Module 1: Vectors Engineering Physics PDF

MODULE 1: VECTORS ENGAGE Where can you apply physical quantities in your daily activities? What is the difference between scalar and vector quantities? Which method is mor accurate in solving vector sums (graphical or analytical method?) What is the difference be...

MODULE 1: VECTORS ENGAGE Where can you apply physical quantities in your daily activities? What is the difference between scalar and vector quantities? Which method is mor accurate in solving vector sums (graphical or analytical method?) What is the difference between DOT and CROSS products? EXPLORE Read Module 1: Introduction and Vectors (pp 24 – 32) EXPLAIN PHYSICS Physics is the most fundamental science - basis or the foundation of other physical sciences like chemistry, geology, and astronomy - most principles or laws of the other sciences are based on the principles of physics Physics is the study of the basic laws of nature - basic concepts and laws of physics govern most of the things that happen around us Physics is an experimental science - physics is a science of measurement Physics and its importance in the field of engineering: Basically, physics is involved with the study of energy and its different forms. It therefore serves as a foundation to engineering which is primarily involved with the design, construction and operation of devices, machines, structures, and systems which in various ways utilize these different forms of energy. 24  Physics is a quantitative science. It involves a lot of measurements and computational analyses. It is therefore imperative that the student of physics should have extensive proficiency with mathematical concepts, principles, and operations. Student should have considerable knowledge on algebra, trigonometry, analytic geometry, and calculus. Basic Concepts of Physics: 1. matter – anything that occupies space and has weight - it possesses inertia - it is subject to gravity 2. mass – the amount or quantity of matter in a body 3. motion – the displacement of a body with reference to another body 4. force – that which is capable of changing the condition of rest or motion of a certain body BRANCHES OF PHYSICS 1. Classical physics - Branch of physics that deals with objects moving less than the speed of light - All branches established before 1900’s Sub-branches: (some) o Mechanics – oldest branch of physics; deals with the behavior of objects subjected to forces and/or motion ▪ Statics – study of matter at rest, and forces in equilibrium ▪ Dynamics – deals with forces and their relation to motion o Acoustics – deals with the behavior and properties of sound waves o Thermodynamics – deals with the relation of heat and other forms of energy 2. Modern physics - Branch of physics that deals with objects at Extreme events Sub-branches: (some) o Nuclear physics – physics of atomic nuclei and their interactions o Theory of relativity o Quantum physics – describes nature at smallest scale of energy of atoms and subatomic particles Physical Quantities: The study of Physics involves