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FORCE

FORCE. DEFINITION OF FORCE NEWTON’S THREE LAWS OF MOTION WEIGHT NORMAL FORCE EQUILIBRIUM FRICTION. FORCE:. Force. is a vector quantity that may be thought of as a push or a. pull. Newton. @. 2. The SI unit of force is the. , N, equal to kg. m/s.

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FORCE

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  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. Newton @ 2 The SI unit of force is the , N, equal to kg m/s . 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.

  5. 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?

  6. The net force will be 20N -10N = 10 N to the left. 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?

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

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

  9. 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.

  10. Vector Form: Component Form:

  11. Force Force Analysis Diagram T 2 T 1 q 2 q 1 F g Basic Eq’s Working Eq’s S Y Y F = 0 -.866T + .743T = 0 T = .858T x 1 2 1 2 S Y Y F = 0 .5T + .669T - 490N = 0 .5(.858T ) +.669T = 490N y 1 2 2 2 ANSWER: T = 446.3N T = 382.9N 2 1

  12. Force Analysis Force Diagram N m = 40kg m q F y q o = 35 F g Basic Eq’s Working Eq’s F x S Y Y 2 F = ma 224.8N = (40kg) a a = 5.6m/s x S Y Y F = 0 N - 321.1N = 0 N = 321.1N y

  13. Force Analysis m Force Diagrams 1 m = 60kg N 1 m = 6kg 2 q o = 30 T 1 T m 2 1 W q 2 Force Analysis m 2 W y W 1 W x

  14. Force Analysis m Force Analysis m 1 2 Basic Eq’s Working Eq’s (1) S Y F = m a 0 - T + 249.0N = (60kg) a x 1 (2) S Y Y F = 0 N + 0 - 509.2N = 0 N = 509.2N y (3) S Y Y F = m a T - 58.8N = (6kg) a T = (6kg)a + 58.8N x 2 Y (1 & 3) [ -(6kg)a -58.8N ] +249.0N = (60kg)a 190.2N = (66kg)a 2 ANSWER: a = 2.9m/s T = 76.1N

  15. Compute the N acceleration of Force Diagram the box. F = 100N a q o = 30 m = 40kg f m k = 0.2 k F g Force Analysis

  16. Basic Eq’s Working Eq’s S Y (1) F = ma 0 + 86.6N +0 - (0.2)N = (40kg)a x S Y Y (2) F = 0 N + 50.0N -392.0N = 0 N = 342.0N y Y 2 (1 & 2) 86.6N - (0.2)(342.0N) = (40kg) a a = 0.46 m/s

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