Chapter 3 Motion in a Plane

Transcription

1 Chapter 3 Motion in a Plane Introduce ectors and scalars. Vectors hae direction as well as magnitude. The are represented b arrows. The arrow points in the direction of the ector and its length is related to the ector s magnitude. Scalars onl hae magnitude. We write = B if the ectors hae the same magnitude and point in the same direction. Scalars can hae magnitude, algebraic sign, and units. dding scalars is er familiar. You add 10 grams to 15 grams and get 5 grams. You hae $0 and gie $5 to friend and ou hae $15 remaining. Vector addition is different since ectors hae direction as well as magnitude. How do we add ectors? We alread know how to add ectors in one dimension (along the -ais for eample). B +B What happened? The ector B is positioned so that the tail of B is positioned at the head of. The ector sum is drawn from the tail of to the head of B. If is 8 m long and B is 10 m long, the magnitude of +B is 18 m. What if B is reersed? B +B What is the magnitude of +B? Here we see a hint of the problem. Vector do not add like scalars. How do we add ectors that do not point along the same direction?

2 1. Draw the first ector in the correct direction and with the appropriate magnitude.. Draw the second ector with the correct direction and magnitude so that its tail is placed at the head of the first ector. 3. If there is a third ector, draw it with the correct direction and magnitude to that its tail is placed at the head of the second ector. 4. When finished with all the ectors, find the ector sum b drawing a ector that starts at the tail of the first ector and ends at the head of the last ector. How do we subtract ectors? Use B = +( B). What is a reasonable definition for B? The negatie of a ector has the same magnitude as the original ector but points in the opposite direction. The idea of ectors is built from the idea of displacement. In the diagram aboe, imagine that ou are in a forest. is our walk to a tree and B is our walk from the first tree to a friend ou see across the forest. +B is our net displacement from our starting point. nother eample: This procedure is called the graphical addition of ectors. You need to understand this procedure. Howeer, it is too slow and imprecise to be used in soling problems.

3 We add ectors b taking their components. The process is summarized in this figure. We are adding two ectors that are not collinear. We replace each ector with two ectors (called its components). We then add like components together, giing the components of the ector sum. What happens net? C C C We add C and C to find C. Since the - and -aes are perpendicular, we can find the magnitude of C from the Pthagorean theorem, C C C The direction is normall measured counterclockwise from the +-ais. For this ector in the nd quadrant, first find arctan C C and then subtract from 180º. (Wh?) How do we find the components of a ector? s an eample suppose has magnitude 0 N and it points at 40º. s the following diagram shows, we are dealing with a right triangle. To find the components we need to use trigonometr. Recall, cos adjacent hpotenuse sin opposite hpotenuse tan opposite adjacent

4 =40 o Using the definitions of cosine and sine, cos sin adjacent hpotenuse cos (0N) cos40 opposite hpotenuse sin (0N)sin N 1.9 N Wh is >? When are the equal? Suppose we had this picture. What would ou do? =50 o (0N)sin50 (0N) cos N 1.9 N Usuall we hae cosine associated with -components and sine associated with -components, but not alwas. You hae to look at the diagram. ( er common remark for this semester!) Problem-Soling Strateg: Finding the - and -components of a Vector from its Magnitude and Direction (page 60) 1. Draw a right triangle with the ector as the hpotenuse and the other two sides parallel to the - and -aes.. Determine one of the unknown angles in the triangle. 3. Use trigonometric functions to find the magnitudes of the components. Make sure our calculator is in degree mode to ealuate trigonometric functions of angles in degrees and in radian mode for angles in radians. 4. Determine the correct algebraic sign for each component.