dealing with a lot of physical quantities.These physical quantities are used to define all physical characteristics of matter such as length, mass and time. In Mechanics, we have the basic quantities and all others are considered as derived quantities because they are obtained or defined by simple relations between the basic ones. 25 Table 1.1 Basic Quantities Basic Length, Mass, Time, Temperature, Luminous Current, Amount of Quantities L m t T Intensity I Substance Metric (SI) m g sec C Cd A Mol English ft, in lbs sec F Mol Table 1.2 Derived Quantities Derived Quantities Combination of two or more basic quantities area Length length acceleration Length time Force Mass acceleration pressure Force area In the proper expression of physical quantities, it should have magnitude. There should at least be a number (to indicate how large or how small the quantity is) and the unit (to indicate the nature and type of the quantity). SCALAR AND VECTOR QUANTITY: Many physical quantities have magnitudes only but no directions. These are called scalar quantity. Examples are mass, time, density, temperature, etc. There are, however, many physical quantities such as force, velocity, displacement, etc. which have directions as well as magnitude and these aspects always have to be indicated when expressing these quantities. They are called vectors. In physical computations and analyses, we have to be aware of the difference between vectors and scalars because the mathematical treatments are not the same. For example, we add scalars arithmetically, but we cannot do the same to vectors. Special methods are used. BASIC CONCEPTS ABOUT VECTORS A. Vector notation Vectors are typically represented by a CAPITAL BOLD LETTER or drawing an ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ARROW above the symbol. The arrow is used to convey direction and magnitude. ForF ⃗ = a vector of magnitude|F ⃗ |or F and in a certain direction ⃗ = 30N due south F Direction Magnitude 26 B. Graphical representation of a vector Vector quantity is represented graphically by an arrow Tail or foot tip or head (direction) Magnitude the length represents magnitude the arrowhead faces the direction of motion C. Specifying directions of vectors – There are two common methods being used: Method 1: Using the angle Ɵ that the vector makes with the “zero-degree reference line”. Ex: A = 10 m 40o A B = 10 m 140o B o 140 40o Method 2: Using geographic directions. N Ex. C = 30 km 500 N of E C D = 10 km 60o W of S F F = 20 km N 40o W 40o 50o W E 60o D S OTHER CONCEPTS ON VECTORS: o COLLINEAR – parallel vectors which lie on the same straight line irrespective of their magnitudes and direction o COPLANAR – parallel to the same plane, eg.two vectors parallel to the x-y plane or any plane o CONCURRENT – vectors which passes through the same point Collinear Coplanar Concurrent 27 VECTOR ADDITION AND SUBTRACTION Vector addition is the process of combining two or more vectors into one. The combination is called the RESULTANT of the vectors. Subtraction is just like addition. In vector