1 / 20

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)

jorden-kennedy
Download Presentation

Physics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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)

More Related