1 / 35

REVIEW FOR MID TERM 1

REVIEW FOR MID TERM 1. English  S.I. 1 inch = 2.540 cm (centimeter = 1/100 meter) 1mile = 1.609 km (kilometer= 1000 meters) 1 pound = 454 g (gram) 1.06 quart = 1 L (litre). Components Of Vectors A = A x + A y. Using Components To Add Vectors. C x = A x + B x

aforeman
Download Presentation

REVIEW FOR MID TERM 1

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. REVIEW FOR MID TERM 1

  2. English  S.I. • 1 inch = 2.540 cm (centimeter = 1/100 meter) • 1mile = 1.609 km (kilometer= 1000 meters) • 1 pound = 454 g (gram) • 1.06 quart = 1 L (litre)

  3. Components Of Vectors A = Ax + Ay

  4. Using Components To Add Vectors Cx = Ax + Bx Cy = Ay + By

  5. Vav,x = (x2-x1)/(t2-t1) = Δx/Δt SI unit: m/s

  6. Average Acceleration aav,x = (v2x – v1x) / (t2 - t1) = Δvx / Δt SI unit: m/s2 Instantaneous Acceleration ax= lim (Δvx/Δt) Δt 0 SI unit: m/s2

  7. Cars Accelerating or Decelerating

  8. Constant-Acceleration Equations of Motion

  9. Constant-Acceleration Equations of Motion in Two-Dimensions

  10. Equations of Motion for projectileax = 0m/s2 ay = -9.81m/s2

  11. Determination of key itemsfor projectiles • x = (vocos o)t •  = tan-1(vy/vx) • y = (vosin o)t - ½gt2 • vx = vocoso • vy = vosino- gt

  12. R = F1 + F2 + F3 + ……..= Σ F, (resultant, and vector sum, of forces) Rx = Σ Fx , Ry = Σ Fy (components of vector sum of forces) Once we have the components Rx and Ry, we can find the magnitude and direction of the vector R.

  13. Newton’s First Law – Figure 4.7 • “Objects at rest tend to stay at rest and objects in motion tend to stay in motion in a straight line unless it is forced to change that state by forces acting on it”

  14. Newton’s Second Law of Motion (Vector Form) The vector sum (resultant) of all the forces acting on an object equals the object’s mass times its acceleration : ΣF = ma The acceleration a has the same direction as the resultant force ΣF.

  15. Newton’s Second Law of Motion (Vector Form) The vector sum (resultant) of all the forces acting on an object equals the object’s mass times its acceleration : ΣF = ma The acceleration a has the same direction as the resultant force ΣF.

  16. Forces are the origin of motion Acceleration a = F/m Forces Velocity v= v0 + at Position x = x0 + v0t + ½ at2

  17. y x positive y positive x

  18. y x negative y positive x

  19. y x x negative y negative

  20. y x x positive y negative

  21. y x positive y positive x ??????

  22. Drawing a FBD of forces on an object (on, not by) 1. Choose the object to analyze. Draw it as a dot. 2. What forces physically touch this object? This object, not some other 3. What “action at a distance” forces act on the object? Gravity is the only one for this PHYS2053 4. Draw these forces as arrows with tails at the dot (object). 5. Forces only! No accelerations, velocities, … Get components of Newton’s 2nd Law Choose a convenient xy coordinate system Find the x and y components of each force in the FBD Add the x and y components separately

  23. In a rescue, the 70.0  police officer is suspended by two cables, as shown in the figure below Find the tension in the cables.

  24. y T2 sin480 T2 T1 T1 sin350 48o 35o x T1 cos350 T2 cos480 w

More Related