l Vectors quantities Fully described by magnitude and

l Vectors quantities – Fully described by magnitude and direction (include force, acceleration, velocity, and displacement) l Scalar quantities – Fully described by magnitude only

Representation of Vector Quantities l Can easily represent vector quantities by using an arrow – Length represents magnitude – Direction points from the tail towards the head

Scaled Vector Diagrams l Scale clearly listed l An arrow is drawn in a specified direction l The magnitude and direction of the vector is clearly labeled

Scaled Vector Diagrams

Representing Vector Direction l Two techniques – Direction indicated by the angle made with the x-axis (horizontal) l 400 N of E l 600 S of W

Representing Vector Direction l 2 nd Technique – Direction expressed as counterclockwise angle of rotation from due east (or horizontal)

Counterclockwise Convention

Vector Addition The rules for summing vectors were applied to freebody diagrams in order to determine the net force (i. e. , The vector sum of all the individual forces). l Rules for summing vectors were simple l

Vector Addition l The task of summing vectors will be extended to more complicated cases in which the vectors are directed in directions other than vertical and horizontal directions

Two Vector Addition Techniques l Algebraically – Pythagorean Theorem – SOH CAH TOA – Law of Cosines – Law of Sines l Graphically – Using scaled diagrams and protractors

Sample Problem – Vector Addition (Right Angles) l. A hiker leaves camp and hikes 11 km, north and then hikes 11 km east. Determine the resulting displacement of the hiker.

Sample Problem - Hiker l Begin by placing the two vectors head-to-tail l The sum of the vectors (indicated by the red line) is called the resultant vector (R)

Sample Problem - Hiker l. Resultant Vector is determined by drawing an arrow at the tail of the first vector and extending to the head of the second vector

Pythagorean Theorem

Determine the Direction l Direction can be determined using trigonometric functions l SOH CAH TOA l Reference Tables

Determine the Direction l. Select one of the two angles other than the 900 angle l. Any of the three trig functions can be used to find the angle

Vector Addition Guideline l Maximum magnitude of 8 N and 6 N forces acting concurrently is 14 N (occurs when angle is 00) l Minimum magnitude is 2 N (occurs when angle is 1800)

Practice l. A hiker walks 10 km north, then walks 5 km west. Determine the displacement and vector direction both graphically and algebraically.

Effect of Two Vectors Angle Other Than 900 l. An object is subject to two concurrent forces F 1 = 50 newtons [east] F 2 = 30 newtons [300 N of east] l. What is the resultant force on the object?

Effect of Two Vectors Angle Other Than 900 l. Graphical solution – Place vectors head-to-tail – Draw resultant – Using a scale, ruler, and protractor, we find the magnitude of R is 77 N, and the angle is 110

Effect of Two Vectors Angle Other Than 900 l. Algebraically – Law of cosines c = a 2 + b 2 – 2(a)*(b)*(cos C) l 2 – Law of sines l a / sin A = b / sin B = c / sin C

Practice Problem l. An object is subject to two concurrent forces: F 1 = 60 N [east] F 2 = 30 N [400 north of east] l. What is the resultant force on the object?

Vector Subtraction l Subtraction is really a special case of addition la – b is equivalent to a + (-b) l Since vectors have magnitude and direction, the negative of a vector is equal in magnitude, but opposite in direction.

Vector Subtraction l We subtract two vectors by adding one of them to the negative of the other V -V

Vector Subtraction l. The following compares vector addition and subtraction A-B A+B A -B

Vector Addition Re-Visited l Is order important? l the order in which two or more vectors are added does not effect the outcome.

Vector Resolution l. Breaking a vector down into a number of components that will add to produce the original vector.

Vector Resolution l. A vector can be resolved into any number of components V A V B F C D E

Vector Resolution l Usually most useful to resolve a vector into two components that lie along the x- and yaxis l The components are perpendicular to each other

Vector Resolution Illustration l Fhoriz is the xcomponent of the Force vector l Fvert is the ycomponent of the Force vector l Theta ( ) is the angle the vector makes with the x-axis ( = 400)

Vector Resolution Illustration l Since the vector and its components form a right triangle, it follows: – Sin = Fvert / F lor – Fvert = F * Sin

Vector Resolution Illustration l. AND l. Cos = Fhoriz / F l. Fhoriz = F * Cos

Reference Tables l. General form of these two equations are in your reference tables (Mechanics) – Ay = A Sin – Ax = A Cos l. Where A = any vector quantity

Component Signs l The x- and y- components of a vector have algebraic signs that depend on the quadrant in which the original vector lies Quadrant II Quadrant I x = negative x= positive y = positive Quadrant III y = positive Quadrant IV x = negative x = positive y = negative

l. A Practice Problem 400 N force is exerted at 600 [S of E] to move a railroad car along a railroad track (top down view in the diagram). Determine the horizontal component of the force that moves the car along the track.