subtraction, the negative of one vector is added to the other. For example, if two vectors A and B are to be added, the operation is indicated as A + B. However, if vector B is to be subtracted from vector A, the operation is indicated as A – B which is the same as A + (-B). The negative of vector B is added to vector A. The negative of a vector is a vector of the same magnitude but in the opposite direction. For example, if vector A = 50 units 30oN of W, its negative or –A = 50 units 30oS of E. METHODS OF DETERMINING RESULTANT: 1. Algebraic Method (for co-linear vectors only). o Resultant vector, R= algebraic sum of the vectors o 𝐑 = ⃗⃗⃗⃗ 𝐕𝟏 + ⃗⃗⃗⃗ ⃗⃗⃗⃗𝐧 = 𝚺 𝐯𝐞𝐜𝐭𝐨𝐫𝐬 𝐕𝟐 +... +𝐕 2. Parallelogram Method o Applied if only 2 vectors are given o 2 vectors are joined tail to tail, forming a parallelogram (such for the name of the method) o a diagonal represents either the resultant or the equilibrant (depending on its direction) ▪ STEPS: 1. Draw vectors on same point of origin (use a scale) 2. Form a parallelogram, identify parallels with (‘) A’ 3. Connect the origin to the intersection of the B parallel lines R 4. The line formed from the origin to the intersection represents the RESULTANT of the vectors B’ 5. The line formed from the intersection to the origin A represents the EQUILIBRANT 6. Measure the length (magnitude), and the angle for the direction ** EQUILIBRANT – vector which when added to a set of vectors will result to zero, thus resulting to an equilibrium or balance, it is exactly equal to the magnitude of the resultant but directly opposite in direction 28 3. The Triangle Method (for two coplanar vectors at a time) ▪ STEPS: 1. Draw the vectors by joining them head to tail 2. Draw the resultant vector by completing the triangle **(direction is from the origin) 3. Determine the value of the included angle of the given vectors 4. Solve for the value of the resultant (magnitude and direction) 5.aIf the triangle formed is a right triangle, solve R by using Pythagorean Theorem and the trigonometric identities. Pythagorean Theorem: c 2 = a2 + b 2 c b b Direction: θ = Tan−1 (a) a 5.b If the triangle formed is not a right triangle, solve R using sine and cosine law. Cosine Law: c 2 = a2 + b2 − 2abcos(C) Sine Law: a b c = = sin (A) sin(B) sin (C) 4. The Polygon Method (graphical method) – Tip to tail method - suggested to be used for two or more vectors which are non-collinear but coplanar The goal is to draw a mini version of the vectors to give an accurate picture of the magnitude and direction. Steps: 1. Pick appropriate scale. 2. Using ruler and protractor, draw the first vector to scale in appropriate direction. 3. Draw the second vector starting from the head of the first vector. 4. All vectors must be connected in head-to-tail fashion. 5. To determine the resultant vector, connect the tail of the first vector to the head of the last vector drawn. 6. Measure the magnitude of R with a ruler and convert this length to its actual amount and unit. 7. Measure the direction of R with a protractor and add this value along with the direction after the magnitude. 