Vectors and Scalars PDF

Vectors and Scalars  A scalar quantity is a quantity that has magnitude only and has no direction in space Examples of Scalar Quantities:  Length  Area  Volume  Time  Mass  A vector quantity is a quantity that has both magnitude and a direction in space Examples of Vector Quan...

Vectors and Scalars  A scalar quantity is a quantity that has magnitude only and has no direction in space Examples of Scalar Quantities:  Length  Area  Volume  Time  Mass  A vector quantity is a quantity that has both magnitude and a direction in space Examples of Vector Quantities:  Displacement  Velocity  Acceleration  Force  Vector diagrams are shown using an arrow  The length of the arrow represents its magnitude  The direction of the arrow shows its direction   The resultant is the sum or the combined effect of two vector quantities Vectors in the same direction: 6N 4N = 10 N 6m = 10 m 4m Vectors in opposite directions: 6 m s-1 10 m s-1 = 4 m s-1 6N 10 N = 4N  When two vectors are joined tail to tail  Complete the parallelogram  The resultant is found by drawing the diagonal  When two vectors are joined head to tail  Draw the resultant vector by completing the triangle 2004 HL Section B Q5 (a) Two forces are applied to a body, as shown. What is the magnitude and direction of the resultant force acting on the body? Solution:  Complete the parallelogram (rectangle)  The diagonal of the parallelogram ac represents the resultant force  The magnitude of the resultant is found using Pythagoras’ Theorem on the triangle abc a 12 N d Magnitude = ac = 12 + 52 2 θ 5N 5 ac = 13 N b c 12 12 Direction of ac : tan  = 5 12   = tan −1 = 67  Resultant displacement is 13 N 67º 5 with the 5 N force Find the magnitude (correct to two decimal places) and direction of the resultant of the three forces shown below. Solution:  Find the resultant of the two 5 N forces first (do right angles first) ac = 52 + 52 = 50 = 7.07 N 5 d 5 c tan  = = 1   = 45 5 Now find the resultant of the 10 N and 5 5N  7.07 N forces  The 2 forces are in a straight line (45º + 90º θ 45º a b 135º = 180º) and in opposite directions 135º 5N  So, Resultant = 10 N – 7.07 N = 2.93 N in the direction of the 10 N force  What is a scalar quantity?  Give 2 examples  What is a vector quantity?  Give 2 examples  How are vectors represented?  What is the resultant of 2 vector quantities?  What is the triangle law?  What is the parallelogram law?  When resolving a vector into components we are doing the opposite to finding the resultant  We usually resolve a vector into components that are perpendicular to each other  Here a vector v is resolved into an x component and a y y component x  a single force can also be resolved or broken up into two components this process is called resolution of forces. Usually, a force is resolved into components which are perpendicular to each other. ❑ The magnitudes of the vertical and horizontal components can be determined with knowledge of simple trigonometry.  Here we see a table y=25 N being pulled by a force of 50 N at a 30º angle 30º x=43.3 N to the horizontal  When resolved we see that this is the same as  We can see that it pulling the table up would be more efficient to pull the with a force of 25 N table with a and pulling it horizontal force of horizontally with a 50 N force of 43.3 N  If a vector of magnitude v and makes an angle θ with the horizontal then the magnitude of the components are:  x = v Cos θ y=v Sin θ y  y = v Sin θ θ x=vx Cos θ  Proof: x y Cos = Sin  = v v x = vCos y = vSin  2002 HL Sample Paper Section B Q5 (a) A force of 15 N acts on a box as shown. What is the horizontal component of the force? Solution: 12.99 N Horizontal Component = x = 15Cos60 = 7.5 N Component Vertical Vertical Component = y = 15Sin 60 = 12.99 N 60º Horizontal 7.5 N Component  2003 HL Section B Q6  A person in a wheelchair is moving up a ramp at constant speed. Their total weight is 900 N. The ramp makes an angle of 10º with the horizontal. Calculate the force required to keep the wheelchair moving at constant speed up the ramp. (You may ignore the effects of friction). Solution: If the wheelchair is moving at constant speed (no acceleration), then the force that moves it up the ramp must be the same as the component of it’s weight parallel to the ramp. Complete the parallelogram. Component of weight parallel to ramp: 10º = 900Sin10 = 156.28 N 80º 10º Component of weight perpendicular to ramp: 886.33 N 900 N = 900Cos10 = 886.33 N  If a vector of magnitude v has two perpendicular components x and y, and v makes and angle θ with the x component then the magnitude of the components are:  x= v Cos θ  y= v Sin θ y=v Sin θ y θ x=v Cosθ