UNIT 1-PHY 131 Chapter 2- Introduction to vectors and Forces-Students Notes PDF

UNIT 1 Chapter 2 Introductions to vectors and Forces Slide 1 SCALARS AND VECTORS A scalar quantity can have magnitude, algebraic sign (positive or negative), and units, but not a direction in space. When scalars are added or subtracted, they do so in th...

UNIT 1 Chapter 2 Introductions to vectors and Forces Slide 1 SCALARS AND VECTORS A scalar quantity can have magnitude, algebraic sign (positive or negative), and units, but not a direction in space. When scalars are added or subtracted, they do so in the usual way of addition algebraic numbers: 3 kg of water plus 2 kg of water is always equal to 5 kg of water. Adding vectors is different. All vectors follow the same rules of addition and takes into account the directions of the vectors being added. SCALARS AND VECTORS: NOTATION An arrow over a boldface symbol indicates a vector quantity: When the symbol for a vector is written without the arrow and in italics rather than boldface ( F ), it represents the magnitude of the vector (which is a scalar). Slide 2 SCALARS AND VECTORS: NOTATION Absolute value bars are also used to represent the magnitude of a vector: The magnitude of a vector may have units and is never negative; it can be positive or zero. VECTOR REPRESENTATION Arrows are used to represent vectors. The direction of the arrow gives the direction of the vector. The length of a vector arrow, when drawn according to scale, is proportional to the magnitude of the vector. 4N 8N The negative of a vector is simply in the opposite direction to the original vector: +𝐹 −𝐹 Slide 3 Slide 4 Often it is necessary to add one vector to another. That is, when the vectors are along the same straight line and pointing in the same directions. 5m 3m 8m Slide 5  GRAPHICAL VECTOR ADDITION Additional graphical vector addition illustrations for vectors along the same line. What if all the vectors start at the same point? How would one draw the resultant of such number of vector? Slide 6  VECTOR ADDITION USING COMPONENTS Components of a Vector Any vector can be expressed as the sum of vectors parallel to the x-, y-, and (if needed) z-axes. The x-, y-, and z- components of a vector indicate the magnitude and direction of the three vectors along the axes. A component has magnitude, units, and an algebraic sign ( + or − ). The sign of a component indicates the direction along that axis. The process of finding the components of a vector is called resolving the vector into its components. Slide 7  VECTOR ADDITION USING COMPONENTS Finding Components of a vector Consider a force vector that has magnitude 9.4 N and is directed 58 ° below the +x-axis ). Think of as the sum of two vectors, i.e., its components, one parallel to the x-axis and the other parallel to the y-axis. The magnitudes of these two vectors. i.e., the components, are the absolute values of the x- and y-components of. Slide 8  REVIEW OF THE TRIGONOMETRIC FUCTIONS side opposite ∠𝜙 𝑎 sin 𝜙 = = PYTHAGOREAN THEOREM: hypotenuse 𝑐 side adjacent ∠𝜙 𝑏 𝑐 2 = 𝑎2 + 𝑏2 cos 𝜙 = = hypotenuse 𝑐 side opposite ∠𝜙 𝑎 tan 𝜙 = = side adjacent ∠𝜙 𝑏 Slide 9 Problem-Solving Strategy: Finding the x- and y- components of a vector from its magnitude and direction 1. Draw a right triangle with the vector as the hypotenuse and the other two sides parallel to the x- and y-axes. 2. Determine one of the angles in the triangle. 3. Use trigonometric functions to find the magnitudes of the components. Make sure your calculator is in “degree mode” to evaluate trigonometric functions of angles in degrees. 4. Determine the correct algebraic sign for each component by looking at the direction the vector arrow representing that component Slide 10 Problem-Solving Strategy: Finding the x- and y- components of a vector from its magnitude and direction 1. Another way to determine the correct algebraic sign for each component by looking at the diagram below: − x-component + x-component + y-component + y-component − x-component + x-component − y-component − y-component Slide 11  VECTOR ADDITION USING COMPONENTS Finding Components of the force vector 𝐅Ԧ We are given that F = 9.4 N Slide 12 Problem-Solving Strategy: Finding the magnitude and direction of a vector from Its x- and y- components 1. Sketch the vector on a set of x- and y-axes in the correct quadrant, according to the signs of the components. 2. Draw a right triangle with the vector as the hypotenuse and the other two sides parallel to the x- and y-axes. 3. In the right triangle, choose which of the unknown angles you want to determine. 