Diese Präsentation wurde erfolgreich gemeldet.
Die SlideShare-Präsentation wird heruntergeladen. ×

# 20221102165222_PPT03.ppt

Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Anzeige
Nächste SlideShare
4_Forces.doc
Wird geladen in …3
×

1 von 33 Anzeige

# 20221102165222_PPT03.ppt

Laboratory physic

Laboratory physic

Anzeige
Anzeige

## Weitere Verwandte Inhalte

Anzeige

### 20221102165222_PPT03.ppt

1. 1. Topic 03 Two Dimensional Motion-2 And Force and Motion -1 Physics
2. 2. Two Dimensional Motion-2
3. 3. Topics 1. Projectile motion 2. Uniform Circular Motion
4. 4. 1. Projectile motion A particle moves in a vertical plane, with the only acceleration equal to the free fall acceleration, g . In projectile motion, the horizontal motion and the vertical motion are independent of each other, that is, neither motion affects the other. The initial velocity of the projectile is: j i yo x v v v ˆ ˆ 0   
5. 5. Horizontal Range, assuming no external forces: The horizontal range of a projectile is the horizontal distance when it returns to its launching height. The distance equations in the x- and y- directions respectively: Eliminating t:
6. 6. Horizontal Motion: no acceleration Vertical Motion; acceleration = g Eliminate time, t: 2 2 ) cos ( 2 ) 0 0 0 tan   ( v gx x y  
7. 7. Example, projectile motion: A rescue plane flies at 198 km/h and constant height h=500 m toward a point directly over a victim, where a rescue capsule is to land. What should be the angle φ of the pilot’s line of sight to the victim when the capsule release is made. (b) As the capsule reaches the water, what is its velocity V in unit-vector notation and in magnitude-angle notation.
8. 8. The speed of the particle is constant A particle travels around a circle/circular arc Uniform circular motion + 2. Uniform Circular Motion
9. 9. As the direction of the velocity of the particle changes, there is an acceleration !!! CENTRIPETAL (center-seeking) ACCELERATION Here v is the speed of the particle and r is the radius of the circle.
10. 10. The aceleration vector always points toward the center The velocity vector is always tangent to the path.
11. 11. Sample problem, top gun pilots
12. 12. We assume the turn is made with uniform circular motion. Then the pilot’s acceleration is centripetal and has magnitude a given by a = v2/R. Also, the time required to complete a full circle is the period given by T =2πR/v. Because we do not know radius R, let’s solve for R from the period equation for R and substitute into the acceleration eqn. Speed v here is the (constant) magnitude of the velocity during the turning.
13. 13. To find the period T of the motion, first note that the final velocity is the reverse of the initial velocity. This means the aircraft leaves on the opposite side of the circle from the initial point and must have completed half a circle in the given 24.0 s. Thus a full circle would have taken T 48.0 s. Substituting these values into our equation for a, we find
14. 14. Reference. Halliday D.; Resniick R. and Walker J. (2010). Principles Of Physics , ninth Edition , John Wiley & SONS Inc, New York, ISBN: 978-0-470-55653-5
15. 15. Force and Motion -1
16. 16. Topics 1. Newton’s Laws 2. Mass and weight 3. Applying Newton’s Laws
17. 17.  Study of relation between force and acceleration of a body: Newtonian Mechanics.  Newtonian Mechanics does not hold good for all situations. Examples: 1.Relativistic or near-relativistic motion 2.Motion of atomic-scale particles
18. 18. If the body is at rest, it stays at rest. If it is moving, it continues to move with the same velocity (same magnitude and same direction). (1) Newton’s First Law: If no force acts on a body, the body’s velocity cannot change; that is, the body cannot accelerate. 1. Newton’s Law
19. 19. a m Fnet    The net force on a body is equal to the product of the body’s mass and its acceleration. In component form, z net,z y y net, x x net, ma F ; ma F ; ma F    The acceleration component along a given axis is caused only by the sum of the force components along that same axis, and not by force components along any other axis. (2) Newton’s second law
20. 20. reaction action F F     (3) Newton’s Third Law When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction.
