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: direction:inclination /6 path:upwards magnitude:5 u direction:inclination /6 path:upwards magnitude:5 u The Polaris Coordinates are (magnitude, inclination) (, ) e.g. (5, /6)

·cos( ) and ·sin( ) are known as Cartesian Components Cartesian Coordinate of a Vector We can use the Cartesians Axes for represent a vector e.g. we plot the previous vector: direction:slope /6 path:upwards magnitude:5 u x O y ·cos( ) ·sin( ) ( ·cos( ), ·sin( ) ) The Cartesians Coordinates are (x-coordinate, y-coordinate) (x, y) e.g. ( ·cos( ), ·sin( ) ) We can write the vector as

Change of Coordinate: Cartesians Polaris x O y ·cos( ) ·sin( ) ( ·cos( ), ·sin( ) ) (x, y) Cartesians Coordinates (, ) Polaris Coordinates From Cartesian to Polar From Polar to Cartesian

Calculation with Vectors Product with a scalar (k ) k=3 a has the same direction 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 k = -1 if k<0 the path of a is the opposite of path of b

Calculation with Vectors Addition of vectors We calculate the magnitude of s with Carnot Theorem The Parallelogram Law + s 2 = a 2 + b 2 – 2ab cos( ) s is known as the Resultant of a and b N.B. We always have s a + b

Calculation with Vectors Addition of vectors, Cartesian Coordinates + x O y a·cos( ) b·sin( ) a·sin( ) b·cos( ) s·cos( ) s·sin( ) We describe the vector sum s by adding the components of a and b

Calculation with Vectors difference between vectors - - d 2 = a 2 + b 2 – 2ab cos( ) In both cases the magnitude of d is We describe vector difference d by subtracting the components of b of a Proof

Calculation with Vectors Comparison between addition and difference of vectors Difference Addition magnitude of d ismagnitude of s is N.B. Observe the position of angle d 2 = a 2 + b 2 – 2ab cos( )s 2 = a 2 + b 2 – 2ab cos( )

Decomposition of a vector To decompose a vector v find two vectors v 1, v 2 in two prefixed direction, whose sum is equal to vector v a1a1 O a2a2 c1c1 O c2c2 d2d2 O d1d1 b1b1 O b2b2 now from v we can draw two lines parallel to the axes in every case we can write:

The vector of body weight is directed downwards Decomposition of a vector: Examples 1 st A body is hung to the ceiling with two different ropes. It forms angles 1, and 2 with the ropes. What are the tensions of the ropes? O The vectors of tensions are direct toward the ropes So we can recognize two particular direction: the lines along the ropes So we draw a system of reference with the AXES parallel to the ropes and ORIGINATING from the body We can decompose v in directions 1, 2 to find 2 vectors, v 1 and v 2, whose sum is equal to v: v = v 1 + v 2 We can observe that T 1 = -v 1 and T 2 = -v 2 Therefore v is balanced by the two vectors T 1, T 2 tensions of ropes: -v = T 1 + T 2

The vector of body weight is directed downwards Decomposition of a vector: Examples 2 nd A body is sliding on an inclined plane. What force pulls down the body? So we can recognize two particular direction: the line along the plane and the line they perpendicular to it Therefore we draw a system of reference with the AXES parallel and perpendicular to the plane We can decompose v in directions 1, 2 to find 2 vectors, v 1 and v 2, whose sum is equal to v: v=v 1 +v 2 In this way we find the force that pulls the body downwards. This is vector v 1 Vector v directed downwards is the Force (or acceleration) of gravity Vector v 1 directed parallel to the plane is the active component of the Force (or acceleration) of gravity responsible of sliding of the body. It is indicated by v // v // = v · sin( ) Vector v 2 directed perpendicularly to the plane is the component of the Force (or acceleration) of gravity that holds the body to the plane. It is indicated by v perp v perp = v · cos( )

Decomposition of a vector: Examples 3 rd How changes the velocity vector for a cannon-ball? We can decompose v on directions 1, 2 for find 2 vectors, v 1 v 2, whose sum is equal to v: v = v 1 + v 2 So we find the force that pulls down the body. It is the vector v 1 Therefore we draw a Cartesian system of reference with the AXES horizontal and vertical We can recognize two particular direction: horizontal shifting and vertical shifting We choose 3 positions and we study the velocity vector v a2a2 a1a1 b2b2 b1b1 c2c2 c1c1 We can repeat this for every point of trajectory. The vectors v 1 v 2 are the velocity with there the cannon-ball moves in horizontal (v 1 ) and vertical (v 2 )

Difference between Vectors Proof - We can translate d 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 of b and arrives at the arrow of a Return

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