4. Use the inverse tangent function to find the angle. The lengths of the sides of the triangle represent Fx and Fy. If θ is opposite the side parallel to the x-axis, then tan θ = opposite/adjacent = Fx / Fy. If θ is opposite the side parallel to the y-axis, then tan θ = opposite/adjacent = Fy / Fx. Slide 13 Problem-Solving Strategy: Finding the magnitude and direction of a vector from its x- and y- components 5. Interpret the angle: specify whether it is the angle below the horizontal, or the angle west of south, or the angle clockwise from the negative y-axis, etc. 6. Use the Pythagorean theorem to find the magnitude of the vector. Slide 14  VECTOR ADDITION USING COMPONENTS Finding Magnitude and Direction We can find a vector’s magnitude and direction from its components given. Fx = 5.0 N, Fy = –8.0 N The magnitude of the force F is 9.4 N directed 58 south of east (or below the +x axis) Slide 15 Example 2.3 Suppose you are standing on the floor doing your daily exercises. For one exercise, you lift your arms up and out until they are horizontal. In this position, assume that the deltoid muscle exerts a force of 270 N at an angle of 15 ° above the horizontal on the humerus. What are the x- and y-components of this force? Slide 16 Example 2.3 Solution Slide 17 Problem-Solving Strategy: Adding Vectors Using Components 1. Find the x- and y-components of each vector to be added. 2. Add the x-components ( with their algebraic signs ) of the vectors to find the x-component of the sum. (If the signs are not correct, the sum will not be correct.) 3. Add the y-components (with their algebraic signs) of the vectors to find the y-component of the sum. 4. If necessary, use the x- and y-components of the sum to find the magnitude and direction of the sum. Slide 18  VECTOR ADDITION USING COMPONENTS Adding Vectors Using Components    C= A+B C x = Ax + Bx C y = Ay + By 𝐶𝑦 Magnitude of vector 𝐶Ԧ is given as 𝐶Ԧ = 𝐶𝑥2 + 𝐶𝑦2 and  given by 𝜃 = 𝑡𝑎𝑛− 𝐶𝑥 Slide 19 Example 2.4 In a traction apparatus, three cords pull on the central pulley, each with magnitude 22.0 N, in the directions shown in the figure. What is the sum of the forces exerted on the central pulley by the three cords? Give the magnitude and direction of the sum. Slide 20  Free-body diagram Reasoning Strategy Select an object(s) [or a point] to which the equations of equilibrium are to be applied. Draw a free-body diagram for each object chosen above. Include only forces acting on the object, not forces the object exerts on its environment. Choose a set of x, y axes for each object and resolve all forces in the free-body diagram into components that point along these axes. Apply the equations and solve for the unknown quantities. Slide 21 Example 2.4 Solution Slide 22 Example 2.4 Solution Slide 23 Example 2.4 Solution The magnitude of the force F is 54.o N directed 21.8 north of east (or above the +x axis) Slide 24  SELF ASSESSMENT (a) Revisit example 2.2 from your text book, and calculate the magnitude and direction of resultant force using vector components. (b) Two ropes are attached to a heavy box to pull it along the floor. One rope applies a force of 475 newton's in a direction due west; the other applies a force of 315 newton's in a direction due south. As we will see later in the text, force is a vector quantity. (a) How much force should be applied by a single rope, and (b) in what direction (relative to due west), if it is to accomplish the same effect as the two forces added together? Slide 25 (c) Answer: 30.2 m at 10.2 south of east Slide 26 (d) Answer: 6.88 km at 26.9 south of east Slide 27 INTERACTIONS AND FORCES Long-range forces sometimes referred to as fundamental forces are contact free forces and there are fours of them identified in Physics: Gravitational force (gravity) Electromagnetic force Strong nuclear force Weak nuclear force All non-fundamental (convenience) forces are contact forces: Normal Forces Forces of Friction (Kinetic and static frictional forces) Tension Note that all forces are vector quantities, i.e. they have both magnitude and direction. In the SI system, the unit of forces is Newton (N) Slide 28 FUNDAMENTAL FORCES N P P P N N P Video on fundamental forces: https://www.youtube.com/watch?v=S1eK_frCTCE  NET FORCE When more than one force acts on an object, the subsequent motion of the object is determined by the net force acting on the object. The net force is the vector sum of all the forces acting on an object. Definition of net force: If all the forces acting on a body are given by: Then the net force, ( ,acting on that object is the vector sum of those forces: Slide 30  Individual Forces Net Force 4N 10 N 6N 5N 37  3N 4N Newton’s First Law of Motion An object’s velocity vector remains constant if and only if the net force acting on the object is zero. Inertia: Newton’s first law is also called the law of inertia. In physics, inertia means resistance to changes in velocity. Slide 32  FREE-BODY DIAGRAMS A free-body diagram (FBD) is a simplified sketch of a single object with force vectors drawn to represent every force acting on that object. It must not include any forces that act on other objects. A free-body diagram is a drawing that consists of all forces that act on the system of interest. No other physical features of the system are included, such as its velocity or acceleration. A coordinate system and labels for angles are often added. To draw a free-body diagram: Draw the object in a simplified way. Identify all the forces that are exerted on the object. Check