29 5. The Component Method (used for any number of vectors which are non- co-linear) Steps: 1. Resolve the vectors into their x- and y-components. y – component When the angle  is measured from the horizontal x component = A cos θ A y component = A sin θ  When the angle  is measured from the vertical  x component = A sin  x – component y component = A cos  2. Add the x- and y-components of each vector to determine the components Rx and Ry of the resultant vector, R. 3. To get the magnitude R of the resultant, use the Pythagorean theorem: R = √R x 2 + R y 2 4. To get the direction of the resultant: Angles: Direction Ry N of E or Nof W or measured from the horizontal θ = tan−1 θ Y of X Rx S of E or S of W Rx W of N or E of N or measured from the vertical ϕ = tan−1 ϕ X of Y Ry W of S or E fo S UNIT VECTORS – is a vector having a magnitude of unity with no units. Its purpose is to describe a direction in space. y +j Let i = unit vector pointing in the + x-axis j = unit vector pointing in the + y-axis x k = unit vector pointing in the + z-axis +k +i z i.e. Vector Ax = Axi Bx = Bxi In terms of its components A = Axi + Ayj B = Bxi + Byj 30 VECTOR SUM: Vector sum of unit vectors 𝐂 = 𝐀 + 𝐁 𝐂 = (𝐀 𝐱 𝐢 + 𝐀 𝐲 𝐣) + (𝐁𝐱 𝐢 + 𝐁𝐲 𝐣) 𝐂 = (𝐀 𝐱 + 𝐁𝐱 )𝐢 + (𝐀 𝐲 + 𝐁𝐲 )𝐣 𝐂 = 𝐂𝐱 𝐢 + 𝐂𝐲 𝐣 If vectors do not lie in the x-y plane, then a third component is needed. Then: 𝐀 = 𝐀 𝐱 𝐢 + 𝐀 𝐲 𝐣 + 𝐀𝐳 𝐤 𝐁 = 𝐁𝐱 𝐢 + 𝐁𝐲 𝐣 + 𝐁𝐳 𝐤 𝐂=𝐀+𝐁 𝐂 = (𝐀 𝐱 𝐢 + 𝐀 𝐲 𝐣 + 𝐀𝐳 𝐤) + (𝐁𝐱 𝐢 + 𝐁𝐲 𝐣 + 𝐁𝐳 𝐤) 𝐂 = (𝐀 𝐱 + 𝐁𝐱 )𝐢 + (𝐀 𝐲 + 𝐁𝐲 )𝐣 + (𝐀𝐳 + 𝐁𝐳 )𝐤 𝐂 = 𝐂𝐱 𝐢 + 𝐂𝐲 𝐣 + 𝐂𝐳 𝐤 PRODUCTS OF VECTORS: Since vectors are not ordinary numbers, ordinary multiplication is not directly applicable to vectors. a) SCALAR PRODUCT – is also called dot product. It is a scalar quantity and it may be positive or negative. If 𝟎° ≤ 𝛟 < 90°, then A ∙ B is 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝟗𝟎° < 𝛟 ≤ 𝟏𝟖𝟎°, then A ∙ B is 𝐧𝐞𝐠𝐚𝐭𝐢𝐯𝐞 B 𝐀. 𝐁 = 𝐀𝐁 𝐜𝐨𝐬 𝛟 𝛟 = 𝟗𝟎° then A ∙ B = 𝟎 (Scalar product of two perpendicular ϕ vectors is always zero) A Using unit vector representation: 𝐀 ∙ 𝐁 = (𝐀 𝐱 𝐢 + 𝐀 𝐲 𝐣 + 𝐀𝐳 𝐤) ∙ (𝐁𝐱 𝐢 + 𝐁𝐲 𝐣 + 𝐁𝐳 𝐤) 𝐀 ∙ 𝐁 = (𝐀 𝐱 𝐢 ∙ 𝐁𝐱 𝐢) + (𝐀 𝐱 𝐢 ∙ 𝐁𝐲 𝐣) + (𝐀 𝐱 𝐢 ∙ 𝐁𝐳 𝐤) + (𝐀 𝐲 𝐣 ∙ 𝐁𝐱 𝐢) + (𝐀 𝐲 𝐣 ∙ 𝐁𝐲 𝐣) + (𝐀 𝐲 𝐣 ∙ 𝐁𝐳 𝐤) + (𝐀𝐳 𝐤 ∙ 𝐁𝐱 𝐢) + (𝐀𝐳 𝐤 ∙ 𝐁𝐲 𝐣) + (𝐀𝐳 𝐤 ∙ 𝐁𝐳 𝐤) If vectors are parallel to each other, ϕ = 0o; cos 0o = 1 If vectors are perpendicular to each other, ϕ = 90o; cos 90o = 0 i∙i=1 i∙j=0 j∙j=1 i∙k=0 𝐀 ∙ 𝐁 = 𝐀 𝐱 ∙ 𝐁𝐱 + 𝐀 𝐲 ∙ 𝐁𝐲 + 𝐀𝐳 ∙ 𝐁𝐳 k∙k=1 j∙k=0 31  b) VECTOR PRODUCT – also called cross product. It is a vector quantity with a direction perpendicular to the plane of the vectors A & B and a magnitude given by ABsinθ. |𝐀 × B| = |A||B| sin θ Using unit vector representation: 𝐀 × 𝐁 = (Ax i + Ay j + Az k) × (Bx i + By j + Bz k) 0 𝐀 × 𝐁 = (Ax i × Bx i)0 + (Ax i × By j)k + (Ax i × Bz k)j + (Ay j × Bx i)k + (Ay j × By j) + (Ay j × Bz k)i + (Az k × Bx i)j + (Az k × By j)i + (Az k × Bz k)0 j i×i=0 i×j=k j × i = −k j×j=0 k×i=j i × k = −j i k×k =0 j×k=i k × j = −i k 𝐀 × 𝐁 = (Ax By ) × k − (Ax Bz ) × j − (Ay Bx ) × k + (Ay Bz ) × i + (Az Bx ) × j − (Az