The Dot Product Pg. 377 # 6ace, 7bdf, 9, 11, 14 Pg. 385 # 2, 3, 4, 6bd, 7, 9b, 10, 14 Sept. 25

Transcription

1 UNIT 2 - APPLICATIONS OF VECTORS Date Lesson TOPIC Homework Sept (11) 7.1 Vectors as Forces Pg. 362 # 2, 5a, 6, 8, 10 13, 16, 17 Sept (12) 7.2 Velocity as Vectors Pg. 369 # 2,3, 4, 6, 7, 9, 11, 12 Sept (13) 7.3/ 7.4 The Dot Product Pg. 377 # 6ace, 7bdf, 9, 11, 14 Pg. 385 # 2, 3, 4, 6bd, 7, 9b, 10, 14 Sept (14) 7.6 The Cross Product Pg. 407 # 3, 4ace, 5, 7, 8a, 9b, 11 Sept (15) 7.5 Scalar and Vector Projections Pg. 398 # 6, 7ac, 8a, 11, 13 Sept (16) 7.7 Applications of The Dot and Cross Product Pg. 414 # 1, 2,3, 5, 6, 8 Sept (17) Reiew for Unit 2 Test Pg. 418 # 1, 3, 4, 6, , 15, 17, 20, 23, 30, 31, 34 Oct (18) TEST- UNIT 2

2 MCV 4U Lesson 2.1 Vectors as Forces A force can be described as a push or a pull on an object. It is measured in Newtons (N). To describe a force it is necessary to state: 1. its direction 2. the point at which it is applied 3. its magnitude 40 N.P 30 N The diagram shows 40 N and 30 N forces acting in opposite directions at point P. The combined effect of these two forces as a 10 N force to the left. This single force that has the same combined effect as two forces combined is called the resultant. The equilibrant is the opposite force that would exactly counterbalance the resultant force. In this case the equilibrant force would be a 10 N force to the right.

3 Ex. 1 Two horses are pulling a load. The chains between them are at an angle of 45 to each other. One horse pulls with a force of 260 N, the other with a force of 320 N. a) What is the magnitude of the resultant force on the load? In what direction is this force? b) What is the magnitude of the equilibrant force on the load? In what direction is this force? Ex. 2 A sign is suspended in the lobby of GHS by two wires making angles with the ceiling of 50 and 30. If the sign weighs 2 kg, find the tensions (forces) acting in the wires. Proide an accurate ector diagram. N.B. The magnitude of a force is measured in Newtons (N). At the earth's surface, acceleration due to graity is approximately 9.8 m/s 2. Force = Mass acceleration due to graity (where mass is measured in kg) When calculating the force we use the constant 9.8.

4 Ex. 3 A lawn mower is pushed with a force of 90 N directed along the handle, which makes an angle of 36 with the ground. a) Determine the horizontal and ertical components of the force on the mower. b) Describe the physical consequence of each component of the pushing force.

5 Ex. Find the resultant of the ectors below. Pg. 362 # 2, 5a, 6, 8, 10 13, 15-17

6 MCV 4U Lesson 2.2 Velocity as a Vector Since elocity has direction it can be represented by a ector. When a plane flies, its elocity relatie to the earth (ground elocity) is the resultant of (1) the plane's elocity through the air (air elocity) and (2) the elocity of the wind. g a w Relatie Velocity The elocity of an object is its elocity relatie to the frame of reference of some obserer in a gien situation. ie: 80 km/h Bus WEST 100 km/h Car EAST 90 km/h Truck To someone in the bus the car is passing at 20 km/h. the elocity of the car relatie to the bus is 20 km/h east. The elocity of the truck relatie to the car is 190 km/h west. Thus we hae the principle: Velocity of A relatie to B = Velocity of A elocity of B Ex. 1 A plane is steering at N60 E at an airspeed of 450 km/h. The wind is from the N50 W at 75km/h. Determine the planes ground elocity.

