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Physics

Physics. PHS 5043 Forces & Energy Newton’s Laws. Force: Any agent capable of changing the shape of an object or changing its state of rest or motion Symbol: F Units: N (newton) N = kg m/s 2. PHS 5043 Forces & Energy Newton’s Laws. Force: Vector quantity (magnitude and direction)

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Physics

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  1. Physics

  2. PHS 5043 Forces & EnergyNewton’s Laws Force: Any agent capable of changing the shape of an object or changing its state of rest or motion Symbol: F Units: N (newton) N = kg m/s2

  3. PHS 5043 Forces & EnergyNewton’s Laws Force: Vector quantity (magnitude and direction) Forces can be added as any other vector (polygon, parallelogram and component method)

  4. PHS 5043 Forces & EnergyNewton’s Laws Force: At any given moment in time, all objects are subjected to the simultaneous action of many forces

  5. PHS 5043 Forces & EnergyNewton’s Laws Equilibrium: Objects are said to be in equilibrium when they are “at rest” or when they have “uniform rectilinear motion”

  6. PHS 5043 Forces & EnergyNewton’s Laws Equilibrium: • When an object is in equilibrium all external forces add up to zero. • If we add all forces vectors (regardless of the method), we should obtain a zero value resultant

  7. PHS 5043 Forces & EnergyNewton’s Laws Practice: Given the following forces values, determine if the object is in equilibrium. Fa = 30N, 270° Fb = 10N, 45° Fc = 20N, 135° _Coordinates: Fa (0, -30); Fb (7.1,7.1); Fc (-20,20) _Resultant Force: Fr (-12.9,-2.9) No, the system is not in equilibrium for the resultant vector (component method) does not have (0.0) coordinates Try with the polygon method!

  8. PHS 5043 Forces & EnergyNewton’s Laws Equilibrant force: • Is the force that balances the resultant force • It brings the system (object) to a state of equilibrium

  9. PHS 5043 Forces & EnergyNewton’s Laws Practice: Given the following forces values, determine the magnitude and direction of the equilibrant force Fa = 3N, 90° Fb = 4N, 0° _Coordinates: Fa (0, 3); Fb (4, 0), therefore Fr (4, 3) _Magnitude Fr = 5N (Pythagoras), direction = 37° _Feq = Fr but in opposite direction Therefore, the equilibrant force would be Feq = 5N, 217°

  10. PHS 5043 Forces & EnergyNewton’s Laws Inertia: Property that causes an object to remain in its state of rest or uniform rectilinear motion, that is, to remain in equilibrium.

  11. PHS 5043 Forces & EnergyNewton’s Laws Newton’s First Law: “If no external or unbalanced force acts on an object, it maintains its state of rest or its constant velocity in a straight line”

  12. PHS 5043 Forces & EnergyNewton’s Laws Newton’s Second Law: “The change of momentum of an object is proportional to the net applied force”

  13. PHS 5043 Forces & EnergyNewton’s Laws Newton’s Second Law: F = m a F = Δp / Δt F: Net force (N) F = mΔv / Δt m: Total mass (kg) F = m a a: acceleration (m/s2)

  14. PHS 5043 Forces & EnergyNewton’s Laws Practice: A 15 kg sled is pulled horizontally with a constant force of 60N. What acceleration would this force impart on the sled if the frictional force (sled-snow) is 10N? F = m a Fa – Ff= m a a = (Fa– Ff) / m a = (60N – 10N) / 15kg a = 3.3 m/s2

  15. PHS 5043 Forces & EnergyNewton’s Laws Practice: A 15 kg sled is pulled horizontally with a constant force of 60N at 30° with respect to ground. What acceleration would this force impart on the sled if the frictional force (sled-snow) is 10N? F = m a Fa – Ff= m a a = (Fax cos 30°– Ff) / m a = (60N)(cos 30°) – 10N / 15kg a = 2.8 m/s2

  16. PHS 5043 Forces & EnergyNewton’s Laws Impulse: • It can be consider the cause of the motion of an object • Its effect is the change of Momentum of a moving object • It is defined as the product of Force and time, or more conveniently as the variation of momentum

  17. PHS 5043 Forces & EnergyNewton’s Laws Practice: A 168 g hockey puck hits the boards of the ice rink perpendicularly at a velocity of 28 m/s and rebounds along same trajectory at 20 m/s. What impulses did the boards impart to the puck? Δp = mΔv Δp = 0.168 kg (-20 m/s – 28 m/s) Δp = 0.168 kg (-48 m/s) Δp = – 8.1 kg m/s Δp = – 8.1 N s

  18. PHS 5043 Forces & EnergyNewton’s Laws Newton's second law & momentum: You (50 kg) ride a bike (10 kg) at 10 m/s and speed up to 20 m/s, after 100 m. If frictional force (air & ground) is 30 N, what force did you apply during the acceleration? F = m a Fa – Ff= m a Fa = m a + FfV22 – V12 = 2ad a = (V22 – V12 ) / 2d a = 1.5 m/s2 Fa= (60 kg) (1.5 m/s2) + 30 N Fa= 120 N

  19. PHS 5043 Forces & EnergyNewton’s Laws Newton’s Third Law: “If object A exerts a force on object B (action), then object B exerts a force equal in magnitude, but opposite in direction, on object A (reaction)”

  20. PHS 5043 Forces & EnergyNewton’s Laws Newton’s Third Law: • Rockets and airplanes are examples of the application of Newton’s third law • Both forces act simultaneously (always)

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