to the mass M. w F y = T w = 0 T = w = Mg 26

27 Example continued: FBD for the pulley: Apply Newton s 2 nd Law: y F x = F cosθ T = 0 T T θ F x F y = F sinθ T = T = F cosθ = F sinθ This statement is true only when θ = 45 and F = 2 T = 2Mg 0 27

28 What is the net force acting on the object shown below? y x 15 N 15 N 10 N a. 40 N b. 0 N c. 10 N down d. 10 N up 28

29 The gravitational field strength of the Moon is about 1/6 that of Earth. If the mass and weight of an astronaut, as measured on Earth, are m and w respectively, what will they be on the Moon? a. b. c. d. m, 1 6 m, 1 6 w m, w 1 6 m, w 1 6 w 29