By ) × i 𝐀 × 𝐁 = (Ay Bz − Az By ) × i + (Az Bx − Ax Bz ) × j + (Ax By − Ay Bx ) × k Another method on how to solve cross product is by using the method of determinants. i j k i j + 𝐀×𝐁 Ax Ay Az Ax Ay Bx - By Bz Bx By 𝐀 × 𝐁 = Ay Bz i + Az Bx j + Ax By k − Bx Ay k − By Az i − Bz Ax j 𝐀 × 𝐁 = (Ay Bz − By Az ) × i + (Az Bx − Bz Ax ) × j + (Ax By − Bx Ay ) × k → same as the above equation 32 ELABORATE Directions: Determine the equivalent directions for the following: 1. 15° S of E 2. 30° N of W 3. Due South Solution: N E W E 15° 75° 60° 30° S W S 75° E of S = S 75° E = E 15° S 60° W of N = W 30° N = N 60° W From E (0°) = 270° From E (0°) = 345° From E (0°) = 150° Algebraic Method: 1. For the given vectors: A = 50 km due east, B = 20 km due west, C = 30 km due west, D = 25 km due east, E = 60 km due west. Determine (a) their resultant, (b) C – D and (c) D – A – B. E C B D A 60 30 20 25 50 Solution: For sign convention of vectors: to the right is positive, to the left is negative. a) R = Σ vectors = A + B + C + D + E R = 50 km + (−20 km) + (−30 km) + 25 km + (−60 km) 𝐑 = −𝟑𝟓 𝐤𝐦 Therefore, R = 35 km due West b) C − D = −30 km − 25 km = −55 km C – D = 55 km due West c) D − A − B = 25 km − 50 km − (−20 km) = −5 km D – A – B = 5 km due West 33 Parallelogram Method: 1. Determine the resultant for the given vectors by parallelogram A: 15 km, 33° N of E B: 10 km, 27° S of E Solution: Scale: 4km: 1cm Measured R = 5.4cm Measured θ = 10° 4 km 5.4 cm × = 21.6 km 1 cm Therefore, 𝐑 = 𝟐𝟏. 𝟔 𝐤𝐦 𝟏𝟎° 𝐍 𝐨𝐟 𝐄 TRIANGLE METHOD: 1. Given two vectors A = 70 m 60o N of E and B =35 m 30o N of W. Determine their resultant (magnitude and direction). Solution: First step is to draw the triangle formed by the two vectors, that is by connecting the two vectors head/tip to tail and by drawing the resultant by completing the triangle (from origin to the head of the second vector). Since the triangle formed by the two vectors is a right Δ, therefore, use Pythagorean theorem to solve for R. R2 = A2 + B 2 B = 35 m R2 = (70m)2 + (35 m)2 30o 𝐑 = 𝟕𝟖. 𝟐𝟔𝟐𝟒 𝐦 For the direction: R θ = α + 60o B 35 A = 70 m Wherein, α = Tan−1 ( ) = Tan−1 ( ) A 70 α = 26.5651o α θ Hence, 𝛉 = 26.5651° + 60° = 𝟖𝟔. 𝟓𝟔𝟓𝟏o 60o Therefore, R = 78.2624 m 86.5651o N of E 34 POLYGON METHOD: 1. Determine the magnitude and direction of the resultant for the given vectors. Use polygon method. A: 2 kph 45° N of E; B: 2.8 kph 58° E of S; C: 5.1 kph W 28° S; D: 3.2 kph N 18° W Solution: Scale: 1kph:1.5cm Measured R ≈2.7 cm Measured θ≈ 20 1kph 2.7 cm × = 1.8 kph 1.5 cm therefore, 𝐑 ≈ 𝟏. 𝟖 𝐤𝐩𝐡, 𝟐𝟎° 𝐍 𝐨𝐟 𝐖 *For polygon method, you may start with any vector and you will still arrive with the same result. For practice, you may try solving this problem with different pattern ☺ COMPONENT METHOD: 1. A sailor in a small boat encounters shifting winds. She sails 8 km south, then 15 km 30o E of N, and then 12 km 25o N of W. Use component method to determine the magnitude and direction of her resultant displacement. SOLUTION: N Bx A = 8 km Ax = 0 By Ay = - 8 km S B = 15 km 30o E C = 12 km N Cy Cx = C x cos 25o = 12 km x cos 25o Bx = B x sin30o = 15 km x sin30o 25O Cx = - 10.8757 km Bx = + 7.5 km W Cx Cy = C x sin 25o = 12 km x sin 25o By = B x cos30o = 15 km x cos30o Cy = + 5.0714 km By = + 12.9904 km 35 To solve for the resultant of the three displacements, simply follow the steps stated above. 