7 Ex. 2 A canoeist who can paddle at 4 km/h in still water wants to cross a 500 m wide rier that has a current of 1 km/h. a) If he steers the canoe in a direction perpendicular to the current, determine theresultant elocity and the point on the opposite bank where he lands. b) If he had wished to trael straight across the rier, determine the direction in which he must head and the time it takes to cross the rier. Pg. 369 # 2,3, 4, 6, 7, 9, 11, 12

8 MCV 4U Lesson 2.3 The Dot Product Ex. 1 Find the angle between ectors u (3, 1) and (2, 5). (2, 5) Pg. 377 # 6ace, 7bdf, 9, 11, 14 Pg. 385 # 2, 3, 4, 6bd, 7, 9b, 10, 14 w u (3, 1) We can define the dot product of two algebraic ectors u and as u u cos, where is the angle between u and and u u 1 1 u 2 2 when in component form. ie: u (u 1, u 2 ) and ( 1, 2 ). Similarly, if u (u 1, u 2, u 3 ) and ( 1, 2, 3 ) then u u 1 1 u 2 2 u 3 3 The dot product of two ectors is a scalar quantity, so it is sometimes called the scalar product. Ex. 2 Find u for each of the following. a) u (1,3), ( 2,5) b) u ( 4,2, 3), (2, 3, 1)

9 If we want to find the angle between two ectors, we can use cos u u. Ex. 3 Find between the following pair of ectors. a) u ( 2,5), (3,6) b) u (1,1,2) ( 1,2,1) Ex. 4 Find u in each of the following cases. a) u 6, 8, 60 b) u 3, 12, 6 c) u 2, 9, 20 OTHER THINGS TO CONSIDER: Properties of the Dot Product: 1. Associatie: a(u ) (au ) u (a ) 2. Distributie: u ( w) u u w 3. u u u 2, since u u u u cos0 u u (1) u 2 Also, consider the dot product of unit ectors. i i 1 j j 1 k k 1 i j j i 0 i k k i 0 j k k j 0 Examples: 3. If u (1,2,3) 2. If a (1,2) b (3,4) c (5,6) 2 u u u Distributie Property:

10 MCV 4U Lesson 2.4 The Cross Product The cross product is a ector quantity unlike the dot product, which is a scalar quantity. As a result, the cross product is also known as the ector product. For ectors u and, we define the cross product as: u u sin n where us the angle between u and, n is a unit ector perpendicular to both u and such that u, and n form a right handed system as indicated below. 1. Place ectors tail to tail. 2. Place right hand along first ector u with fingers extended in the direction of the ector. 3. Curl the fingers in the direction of the angle that makes with u 4. Thumb gies direction of u. The magnitude of the cross product of two ectors u and is u u sin. For non-zero ectors u and, u and are collinear iff u 0 u The properties of the cross product are similar to those of the dot product. 1. u ( w ) u u w 2. (u ) w u w w 3. ku k(u ) u (k ) 4. a a 0 The cross product in component form, where u = u, u, ) and =,, ) is gien by: u (u 2 3 u 3 2,u 3 1 u 1 3,u 1 2 u 2 1 ) ( 1 2 u3 ( If u and are two non-collinear ectors in three-dimensional space, then eery ector perpendicular to both u and is in the form k(u ), k R. For the unit ectors i, j, and k, we hae the following results. i i 0 i j k j i k j j 0 j k i k j i k k 0 k i j i k j NOTE: u u

11 Ex. 1 For the following pairs of ectors, state whether u is directed into or out of the page. u u u Ex. 2 Find a ector which is perpendicular to both u = (1, 2, 3) and = (3, 4, 5). Ex. 3 u = 5 and = 11 and the angle between them is 85. Determine u. Ex. 4 Find the cross product of a 2i 4 j 3k and b 7i 3 j k. Pg. 407 # 3, 4ace, 5, 7, 8a, 9b, 11

12 MCV 4U Lesson 2.5 Scalar and Vector Projections I. Scalar Projections: For ectors u and, 0, proj u u cos and u scalar proj(u onto ) ie: cos adj hyp cos Projection u Projection = u cos We know that u u cos. We can rewrite this formula as u u cos Soling for u cos gies u cos = u. So, the Scalar Projection of u onto = u and the Scalar Projection of onto u = u u In general, u u u II. Vector Projections: A projection is formed by dropping a perpendicular from each point in an object onto a line or plane. A shadow is a projection if the light rays forming the shadow meet a line or plane at right angles. A projection is a ector. Used in physics, computer programming and astronomy to determine the lengths of "shadows" projected onto a surface. The ector we obtain by projecting u perpendicularly onto the line through is called the ector projection of u onto and is denoted proj(u onto ) or proj u. For ectors u and, 0, proj u u cos ( ) u where is a unit ector in the direction of and Vector proj(u onto )