5 Problem-Soling Strateg: Finding the Magnitude and Direction of a Vector from its - and -components (page 60) 1. Sketch the ector on a set of - and -aes in the correct quadrant, according to the signs of the components.. Draw a right triangle with the ector as the hpotenuse and the other two sides parallel to the - and -aes. 3. In the right triangle, choose which of the unknown angles ou want to determine. 4. Use the inerse tangent function to find the angle. The lengths of the sides of the triangle represent and. If is opposite the side parallel the side perpendicular to the - ais, then tan = opposite/adjacent = /. If is opposite the side parallel the side perpendicular to the -ais, then tan = opposite/adjacent = /. If our calculator is in degree mode, then the result of the inerse tangent will be in degrees. [In general, the inerse tangent has has two possible alues between 0 and 360º because tan = tan ( + 180º). Howeer, when the inerse tangent is used to find one of the angles in a right triangle, the result can neer be greater than 90º, so the alue the calculator returns is the one ou want. 5. Interpret the angle: specif whether it is the angle below the horizontal, or the angle west of south, or the angle clockwise from the negatie -ais, etc. 6. Use the Pthagorean theorem to find the magnitude of the ector. Problem-Soling Strateg: dding Vectors Using Components (page 61) 1. Find the - and -components of each ector to be added.. dd the -components (with their algebraic signs) of the ectors to find the -component of the sum. (If the signs are not correct, the sum will not be correct. 3. dd the -components (with their algebraic signs) of the ectors to find the -component of the sum. 4. If necessar, use the - and -components of the sum to find the magnitude and direction of the sum. Een when using the component method to add ectors, the graphical method is an important first step. Graphical addition gies ou a mental picture of what is going on. problem can be made easier to sole with a good choice of aes. Common choices are -ais horizontal and -ais ertical, when the ectors all lie in the ertical plane; -ais east and -ais north, when the ectors lie in a horizontal plane; and -ais parallel to an inclined surface and -ais perpendicular to it. Read the section on unit ectors on pages We will not use the unit ector approach, but ou ma be familiar with it. Now let s use the concept of ectors to etend the kinematical ariables to more dimensions.

6 erage elocit is the displacement oer the time, a r t Instantaneous elocit is r lim t 0 t The elocit is tangent to the path of the particle. The aerage acceleration is a a t Instantaneous acceleration is a lim t 0 t For straight-line motion the acceleration is alwas along the same line as the elocit. For motion in two dimensions, the acceleration ector can make an angle with the elocit ector because the elocit ector and change in magnitude, in direction, or both. The direction of the during a er short time. The aboe definitions look good, but the are not useful. We call these formal definitions. The are not used in soling problems. Instead we need a set of definitions for the - and - components. The basic rule is WE DO NOT DEL WITH VECTORS. WE DEL WITH THEIR COMPONENTS. For the elocit, we hae

7 , a, t a t with similar definitions for the other parameters. (See pages ) We can now generalize the equations at the top of these notes to two dimensions. It is generall easiest to choose the aes so that the acceleration has onl one non-zero component. We choose the -ais along the direction of acceleration. This means a = 0. f i a t becomes 0 and f i a t 1 ( ) t f i becomes 1 t and ( f i) t t i 1 ) a ( t becomes t and t 1 ) a ( t f a i becomes f 0 and a i Summar: -ais : a = 0 -ais: constant a Equation 0 a t (3-19) f t ( f i) t (3-0) 1 ) t a ( t (3-1) i f i a (3-) f i

8 Projectiles are a good eample of this tpe of motion. Here a = g. This motion is simultaneous constant elocit in the -direction and constant acceleration in the -direction. Relatie elocit is a great eample of adding ectors. Hae ou eer had this happen to ou? While sitting in our car at a red traffic light, the car beside ou slowl drifts forward. You mash on the brake to stop our car from rolling backwards, but our car is not moing. Within our enironment, there is no wa to distinguish between our car moing backwards and the car besides ou moing forward. The elocit is relatie. We need a reference frame (the traffic light, for eample) to define who is moing.

9 The train moes at 10 m/s and Wanda can walk at 1 m/s. How fast will Greg see Wanda walk? Wanda s elocit relatie to Greg is the sum of the elocit of the Wanda relatie to the train plus the elocit of train relatie to Greg. WG Notice the order of the subscripts. We hae the Ts cancelling from the two terms on the right. This equation will alwas hold, but how do we use it? What is our rule about ectors? WT WE DO NOT DEL WITH VECTORS. WE DEL WITH THEIR COMPONENTS. Take the -component: WG WT 11m/s TG TG ( 1m/s) ( 10m/s) Greg sees Wanda walking to the right at 11 m/s. What happens when she walks back to her seat? WG WT ( 1m/s) ( 10m/s) 9m/s ccording to Greg, Wanda is walking at 9 m/s to the right. Hopefull, this is prett eas. But what about this? From Eample Jack wants to row directl across the rier from the east shore to a point on the west shore. The current 0.60 m/s and Jack can row at 0.90 m/s. What direction must he point the boat and what is his elocit across the rier? The elocit of the rowboat relatie to the shore is equal to the elocit of the rowboat relatie to the water plus the elocit of the water relatie to the shore. RS The rowboat is to head directl to the west. WS TG

10 Take components. RS and WS RS WS The diagram is the ke to soling relatie elocit problems. For the -component, RS RS WS cos 0 cos The -component, RS 0 WS WS sin sin WS Our unknowns are and RS. From the -component equation, WS sin WS 0.6 m/s 0.9 m/s sin From the -component equation. RS cos (0.90m/s)cos m/s

11 The boat must point 41.8º N of W upstream. Its speed across the water is 0.67 m/s.