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VECTORS AND SCALARS

VECTORS AND SCALARS A scalar quantity has only magnitude and is completely specified by a number and a unit. Examples are mass (a stone has a mass of 2 kg), volume (1.5 L), and frequency (60 Hz). Scalar quantities of the same kind are added by using ordinary arithmetic.

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VECTORS AND SCALARS

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  1. VECTORS AND SCALARS A scalar quantity has only magnitude and is completely specified by a number and a unit. Examples are mass (a stone has a mass of 2 kg), volume (1.5 L), and frequency (60 Hz). Scalar quantities of the same kind are added by using ordinary arithmetic.

  2. A vector quantity has both magnitude and direction. Examples are displacement (an airplane has flown 200 km to the southwest), velocity (a car is moving at 60 km/h to the north), and force (a person applies an upward force of 25 N to a package). When vector quantities are added, their directions must be taken intoaccount.

  3. A vectoris represented by an arrowed line whose length is proportional to the vector quantity and whose direction indicates the direction of the vector quantity.

  4. The resultant, or sum, of a number of vectors of a particular type (force vectors, for example) is that single vector that would have the same effect as all the original vectors taken together.

  5. GRAPHICAL ADDITION OF VECTORS PARALLELOGRAM METHOD The resultant of two vectors acting at any angle may be represented by the diagonalof a parallelogram. The two vectors are drawn as the sides of the parallelogram and the resultant is its diagonal. The directionof the resultant is away from the origin of the two vectors.

  6. 3.1 Find the magnitude and direction of the resultant force produced by a vertically upward force of 40 N and a left horizontal force of 30 N. A (40 N, up) B (30 N, left) Scale: 1 cm = 10 N R A R (50 N, 127) B

  7. 3.2 Using the graphical method, find the resultant of the following two displacements: A (2.0 m, 40) B (4.0 m, 127) Scale 1 cm = 1 m R B A R (4.5 m, 101)

  8. POLYGON or TIP-TO-TAIL METHOD This method for finding the resultantR of several vectors (A, B, C) consists in beginning at any convenient point and drawing (to scale in the proper directions) each vector arrow in turn. They may be taken in any order of succession A + B + C = C + A + B. The tail of each arrow is positioned at the head of the preceding one.

  9. The resultant is represented by an arrow with its tail end at the starting point and its head at the tip of the last vector added. The size or magnitude of R is given by: R = R

  10. TIP-TO-TAIL METHOD PARALLELOGRAM METHOD

  11. 90 0 180 360 270

  12. 3.3 Add the following two forces by use of the tip-to-tail method: A (30 N at 30) B (20 N at 140) Scale: 1 cm = 10 N B R R (30 N, 72) A The resultant is drawn from tail end at the starting point to thehead of the last vector added.

  13. 3.4 Three ropes are tied to a stake. Find the resultant force. A (20 N, 0) B (30 N, 150) C (40 N, 230) Scale: 1 cm = 10 N B C A R R (35 N, 206)

  14. VECTOR COMPONENTS A component of a vector is its effective value in a given direction. For example, the x-component of a displacement is the displacement parallel to the x-axis caused by the given displacement. A vector in two dimensions may be resolved into two component vectors acting along any two mutually perpendicular directions. Rx = R cos Ry = R sin 

  15. 3.5 Find the components of the vector: F (250 N, 235o) Fx = F cos  = 250 cos (235o) = - 143.4 N Fy = F sin  = 250 sin (235o) = - 204.7 N Fx Fy

  16. VECTOR ADDITION: COMPONENT METHOD To add two or more vectors A, B, C,… by the component method, follow this procedure: 1. Resolve the initial vectors into components x and y. 2.Add the components in the x direction to give Rxand add the components in the y direction to give Ry. That is, the magnitudes of Rx and Ry are given by, respectively: Rx = Ax + Bx + Cx… Ry = Ay + By + Cy… 3. Calculate the magnitude and direction of the resultant R from its components by using the Pythagorean theorem: and

  17. 3.6 Four coplanar forces act on a body at point O as shown in the figure. Find their resultant graphically. A (80 N, 0) B (100 N, 45) C (110 N, 150) D (160 N, 200) Scale: 1 cm = 20 N D C R B A R (119 N, 143)

  18. 3.7 Solve problem 3.6 by use of the component method. A (80 N, 0)B (100 N, 45) C (110 N, 150)D (160 N, 200) x-component y-component 80 cos 080 sin 0 100 cos 45100 sin 45 110 cos 150110 sin 150 160 cos 200160 sin 200 Σx = -95 N Σy = 71 N = 118.6 N

  19. = 36.7 Since Σx = (-) and Σy = (+) R is in the III Quadrant, therefore: 180 - 36.7= 143.3 R (118.6 N, 143.3)

  20. 3.8 The five coplanar forces shown in the figure act on an object. Find their resultant with the component method. A (19 N, 0) B (15 N, 60) C (16 N, 135) D (11 N, 210) E (22 N, 270) x-component y-component 19 cos 019 sin 0 15 cos 6015 sin 60 16 cos 13516 sin 135 11 cos 21011 sin 210 22 cos 27022 sin 270 Σx = 5.7 N Σy = -3.2 N

  21. = 6.5 N = 29 Since Σx = (+) and Σy = (-) R is in the IV Quadrant, therefore: 360 - 29= 331 R (6.5 N, 331)

  22. 3.9 A force of 100 N makes an angle with the x-axis and has a y-component of 30 N. Find both the x-component and the angle . A (100 N,  ) Ay = 30 N A Ay  Ax = A cos  = 100 cos 17.5 = 95 N = 17.5

  23. 3.10 A child pulls on a rope attached to a sled with a force of 60 N. The rope is at 40with respect to the ground. a. Calculate the effective value of the pull tending to move the sled along the ground. Fx = F cos  = 60 cos 40 = 46 N F = 60 N θ = 40 b. Calculate the force tending to lift the sled vertically. Fy = F sin  = 60 sin 40 = 39 N

  24. 3.11 A plane is traveling eastward at a speed of 500 km/h. A 90 km/h wind is blowing southward. What are the direction and speed of the plane relative to the ground? A (500 km/h, E) B (90 km/h, S) = 508 km/h = 10 360 - 10 = 350 R (508 km/h, 350)

  25. PROJECTILE MOTION An object launched into space without motive power of its own is called a projectile. If we neglect air resistance, the only force acting on a projectile is its weight, which causes its path to deviate from a straight line. The projectile has a constant horizontal velocity and a vertical velocity that changes uniformly under the influence of gravity.

