1. FORCE DEFINITION OF FORCE NEWTON’S THREE LAWS OF MOTION WEIGHT NORMAL FORCE EQUILIBRIUM FRICTION

2. FORCE: Force is a vector quantity that may be thought of as a push or a pull. The SI unit of force is the Newton, N, equal to kg @ m/s 2. Forces may be classified as contact forces or forces at a distance. Contact forces require interacting objects to be in physical contact. Forces at a distance are interactions between objects separated by space. Examples of contact forces: Tension, Compression, Spring force, kicking a football. Examples of forces at a distance: Gravity, electrostatic force, and magnetic force.

3. NEWTON’S FIRST LAW: An object moving with a constant velocity, or at rest, will continue in that state of motion unless acted upon by an external and unbalanced force. Inertia is what we call an object’s resistance to a change in its motion. Inertia is measured by mass. An object with 2kg of mass has 2kg of inertia.

4. NEWTON’S SECOND LAW: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. F = magnitude of net force acting on object. m = mass of object. a = resulting acceleration of object.

7. EXAMPLE:A push from the right of 20 N and a push from the left of 10 N acting on a 40 kg mass will produce what acceleration?

8. ANSWER :A push from the right of 20 N and a push from the left of 10 N acting on a 40 kg mass will produce what acceleration? The net force will be 20N -10N = 10 N to the left.

10. EXAMPLE:What is the mass of an object weighing 750N?

11. ANSWER :What is the mass of an object weighing 750N? This is the typical weight and mass of a person.

12. All forces may be classified as acting internally with a system of interacting objects or acting externally upon an object or system. Internal forces, forces acting with a system, consist of equal and opposite action and reaction pairs. The sum of all internal forces is always zero.

14. Vector Form: Component Form:

18.  1 F g T 1 T 2  2  F x = 0 Y -.866T 1 +.743T 2 = 0 Y T 1 =.858T 2  F y = 0 Y.5T 1 +.669T 2 - 490N = 0 Y.5(.858T 2 ) +.669T 2 = 490N ANSWER: T 2 = 446.3N T 1 = 382.9N Force Diagram Force Analysis Basic Eq’s Working Eq’s XY T1T1 -T 1 cos30 o -.866T 1 T 1 sin30 o.500T 1 T2T2 T 2 cos42 o.743T 2 T 2 sin42 o.669T 2 FgFg (mg)cos270 o 0 (mg)sin270 o -490N

20.  F g m F y F x N Force Diagram Force Analysis  F x = ma Y 224.8N = (40kg) a Y a = 5.6m/s 2  F y = 0 Y N - 321.1N = 0 Y N = 321.1N Basic Eq’s Working Eq’s  = 35 o m = 40kg XY NNcos90 o 0 Nsin90 o N FgFg (mg)sin35 o 224.8N -(mg)cos35 o -321.1N

22. T 1 N W 1  m 1 W x W y T 2 W 2 Force Diagrams Force Analysis m 1 2 m 1 = 60kg m 2 = 6kg  = 30 o XY NNcos90 o 0 Nsin90 o N T1T1 Tcos180 o -T Tsin180 o 0 F g1 (m 1 g)sin30 o 294.0N -(m 1 g)cos30 o -509.2N XY T2T2 T0 F g2 -m 2 g -58.8N 0

23.  F x = m 1 a Y 0 - T + 249.0N = (60kg) a  F y = 0 Y N + 0 - 509.2N = 0 Y N = 509.2N  F x = m 2 a Y T - 58.8N = (6kg) a Y T = (6kg)a + 58.8N (1 & 3)[-(6kg)a -58.8N] +249.0N = (60kg)a Y 190.2N = (66kg)a ANSWER: a = 2.9m/s 2 T = 76.1N Force Analysis m 1 2 Basic Eq’s Working Eq’s XY NNcos90 o 0 Nsin90 o N T1T1 Tcos180 o -T Tsin180 o 0 F g1 (m 1 g)sin30 o 294.0N -(m 1 g)cos30 o -509.2N XY T2T2 T0 F g2 -m 2 g -58.8N 0

24. F a = 100N  = 30 o  k = 0.2 N F g f k m = 40kg Compute the acceleration of the box. Force Diagram Force Analysis XY NNcos90 o 0 Nsin90 o N FaFa (100N)cos30 o 86.6N (100N)sin30 o 50.0N FgFg (mg)cos270 o 0 (mg)sin270 o -392.0N f kf k -(0.2)N0

25. (1)  F x = ma Y 0 + 86.6N +0 - (0.2)N = (40kg)a (2)  F y = 0 Y N + 50.0N -392.0N = 0 Y N = 342.0N (1 & 2) 86.6N - (0.2)(342.0N) = (40kg) a Y a = 0.46 m/s 2 Basic Eq’s Working Eq’s XY NNcos90 o 0 Nsin90 o N FaFa (100N)cos30 o 86.6N (100N)sin30 o 50.0N FgFg (mg)cos270 o 0 (mg)sin270 o -392.0N f kf k -(0.2)N0