13 If the ectors are not tail to tail, drop perpendiculars from the tail and the tip of u to meet at O and P and OP is the projection of u onto. proj u u cos or we hae: O Magnitude ector proj(u onto ) P u 2 Vector proj( u onto ) = [ Scalar proj( u onto )]( ector ˆ )] u = = ( u ) (u ) 2 = u III. Direction Cosines Any ector u in space can be written as an ordered triple where u OP (a,b,c) and where a, b, and c are its x-, y-, and z-components. The ector u in space can also be written using the ectors i ˆ, jˆ, and k, where i j and k ˆ, ˆ, are unit ectors along the x-, y-, and z-axes, respectiely. î = (1, 0, 0) ĵ = (0, 1, 0) k = (0, 0, 1) u OP = a î + b ĵ + c k Its magnitude is gien by u a b c To plot a ector in space, you moe a units along the x-axis, then b units parallel to the y-axis, and then c units parallel to the z-axis. Drawing a rectangular prism (box) is often helpful when doing this. The direction angles of a ector (a, b, c) are the angles,, and that the ectors make with the positie x-, y-, and z-axes, respectiely, where 0,, 180 The direction cosines of a ector u are the cosines of the direction angles,, and where cos = a u, cos = b u, cos = c u., where u a b c. Thus the direction cosines are the components of a unit ector. cos 2 + cos 2 + cos 2 = 1

14 Ex. 1 Find the ector proj(b onto a ) gien: a) a ( 4, 2) b (3,1) b) a ( 1,3,6) b ( 2,1, 3) Ex. 2 Find the scalar projections of a onto b and b onto a if a = (2, 1, 9) and b = ( 3, 0, 4). Ex 3. Find the direction cosines and the direction angles of the ector u - (2, 3, -4). Pg. 398 # 6, 7ac, 8a, 11, 13

15 MCV 4U Lesson 2.6 Applications of the Dot Product and Cross Product I. Forces A: Work B: Torque (Moment of force) - is a measure of how much a force acting on an object causes that object to rotate. The turning effect of a force is called torque and is defined by the cross product. T r F r F sin (n ) where F is the applied force, r is the ector determined by the leer arm acting from the axis of rotation, is the angle between the force and the leer arm and n is a unit ector perpendicular to both r and F. Torque is a ector quantity and is measured in Newton-metres (N m). While it is dimensionally correct to express joules as Newton-metres or N m, such use is discouraged by the SI authority to aoid confusion with torque. Torque and energy are fundamentally different physical quantities. For example, adding 1 N m of torque to 1 N m of energy gies a dimensionally consistent result of 2 N m, but this quantity is physically meaningless.

16 One joule in eeryday life is approximately: the energy required to lift a small apple one meter straight up. the energy released when that same apple falls one meter to the ground. the energy released as heat by a person at rest, eery hundredth of a second. one hundredth of the energy a person can receie by drinking a drop of beer. the kinetic energy of an adult human moing at a speed of about a handspan eery second. the kinetic energy of a tennis ball moing at 23 km/h (14 mph).

17 III. Area and Volume of a parallelogram or triangle formed by ectors a and b. Area of a Parallelogram Area of a Triangle = 1 (Area of a Parallelogram) 2 a h a sin a b Area = base x height b a sin b a b = 1 2 b a Volume of a Parallelepiped c b a A parallelepiped is a six faced solid where the opposite faces are congruent parallelograms. Let a, b, and c be the three sides as shown. Volume area of base height a b height This is a triple scalar product. If the triple scalar product of three ectors is zero the ectors are coplanar. Therefore the parallelepiped is flat and has a zero olume. Now the height of the figure is the distance between the upper and lower faces. a b will gie us a ector perpendicular to the plane containing a and b. The height will be the magnitude of the projection of c onto ( a b ). Volume a b height a b proj[ c onto ( a b)] c ( a b) a b a b c ( a b)

18 Ex. 1 A crate on a ramp is hauled 10 m up the ramp under a constant force of 35 N applied at an angle of 25 to the ramp. Find the work done. Ex. 2 A 75 N force is applied to the end of a 0.3 m long wrench and makes an angle of 70 with the handle. What is the torque on the bolt at the other end of the wrench? Pg. 414 # 1, 2,3, 5, 6, 8