𝐑 𝐱 = Σ X − comp = Ax + Bx + Cx 𝐑 𝐲 = Σ Y − comp = Ay + By + Cy R x = 0 + 7.5 km − 10.8757 R y = −8 km + 12.9904 km + 5.0714 𝐑 𝐱 = −𝟑. 𝟑𝟕𝟓𝟕 𝐤𝐦 𝐑 𝐲 = 𝟏𝟎. 𝟎𝟔𝟏𝟖 𝐤𝐦 Therefore, by using Pythagorean theorem: R2 = R x 2 + R y 2 = (−3.3757 km)2 + (10.0618km)2 𝐑 = 𝟏𝟎. 𝟔𝟏𝟑 𝐤𝐦 For the direction: −1 Ry 10.0618 Ry R θ = tan ( ) = tan−1 ( ) Rx 3.3757 θ = 71.4536o Therefore 𝐑 = 𝟏𝟎. 𝟔𝟏𝟑𝐤𝐦 𝟕𝟏. 𝟒𝟓𝟑𝟔𝐨 𝐍 𝐨𝐟 𝐖 θ Or this may also be written as 10.613km 18.5464 W of N Rx NOTE: for uniformity, let us always indicate directions in terms of the horizontal ( W or E) ❖ The solution for component method may also be presented on a tabular form. Take note that angle θ should be the angle the vector makes with zero-degree reference line. Solution 2: Vector Angle, θ X-comp = vector × cos θ Y-comp = vector × sin θ A = 8 km 270o 0 -8 km B = 15 km 60o 7.5 km 12.9904 km C = 12 km 155o -10.8757 km 5.0714 𝐑 𝐱 = Σ X − comp 𝐑 𝐲 = Σ Y − comp 𝑅𝑥 = −3.3757 𝑘𝑚 𝑅𝑦 = 10.0618 𝑘𝑚 R2 = R x 2 + R y 2 Ry θ = tan−1 ( ) 𝐑 = 𝟏𝟎. 𝟔𝟏𝟑 𝐤𝐦 Rx 𝛉 = 𝟕𝟏. 𝟒𝟓𝟑𝟔o Therefore 𝐑 = 𝟏𝟎. 𝟔𝟏𝟑𝐤𝐦 𝟕𝟏. 𝟒𝟓𝟑𝟔𝐨 𝐍 𝐨𝐟 𝐖 36 2. A spelunker is surveying a cave. He follows a passage 180 m straight west, then 210 m in a direction 45o E of S, and then 280 m at 30o E of N. After a fourth unmeasured displacement, he finds himself back where he started. Determine the magnitude and direction of the fourth displacement by using component method. SOLUTION: From the statement “he finds himself back where he started” means that the resultant of the four displacements is equal to zero. To solve for the fourth unmeasured displacement, D, simply use the tabular solution. Vector Angle, θ X-comp = vector x cos θ Y-comp = vector x sin θ A = 180 m 180o -180 m 0m B = 210 m 315o 148.4924 m -148.4924 m C =280 m 60o 140 m 242.4871 m D =? θ =? Dx =? Dy =? R=0m 0o 0 km 0 km To solve for D, apply Pythagorean theorem: D2 = Dx 2 + Dy 2 But, solve first for Dx and Dy. From the equation: 𝐑 𝐱 = Σ X − comp 𝐑 𝐲 = Σ Y − comp 𝐑 𝐱 = Ax + Bx + Cx + Dx 𝐑 𝐲 = Ay + By + Cy + Dy 0 = −180m + 148.4924 m + 140m + Dx 0 = 0 − 148.4924m + 242.4871m + Dy 𝐃𝐱 = −𝟏𝟎𝟖. 𝟒𝟗𝟐𝟒 𝐦 (means West) 𝐃𝐲 = −𝟗𝟑. 𝟗𝟗𝟒𝟕 m (means South) Hence: D2 = (−108.4924m)2 + (−93.9947m)2 Dx 𝐃 = 𝟏𝟒𝟑. 𝟓𝟒𝟔𝟓 𝐦 θ D 93.9947 For the direction: θ = tan−1 (Dy ) = tan−1 (108.4924) x Dy D 𝛉 = 𝟒𝟎. 𝟗𝟎𝟒𝟕𝟏o Therefore: 𝐃 = 𝟏𝟒𝟑. 𝟓𝟒𝟔𝟓 𝐦 𝟒𝟎. 𝟗𝟎𝟒𝟕𝟏o S of W 37 UNIT VECTORS: 1. Given three vectors A = -2i + 3j + 4k, B = 3i + 1j – 3k and C = 3i -4j + 2k, do the following: a) Find the magnitude of each vector b) Write an expression for the vector sum A + B using unit vectors c) Write an expression for the vector difference A – C using unit vectors d) Find the scalar product A ∙ B e) Find the cross product A × B f) Find the angle between A and B g) Find (A x C) ∙ B GIVEN: 3i B 4k 3j -3k -4j 1j A 3i 2i C 2k -2i SOLUTION: A = -2i + 3j + 4k → Ax = -2; Ay = 3; Az = 4 B = 3i + 1j – 3k → Bx = 3; By = 1; Bz = -3 C = 3i -4j + 2k → Cx = 3; Cy = -4; Cz = 2 a) Magnitude of each vector To solve for the magnitude of each vector, simply apply Pythagorean theorem A2 = Ax 2 + Ay 2 + Az 2 = (−2)2 + (3)2 + (4)2 𝐀 = 𝟓. 