Solution l Fx = F * Cos Fy = F * Sin q

Using Resolution to Add Vectors l Graphical solutions are easy, but not very precise. l Laws of Cosines and Sines are tedious and difficult l Need a technique that is simple yet precise, and works for more than two vectors

Sample Problem l Find the resultant of the following force vectors: – F 1 = 50 N [350 N of E] – F 2 = 30 N [450 S of E] – F 3 = 20 N [W]

Step by Step Technique l Sketch the problem l Resolve each of the vectors to be added into its x- and y- components. Remember to include the proper sign, depending on the quadrant of the vector l Be aware that if the vector lies on the x-axis, its y- component is zero; if the vector lies on the y- axis, its x- component is zero

Step by Step Technique (con’t) l Add the x- components together to produce the x- component of the resultant (Rx). l Add the y- components together to produce the y- component of the resultant (Ry). l Calculate the magnitude of the resultant vector by means of the Pythagorean Theorem: R = Sqr Rt (Rx 2 + Ry 2).

Step by Step Technique (con’t) l Find the angle that the resultant vector makes with the x- axis from this relationship: – Tan = Ry / Rx

Practice Problem l Find the resultant of the following force vectors: § A = 40 N [400 N of W] § B = 15 N [W] § C = 20 N [700 S of E]

Re-Visiting Some Terms l Equilibrium. – When all the forces which act upon an object are balanced (i. e Fnet=0), then the object is said to be in a state of equilibrium. l If an object is at rest and is in a state of equilibrium, then we would say that it is at "static equilibrium. "

Tension Defined l Consider a 500 N block hanging from the ceiling by a cable. What other force besides that due to gravity acts on the block? l The term Tension is used to describe the force that causes the cable to be taut. The direction of the tension is upward in this case, because it supports the 500 N block.

Applying What We’ve Learned - Statics l Consider the following picture hanging on a wall in the White House. l What is the Fnet acting on the picture?

Applying What We’ve Learned - Statics l Since it is at rest (static equilibrium), the net force must be zero. l Therefore, the resultant of all vectors must be zero. l Therefore, the x- and ycomponents of the resultant must be zero.

Applying What We’ve Learned - Statics l Suppose the tension in both cables is measured to be 50 N and that the angle which each cable makes with the horizontal is known to be 30 degrees. What is the weight of the sign?

l Fnet = 0 (static equilibrium) l Fg - Fy, tens = 0 l Fg = Fy, tens + Fy, tens l Fg = 25 N + 25 N l Fg = 50 N = weight

Practice Problem - Statics l The sign at left hangs outside the physics classroom, advertising the most important truth to be found inside. If the sign has a mass of 50 kg, then determine the tension in the single cable which supports its weight.

Physics Video Sliding Rock

Next Application – Inclined Planes Objects are known to accelerate down inclined planes because of an unbalanced force l As shown in the diagram, there always at least two forces acting upon any object that is positioned on an inclined plane the force of gravity and the normal force. l

l The first peculiarity of inclined plane problems is that the normal force is not directed in the direction which we are accustomed to.

Determining Net Force On An Inclined Plane l The process of analyzing the forces acting upon objects on inclined planes will involve resolving the weight vector (Fgrav) into two perpendicular components.

Inclined Plane – No Friction l The perpendicular component of the force of gravity balances the normal force. l The parallel component of the force of gravity is not balanced by any other force (assuming no friction) – it causes the acceleration.

Calculating Forces – No Friction = Fgrav Sin = (m * g) Sin l Fperpendicular = Fgrav Cos = (M * g) Cos l Fparallel

Calculating Acceleration – No Friction l Fparallel l Fnet = (m * g) Sin = m * a = Fparallel = (m * g) Sin l Therefore a = g Sin In the absence of friction and other forces

Inclined Planes With Friction l In the presence of friction or other forces (applied force, tensional forces, etc. ), the situation is slightly more complicated. Consider the diagram shown at the right.