  26. THE MONKEY AND THE ZOOKEPER The monkey spends most of its day hanging from a branch of a tree. The zookeeper feeds the monkey by shooting bananas from a banana cannon to the monkey in the tree. The monkey usually drops from the tree the moment that the banana leaves the muzzle of the cannon. The zookeeper is faced with the dilemma of where to aim the banana cannon in order to feed the monkey. If the monkey lets go of the tree the moment that the banana is fired, then where should he aim the banana cannon? http://www.fisica.uniud.it/~deangeli/applets/Multimedia/ExplrSci/dswmedia/monkey.htm

  27. HORIZONTAL PROJECTION  If an object is projected horizontally, its motion can best be described by considering its horizontal and vertical motion separately. In the figure we can see that the vertical velocity and position increase with time as those of a free-falling body. Note that the horizontal distance increases linearly with time, indicating a constant horizontal velocity.

  28. 3.14 A stunt flier is moving at 15 m/s parallel to the flat ground 100 m below. How large must the distance x from the plane from the target be if a sack of flour released from the plane is to strike the target? vx = 15 m/s y = 100 m vy = 0 y = ½ gt2 = 4.52 s x = vx t = 15(4.52) = 67.8 m

  29. 3.15 A person standing on a cliff throws a stone with a horizontal velocity of 15.0 m/s and the stone hits the ground 47 m from the base of the cliff. How high is the cliff? vx = 15 m/s x = 47 m vy = 0 = 3.13 s y = ½ gt2 = ½ (9.8)(3.13)2 = 48 m

  30. 3.16 A cannonball is projected horizontally with an initial velocity of 120 m/s from the top of a cliff 250 m above a lake. a. In what time will it strike the water at the foot of the cliff? vx = 120 m/s y = 250 m vy = 0 = 7.14 s b. What is the x-distance (range) from the foot of the cliff to the point of impact in the lake? x = vx t = 120(7.14) = 857 m

  31. c. What are the horizontal and vertical components of its final velocity? vx = 120 m/s vy = voy + gt = 9.8(7.14) = 70 m/s d. What is the final velocity at the point of impact and its direction? = 139 m/s = 30.2 below horizontal v (139 m/s, 30.2)

  32. PROJECTILE MOTION AT AN ANGLE The more general case of projectile motion occurs when the projectile is fired at an angle.

  33. Problem solution: 1. Upward direction is positive. Acceleration (g) is downward thus negative. 2. Resolve the initial velocity vo into its x and y components: and 3. The horizontal and vertical components of its position at any instant is given by: and 4. The horizontal and vertical components of its velocity at any instant are given by: and 5. The finalposition and velocity can then be obtained from their components.

  34. 3.17 A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40. How far above or below its original level will the ball strike the opposite wall? x = 50 m vo = 20 m/s, 40 g = - 9.8 m/s2 vox = 20 cos 40 = 15.3 m/s voy = 20 sin 40 = 12.9 m/s = 3.27 s y = voy t + ½ gt2 = 12.9(3.27) + ½ (-9.8)(3.27)2 = - 10.2 m or 10.2 m below its original level

  35. 3.18 An artillery shell is fired with an initial velocity of 100 m/s at an angle of 30 above the horizontal. Find: a. Its position and velocity after 8 s vo = 100 m/s, 40 t = 8 s g = - 9.8 m/s2 vox = 100 cos 30 = 86.6 m/s voy = 100 sin 30 = 50 m/s x = vox t = 86.6(8) = 692.8 m y = voy t + ½ gt2 = 50(8) + ½ (-9.8)(8)2 = 86.4 m vx = vox = 86.6 m/s vy = voy + gt = 50 + (-9.8)(8) = - 28.4 m/s

  36. b. The time required to reach its maximum height At top vy = 0vy = voy + gt = 5.1 s c. The horizontal range R Total time T = 2t = 2(5.1) = 10.2 s x = vox t = 86.6(10.2) = 883.7 m

  37. 3.19 A baseball is thrown with an initial velocity of 120 m/s at an angle of 40above the horizontal. How far from the throwing point will the baseball attain its original level? vox = 120 cos 40 = 91.9 m/s voy = 120 sin 40 = 77.1 m/s vo = 120 m/s, 40 g = - 9.8 m/s2 At top vy = 0 = 7.9 s x = vox (2t) = 91.9(2)(7.9) = 1452 m

  38. 3.20 a. Find the range of a gun which fires a shell with muzzle velocity v at an angle . What is the maximum range possible? At top vy = 0 vy = voy + gt = vo sin θ - gt Total time = 2t x = vxt

  39. sin θcos θ= ½ sin 2θ Maximum range is 45 since 2θ = 90

  40. b. Find the angle of elevation  of a gun that fires a shell with muzzle velocity of 120 m/s and hits a target on the same level but 1300 m distant. vo = 120 m/s x = 1300 m = 0.885 sin-1(2θ)= 62 θ = 31

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