your list of forces to make sure that each force is exerted on the object of interest by some other object. Make sure you have not included any forces that are exerted on other objects. Draw vector arrows representing all the forces acting on the object. Slide 33 From the figure below calculate the net force on the car N W The net force in this case is: 275 N + 395 N – 560 N = +110 N and is directed along the + x axis of the coordinate system. Example Draw a free body diagram of a book at rest on a table. Example Draw the free body diagram of a book pushed against the wall by your hand. Example Draw the free body diagram for a heavy box at rest that is sitting on a rough inclined surface. For an object in equilibrium, Definition of Equilibrium An object is in equilibrium when it has zero acceleration. Slide 38  2.7 The forces on an airplane in flight heading eastward are as follows: gravity = 16.0 kN, downward lift = 16.0 kN, upward thrust = 1.8 kN, east and drag = 0.8 kN, west. Calculate the net force on the plane. Strategy After drawing the forces in the FBD for the plane, add the forces to find the net force. To resolve the force vectors into components, as a recommendation choose positive (+) x- and positive (+) y-axes pointing east and north, respectively. Generally, if there is motion, choose the direction of motion of an object as positive. Slide 39  2.7 Solution Slide 40  2.8 To slide a chest that weighs 750 N across the floor at constant velocity, you must push it horizontally with a force of 450 N. Find the contact force that the floor exerts on the chest. Strategy: The chest moves with constant velocity, so it is in equilibrium. The net force acting on it is zero. Identify all the forces acting on the chest, draw an FBD, do a graphical addition of the forces, choose x- and y-axes, resolve the forces into their x- and y-components, and then set Slide 41  2.8 Solution Slide 42 Newton’s Third Law of Motion In an interaction between two objects, each object exerts a force on the other. These two forces are equal in magnitude and opposite in direction. Equivalently, we can write Slide 43 Force of Gravity Slide 44 Example Two spheres, each having 20 kg mass, are placed at the centre-to-centre distance of 3 cm from each other. Calculate the force of gravity acting on each sphere. 2 IMPORTANT NOTES FROM SELF STUDY CHAPTER ONE In the language of Physics, the word factor is used frequently. For example, if the power emitted by a radio transmitter has doubled, we say the power has “increased by a factor of 2.” If the concentration of sodium ions in the bloodstream is half of what it was previously, we might say the concentration has “decreased by a factor of 2.” The factor is the number by which a quantity is multiplied or divided when is changes from one value to another. Generally, a factor is really a ration. In the case of the radio transmitter, if P0 represents the initial power and P represents the power after a new equipment is installed we write: 𝑃 =2 𝑃0  It is also common in Physics to talk about increasing by “5% or decreasing by 20%.” For example if the energy E0 increases by n%, then the value of the new energy Ef is calculated from: 𝑛 𝐸𝑓 = 𝐸0 1 + 100 If the velocity of a moving object v0 decreases by n%, then the value of the new velocity vf is calculated from: 𝑛 𝑣𝑓 = 𝑣0 1 − 100 From the two equations above n is the percentage increase or decrease of a quantity. Example When you are in a commercial airliner cruising at an altitude of 6.4 km. Calculate by what percentage has your weight changes compared with your weight on the ground. Example When you are in a commercial airliner cruising at an altitude of 6.4 km. Calculate by what percentage has your weight changes compared with your weight on the ground. SOLUTION Example When you are in a commercial airliner cruising at an altitude of 6.4 km, by what percentage has you weight changed compared with your weight on the ground? Solution  2.10 Solution Slide 50 Example Coccal Bacteria are spherical shape with mass of approximately 9.5 x 10-13 g and radius of approximately 0.5m. Estimate the force of gravity between two adjacent Coccal bacteria that make Diplococcal bacteria. Solution m1m2 F =G 2 r −16 −16 −11 (9.5 10 kg )(9.5 10 kg ) F = 6.67 10 2 Nm / kg2 (2  0.50 10 −6 m) 2 −29 F = 6.0 10 N Weight mM E  ME  w  Fg = G 2 = m G 2   mg RE  RE  ME g G 2 RE For an object on the Earth’s surface, g, free fall acceleration, is approximately constant everywhere on the planet.  