1. AP Chapter 4

2. Force - a push or pull Contact Force – Noncontact Force – mass

3. Newton’s First Law An object continues it’s motion unless acted upon by a net force Net force is the vector sum of all forces acting on the object

4. Inertia – an objects tendency to resist a change in motion Mass – the quantitative measure of mass

5. Newton’s Second Law A net force on an object results in an acceleration directly proportional to the net force and inversely to it’s mass. F = ma unit of force is N = kgm/s 2

6. Free body diagrams Show the forces acting on an object for easy viewing and analyzing. Ex. Lifting a text book

7. Force Vectors Newton’s Third Law Every action has an equal and opposite reaction (forces)

8. Fundamental Forces Gravitational, Strong Nuclear, Electroweak Nonfundamental Forces Friction, Tension,

9. Newton’s Law of Universal Gravitation G = 6.673 x 10 -11 Nm 2 /kg 2

10. Ex. How much gravitational force do you experience on top of Mount McKinley? R E = 6380000 m H m.m. = 5938 m

11. F = ?

12. Weight? Depends on the distance the mass is from the object’s (Earth’s) center of mass! Apparent Weight?

13. Normal Force The force exerted by the object in contact with another object perpendicular to the surface. Ex. Your book on the table, the gravitational force is downward on the book but the normal force is upward from the table against the book. What is the sum of those forces?

14. Frictional Forces Static vs Kinetic F s is not in the direction of F n, they are perpendicular or at an angle to eachother.

15. Kinectic Friction Always less than static friction μ s > μ k why? Both frictional forces oppose the motion of the object across the other object

16. Ex. If my shoes have a static friction coefficient of 0.6 with the floor and I have a mass of 92 kg, what is the frictional force resisting my movement?

18. Tension Force – an applied force often to a rope or cable. Equilibrium a=0 So the sum of the forces = 0 on that axis ΣF y = 0 ΣF x = 0

19. EX. Hanging mass 1000 g by two ropes One at 75 o above the horizontal and the other at 50 o above the horizontal from the opposite side. Determine the tension force on each rope. 1 st Draw a free-body diagram 2 nd Choose your axis orientation 3 rd Use the equations ΣF y = 0 ΣF x = 0 or ma 4 th Solve

20. 75 o 50 o T1T2 FgFg 1000 g object X and Y axis will be standard so Fg is all on the y axis. F y = T 1 sin(75 o ) + T 2 sin(50 o ) – Fg = 0 T 1 sin(75 o ) + T 2 sin(50 o ) – 1kg(9.80m/s 2 ) = 0 Fx = -T 1 cos(75 o ) + T 2 cos(50 o ) = 0 -T 1 cos(75 o ) + T 2 cos(50 o ) = 0 T 2 = T 1 cos(75 o )/cos(50 o ) Substitute into the other eq. T 1 sin(75 o ) + T 1 (cos(75 o )/cos(50 o )) sin(50 o ) – 1kg(9.80m/s 2 ) = 0 0.966T 1 + 0.308T 1 – 9.80 N = 0 1.274T 1 = 9.80N T 1 =7.69 N Then put back in the other eq and find T 2. T 2 = 7.69 N (cos(75 o )/cos(50 o )) = 3.10 N

21. Nonequilibrium there is a net force and an acceleration