“Tilting The Head” l If you tilt your head, the problem should look familiar

Practice Problem – Inclined Plane l The free-body diagram shows the forces acting upon a 100 -kg crate which is sliding down an inclined plane. The plane is inclined at an angle of 30 degrees. The coefficient of friction between the crate and the incline is 0. 3. Determine the net force and acceleration of the crate.

Next Application – Projectile Motion l. What is a projectile? –By definition, a projectile is an object upon which the only force acting is gravity.

l If we neglect air resistance, the following are projectiles – An object dropped from rest – An object thrown vertically upwards – An object thrown upwards at an angle

Projectile Free Body Diagrams l Draw a free body diagram for each – An object dropped from rest – An object thrown vertically upwards – An object thrown upwards at an angle

l Suppose the “gravity switch” could be turned off…. . l What would happen to a projectile fired horizontally from a cannon?

Turn Gravity Back ON l Will the cannonball travel a greater (or shorter) distance under the influence of gravity? l Will gravity affect the cannonball’s horizontal motion?

Projectile Motion l Key Concept – Perpendicular components of motion are independent of each other

Challenge Question l. A cannonball is fired horizontally off a cliff. At the exact same moment, an identical cannonball is dropped from the cliff. Which will hit the ground first and why? l Both will hit at the same time. Perpendicular components are independent of each other. In this example, the vertical components are the same. Therefore, they will strike the ground at the same time.

Perpendicular Components of Motion

Projectile Motion Perpendicular Components of Motion l The vertical force acts perpendicular to the horizontal motion and will not effect it. l Thus, the projectile travels with a constant horizontal velocity and a downward vertical acceleration.

Projectile Motion Perpendicular Components of Motion l In the y- direction, both objects can be considered as falling freely, and are governed by equations previously developed: dy = viyt + ½ ay t 2 vfy 2 = viy 2 + 2 aydy ay = (vfy – viy) / t

Projectile Motion Perpendicular Components of Motion l In the xdirection, the object travels at constant velocity and its motion is described by vx = dx / t

Sample Problem – Projectile Motion l An object is thrown outward from a cliff with a horizontal velocity of 20 m/s. With air resistance ignored, the object takes 15 seconds to reach the bottom of the cliff. l Calculate a) the height of the cliff, and b) the horizontal distance that was traveled by the object by the time it reaches the ground

Practice Problem – Projectile Motion l An object is thrown outward from a cliff with a horizontal velocity of 15 m/s. The cliff is 100 meters high. How far will the object travel horizontally by the time it reaches the ground [ignore air resistance]?

Projectile Motion – Objects Launched At An Angle To The Ground l If a projectile is launched at an angle to the horizontal, then the initial velocity of the projectile has both a horizontal and a vertical component.

Initial Velocity Components l vix = vi cos l viy = vi sin

Practice Problem l An object is fired from the ground at 100 m/s at an angle of 300 with the horizontal. A. Calculate the horizontal and vertical components of the initial velocity. B. After 2. 0 seconds, how far has the object traveled in the horizontal direction? C. How high is the object at this point (i. e. after 2. 0 seconds)?

Time of Flight l ay = g = (vfy – viy) / t l At the peak, vfy = 0 l So, time to peak is l g = -9. 8 m/s 2 = -viy / t l or l tup = viy / g l ttotal = 2 * tup

Determination of Horizontal Displacement l The horizontal displacement of a projectile is dependent upon the horizontal component of the initial velocity. l dx l If = vx t t = ttotal , then dx is the maximum horizontal displacement in that instance.

Determination of the Peak Height l Peak height is determined by dy = viyt + ½ ay t 2 l Where t = time to reach peak (tup) viy is the initial vertical velocity ay = acceleration due to gravity t = time

Practice Problem (con’t) l An object is fired from the ground at 100 m/s at an angle of 300 with the horizontal. – Determine the time to the peak, and the total time of flight. – Determine the total horizontal displacement. – Determine the peak height of the object.

The Monkey and the Zookeeper

Throw at Monkey - Gravity Free Environment

Throw Above Monkey Gravity On

Throw at Monkey Fast Speed Gravity On

Throw at Monkey Slow Speed Gravity On

Projectile Motion Lab l Write up must include – Procedure (what did you do, what did you measure) – List of Materials (what did you need) – Analysis (what equations did you use, show your calculations)