1 g 2 r As you rise above the Earth’s surface the gravitational field weakens. The gravitational field is proportional to the inverse square of the distance from Earth’s center. Example Calculate the weight of the 20 000 g sphere at Earth’s surface, where g = 9.8 m/s2. w = mg = (20kg )(9.8m / s ) = 196kgm / s = 196 N 2 2 Example Coccal Bacteria are spherical shape with mass of approximately 9.5 x 10-13 g and radius of approximately 0.5µm. Calculate the ratio of the gravitational force between two coccal bacteria to the weight of a coccal bacterium. Solution m1m2 F =G 2 r −16 −16 −11 (9.5  10 kg )(9.5  10 kg ) F = 6.67 10 Nm / kg 2 2 (2  0.50 10 −6 m) 2 −29 F = 6.0 10 N 6.0 6.45 ⨯  Normal Force The normal force is one component of the force that a surface exerts on an object with which it is in contact – namely, the component that is perpendicular to the surface. Slide   𝑵 = 𝑾 𝑵 = 𝑾 𝒄𝒐𝒔∅ 𝑵 = 𝑾 + 𝑭 Slide FN − 11 N − 15 N = 0 FN = 26 N FN + 11 N − 15 N = 0 FN = 4 N Slide  Force of Friction The component of the surface force that is parallel to the contact surface is the force of friction. The force of friction acts between two objects while their surfaces are in contact, opposing the motion of one object slipping along the other. The force of friction is caused by irregularities in the surfaces that are in contact with each other. Slide  Static Friction Static friction is the force that prevent objects from moving with respect to a contacting surface, and allows them to maintain their stationary status. f sMax =  s N In which S is the coefficient of static friction, and N is the normal force. Slide The magnitude of the static frictional force can have any value from zero up to a maximum value. fs  f s MAX fsMAX =  s FN 0  s  1 Slide  Kinetic Friction f k = k N Where k is the coefficient of kinetic friction k  s Slide Note that the magnitude of the frictional force does not depend on the contact area of the surfaces. Slide 64 Concept Question 1 A box is pushed across a horizontal floor. What effect does the weight of the box have on the kinetic friction between the box and the floor. REASONING: More weight on the box requires a larger normal force from the floor on it. A larger normal force produce a larger frictional force on it. Slide 65 Concept Question 2 A car is parked on a ramp that makes an angle  with the ground. (A) What forces act on the car? (B) What force keeps it stationary on the ramp? REASONING (A) The forces are: weight, normal force, and friction. (B) Force of friction (static) keeps the car on the ramp. Slide 66  2.12 Example 2.8 involved sliding a 750-N chest to the right at constant velocity by pushing it with a horizontal force of 450 N. We found that the contact force on the chest due to the floor had components Cx = −450 N and Cy = +750 N, where the x-axis points to the left and the y-axis points up. Calculate the coefficient of kinetic friction for the chest-floor surface? Slide 67  2.12 Strategy To find the coefficient of friction, we need to know what the normal and frictional forces are. They are the components of the contact force that are perpendicular and parallel to the contact surface. Since the surface is horizontal (in the x-direction), the x- component of the contact force is friction and the y- component is the normal force. Slide 68  2.12 Solution Slide 69  2.14 A safe is to be moved up a ramp to a height of 1.5 m above the floor. The mass of the safe is 510 kg, the coefficient of static friction along the incline is s = 0.42, and the coefficient of kinetic friction along the incline is k = 0.33. The ramp forms an angle θ = 15° above the horizontal. (a) Calculate how hard do the movers have to push to start the safe moving up the incline. Assume that they push in a direction parallel to the incline. (b) To slide the safe up at a constant speed, with what magnitude force must the movers push? Slide 70  2.14 Solution Slide 71  2.14 Solution (a) Slide 72  2.14 Solution Slide 73  2.14 Solution (b) The movers push with a force of magnitude 2900 N directed up the incline. Slide 74  Tension Force 𝑇 𝑇 An object can pull on another through the use of a string, cord, rope, chain, wire, tendon, cable or other such object An ideal cord (rope, string, tendon, cable, or chain) pulls in the direction of the cord with forces of equal magnitude on the objects attached to its ends as long as no external force is exerted on it anywhere between the ends. An ideal cord has zero mass and zero weight. Slide 75 APPLICATION: IDEAL PULLEYS A pulley can change the direction of the force exerted by a cord under tension. An ideal pulley has no mass and no friction. An ideal pulley exerts no forces on the cord that are tangent to the cord—it is not pulling in either direction along the cord. As a result, the tension of an ideal cord that runs through an ideal pulley is the same on both sides of the pulley. Slide 76  2.15 Figure 2.37 shows the bowstring of a bow and arrow just before it is released. The archer is pulling back on the midpoint of the bowstring with a horizontal force of 162 N. What is the tension in the bowstring? Slide 77  2.15 Solution Slide 78  2.15 Solution Slide 79  2.16 A 1804-N engine is hauled upward at constant speed. What are the tensions in the three ropes labeled A, B, and C? Assume the ropes and the pulleys labeled L and R are ideal. Slide 80  2.16 Solution There are two forces acting on the engine: the gravitational force (1804 N, downward) and the upward pull of rope A. These must be equal and opposite, since the net force is zero. Therefore TA = 1804 N. Slide 81  2.16 Solution Slide 82  2.16 Solution Slide 83