21. 21. mg F or g m F     Gravitational Force: A gravitational force on a body is a certain type of pull that is directed toward a second body. Suppose a body of mass m is in free fall with the free-fall acceleration of magnitude g. The force that the body feels as a result is: The weight, W, of a body is equal to the magnitude Fg of the gravitational force on the body. W = mg (weight), 2. Mass and weight
22. 22.  In a free-body diagram, the only body shown is the one for which we are summing forces.  Each force on the body is drawn as a vector arrow with its tail on the body. A coordinate system is usually included, and the acceleration of the body is sometimes shown with a vector arrow (labeled as an acceleration). Newton’s second law; drawing a free-body diagram The figure here shows two horizontal forces acting on a block on a frictionless floor. 3. Applying Newton’s Laws
23. 23. System Force Mass Acceleration SI Newton ( N ) kilogram (kg) m/s2 Cgs dyne gram (g) cm/s2 British pound (lb) slug ft/s2 Units in Newton’s Second Law 1 dyne = 1 g.m/s2 1 lb = 1 slug.ft/s2 1 N = 105 dyne = 0,2248 lb
24. 24. Sample problem 2/3/2023 A block S (the sliding block) with mass M =3.3 kg. The block is free to move along a horizontal frictionless surface and connected, by a cord that wraps over a frictionless pulley, to a second block H (the hanging block), with mass m 2.1 kg. The cord and pulley have negligible masses compared to the blocks (they are “massless”). The hanging block H falls as the sliding block S accelerates to the right. block S M m gH F  gS F  N F  T  T  Find (a) the acceleration of block S, (b) the acceleration of block H, and (c) the tension in the cord.
25. 25. a m F    Key Ideas: 1. Forces, masses, and accelerations are involved, and they should suggest Newton’s second law of motion: 2. The expression is a vector equation, so we can write it as three component equations. 3. Identify the forces acting on each of the bodies and draw free body diagrams a m F   
26. 26. Thus, for the block we can write Newton’s second law for a positive- upward y axis, (Fnet, y= may), as: for any vertical acceleration ay of the table and block
27. 27. y y Hanging block H m a Sliding Block S a m (a) (b) (a) A free-body diagram for block S (b) A free-body diagram for block H N F  gH F  gS F  T  T  From the free body diagrams, write Newton’s Second Law in the vector form, assuming a direction of acceleration for the whole system. Identify the net forces for the sliding and the hanging blocks: Fnet,x= max ; Fnet,y= may Fnet,z= maz
28. 28. For the sliding block, S, which does not accelerate vertically. Also, for S, in the x direction, there is only one force component, which is T. gS N gS N y y net, F F or 0 F F ma F      ma T ma F x x net,    For the hanging block, because the acceleration is along the y axis. We eliminate the pulley from consideration by assuming its mass to be negligible compared with the masses of the two blocks. With some algebra,
29. 29. Sample problem A cord pulls on a box up along a frictionless plane inclined at q= 300.The box has mass m =5.00 kg, and the force from the cord has magnitude T =25.0 N. What is the box’s acceleration component a along the inclined plane?
30. 30. For convenience, we draw a coordinate system and a free-body diagram as shown in Fig. b. The positive direction of the x axis is up the plane. Force from the cord is up the plane and has magnitude T=25.0 N. The gravitational force is downward and has magnitude mg =(5.00 kg)(9.8 m/s2) =49.0 N. Also, the component along the plane is down the plane and has magnitude mg sinθ as indicated in the following figure. To indicate the direction, we can write the down-the-plane component as -mg sin θ. The positive result indicates that the box accelerates up the plane. which gives: Using Newton’s Second Law, we have :
31. 31. 4. Normal Force: When a body presses against a surface, the surface deforms and pushes on the body with a normal force, FN, that is perpendicular to the surface. In the figure, forces Fg and FN and are the only two forces on the block and they are both vertical. (a)A block resting on a table experiences a normal force perpendicular to the tabletop. (b) The free-body diagram for the block.
32. 32. Reference. Halliday D.; Resniick R. and Walker J. (2010). Principles Of Physics , ninth Edition , John Wiley & SONS Inc, New York, ISBN: 978-0- 470-55653-5
33. 33. Thank You