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Polaris Coordinates of a Vector

5. 4. 3. 2. 1. p/6. Polaris Coordinates of a Vector. How can we represent a vector?. We plot an arrow: • the length proportional to magnitude of vector • the line represents the vector direction • the point represents the vector path. . e.g. we plot a vector with:

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Polaris Coordinates of a Vector

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  1. 5 4 3 2 1 p/6 Polaris Coordinates of a Vector How can we represent a vector? We plot an arrow: • the length proportional to magnitude of vector • the line represents the vectordirection • the point represents the vectorpath  e.g. we plot a vector with: direction: inclination p/6 path: upwards magnitude: 5 u direction: inclination p/6 path: upwards magnitude: 5 u The Polaris Coordinates are (magnitude, inclination) (r, a) e.g. (5, p/6) 
  2. y x O r·sin(a) r·cos(a) a=p/6 Cartesian Coordinate of a Vector We can use the Cartesians Axes for represent a vector e.g. we plot the previous vector: direction: slope p/6 path: upwards magnitude: 5 u (r·cos(a), r·sin(a)) We can write the vector as The Cartesians Coordinates are (x-coordinate, y-coordinate) (x, y) (r·cos(a), r·sin(a)) e.g. r·cos(a)andr·sin(a) are known as Cartesian Components 
  3. y (r·cos(a), r·sin(a)) r·sin(a) x (x, y) (r, a) O Cartesians Coordinates Polaris Coordinates r·cos(a) a=p/6 5 4 3 2 1 p/6 Change of Coordinate: Cartesians  Polaris From Cartesian toPolar    From Polar toCartesian 
  4. Calculation with Vectors • Product with a scalar (k) k=3  a has the samedirection and path as b and magnitude of a is three times greater than the magnitude of b if 0<k<1 the magnitude of a is smaller than the magnitude of b if k<0 the path of a is the opposite ofpath of b k = -1  
  5. + Calculation with Vectors The Parallelogram Law • Addition of vectors  We calculate the magnitude of s with Carnot Theorem s2 = a2 + b2 – 2ab cos(a) a N.B.We alwayshave sa+b sis known as the Resultantofaandb 
  6. y x s·cos(a) O + a·cos(a) b·cos(b) s·sin(a) a·sin(a) b·sin(b) Calculation with Vectors • Addition of vectors, Cartesian Coordinates  We describe the vectorsums by adding the components of a and b 
  7. -  -  Calculation with Vectors • difference between vectors Proof We describe vector differencedby subtracting the components of b ofa In both cases the magnitude of d is a  d2 = a2 + b2 – 2ab cos(a)
  8. • Difference • Addition a a Calculation with Vectors Comparison between addition and difference of vectors magnitude of s is magnitude of d is s2 = a2 + b2 – 2ab cos(a) d2 = a2 + b2 – 2ab cos(a) N.B.Observe the position of a angle 
  9. d2 a2 b2 c2 O O c1 O a1 b1 O d1 Decomposition of a vector To decompose avector vfindtwo vectorsv1,v2in two prefixed direction,whose sumis equaltovector v now from v we can draw two lines parallel to the axes in every case we can write: 
  10. O z1 z1 z1 z1 z2 z2 z2 z2 Decomposition of a vector: Examples 1stA body is hung to the ceiling with two different ropes. It forms angles a1, and a2 with the ropes. What are the tensions of the ropes?  We can decompose v in directions z1, z2 to find 2 vectors, v1 and v2, whose sum is equal to v: v= v1 +v2 The vector of bodyweight is directed downwards The vectors of tensions are direct toward the ropes So we can recognize two particular direction: the lines alongthe ropes We can observe that T1= -v1 and T2= -v2 Therefore v is balanced by the two vectors T1,T2tensions of ropes: -v= T1+T2  So we draw a system of reference with the AXES parallel to the ropes and ORIGINATING from the body 
  11. z2 z2 z2 z2 z1 z1 z1 z1 a Decomposition of a vector: Examples  2ndA body is sliding on an inclined plane. What force pulls down the body? Vector vdirected downwardsis theForce(oracceleration)of gravity Vector v1directedparalleltotheplaneis theactive componentof theForce(oracceleration) of gravityresponsible of sliding of the body.It is indicated byv// v// = v· sin(a) Vector v2directedperpendicularlyto the plane is thecomponentof theForce(oracceleration)of gravitythat holds the body to the plane. It is indicated byvperp v perp = v· cos(a) The vector of bodyweight is directed downwards So we can recognize two particular direction: the line alongthe plane and the line theyperpendicular to it  Therefore we draw a system of reference with the AXES parallel and perpendicular to the plane  In this way we find the force that pulls the body downwards. This is vector v1 We can decompose v in directions z1, z2 to find 2 vectors, v1 and v2, whose sum is equal to v: v=v1+v2
  12. b2 a2 c2 b1 a1 c1 a Decomposition of a vector: Examples 3rdHow changes the velocity vector for a cannon-ball? We choose 3 positions and we study the velocity vector v We can recognize two particular direction: horizontal shifting and vertical shifting  Therefore we draw a Cartesiansystem of reference with the AXES horizontal and vertical We can decompose v on directions 1, 2 for find 2 vectors, v1v2, whose sum is equal to v: v = v1 + v2 We can repeat this for every point of trajectory. The vectors v1v2 are the velocity with there the cannon-ball moves in horizontal (v1) and vertical (v2) So we find the force that pulls down the body. It is the vector v1 
  13. - + Return Difference between Vectors • Proof We can use only the addition of vectors and multiplication with a number We can use the parallelogram law The vector d, difference a-b, is a vector that start from the arrow ofb and arrives at the arrow ofa We can translate d

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