𝟑𝟖𝟓𝟐 𝐮𝐧𝐢𝐭 B 2 = Bx 2 + By 2 + Bz 2 = (3)2 + (1)2 + (−3)2 𝐁 = 𝟒. 𝟑𝟓𝟖𝟗 𝐮𝐧𝐢𝐭 C 2 = Cx 2 + Cy 2 + Cz 2 = (3)2 + (−4)2 + (2)2 𝐂 = 𝟓. 𝟑𝟖𝟓𝟐 𝐮𝐧𝐢𝐭 a) Vector sum A + B using unit vector A + B = (−2i + 3j + 4k) + (3i + 1j – 3k ) 𝐀 + 𝐁 = 𝐢 + 𝟒𝐣 + 𝐤 38  b) Expression for the vector difference A – C using unit vectors A − C = (−2i + 3j + 4k) − (3i − 4j + 2k) 𝐀 − 𝐂 = −𝟓𝐢 + 𝟕𝐣 + 𝟐𝐣 c) Find the scalar product A. B There are two equations to solve for 𝐀 ∙ 𝐁, these are 𝐀 ∙ 𝐁 = 𝐀𝐁 𝐜𝐨𝐬 𝛟and 𝐀 ∙ 𝐁 = 𝐀 𝐱 𝐁𝐱 + 𝐀 𝐲 𝐁𝐲 + 𝐀𝐳 𝐁𝐳 Use the second equation since angle ϕbetween the two vectors is not given. A ∙ B = AxBx + AyBy + AzBz = (−2)(3) + (3)(1) + (4)(−3) 𝐀 ∙ 𝐁 = −𝟏𝟓 𝐬𝐪. 𝐮𝐧𝐢𝐭 d) Cross product A x B There are two equations to solve for A x B, these are 𝐀 × 𝐁 = 𝐀𝐁 𝐬𝐢𝐧 𝛟and 𝐀 × 𝐁 = (𝐀 𝐲 𝐁𝐳 − 𝐀𝐳 𝐁𝐲 ) 𝐢 + ( 𝐀𝐳 𝐁𝐱 – 𝐀 𝐱 𝐁𝐳 ) 𝐣 + (𝐀 𝐱 𝐁𝐲 − 𝐀 𝐲 𝐁 )𝐤 Use the second equation since angle ϕbetween the two vectors is not given. Solution 1: A × B = (Ay Bz − Az By ) × i + (Az Bx – Ax Bz ) × j + (Ax By − Ay B ) × k AxB = {(3)(−3) − (4)(1)} × i + {(4)(3) − (−2)(−3)} × j + {(−2)(1) − (3)(3)} × k 𝐀𝐱𝐁 = −𝟏𝟑𝐢 + 𝟔𝐣 − 𝟏𝟏𝐤 Solution 2: i j k i j + 𝐀×𝐁 -2 3 4 -2 3 3 - 1 -3 3 1 A x B = −9i + 12j − 2k − 9k − 4i − 6j 𝐀 𝐱 𝐁 = −𝟏𝟑𝐢 + 𝟔𝐣 − 𝟏𝟏𝐤 Magnitude of 𝐀 𝐱 𝐁 = √(−𝟏𝟑)𝟐 + (𝟔)𝟐 + (−𝟏𝟏)𝟐 Magnitude of 𝐀 𝐱 𝐁 = 𝟏𝟖. 𝟎𝟓𝟓𝟓 𝐬𝐪. 𝐮𝐧𝐢𝐭 39  e) Angle ϕ between A and B There are two equations to solve for the angle between the two vectors A and B. These are: 𝐀 ∙ 𝐁 = 𝐀𝐁 𝐜𝐨𝐬 𝛟 and 𝐀 × 𝐁 = 𝐀𝐁 𝐬𝐢𝐧 𝛟 Using A. B = AB cos∅ Using A x B = AB sin∅ −15 = (5.3852 )( 4.3589)cos∅ 18.0555 = (5.3852 )( 4.3589)sin∅ 𝛟 = 𝟏𝟐𝟗. 𝟕𝟏𝟖𝟔 o ∅ = 50.2808𝐨 Final ϕ = 1800 − 50.2808o → 𝛟 = 𝟏𝟐𝟗. 𝟕𝟏𝟗𝟐o AxB B B A θ A ❖ To solve for the angle between the two vectors, it is better to use the dot product equation because it will give the exact answer for the angle ∅. The angle obtained from the cross product equation can sometimes be the exact angle, but in some cases it should be deducted from 180° (i.e. if 𝐀 ∙ 𝐁 is negative) because the final answer is the supplementary angle of the angle that is obtained from cross product equation. EVALUATE 1. Determine the equivalent directions for the given vectors. a. 35° N of E c. S 40° E b. 20° W of S d. N 65° W 2. Determine the equilibrant of the given vectors by (a) parallelogram method and (b) triangle method. Use the given scale for your answer. C = 10 kph, 12 N of W E = 15 kph, 83 S of E 3. Determine the resultant of the given vectors by component method. Summarize your answers in the given table. VECTORS  from 0° X – component Y - component E 23 km 11° N of E N 25 km 24° E of S G 19 km 18° S of W R 27 km 58° W of N 40 4. Given are the vectors A = 30 N 45o N of W and B = 45 N 75o S of E. Determine a) their resultant and b) their vector difference B – A. 5. A sailor in a small boat encounters shifting winds. She sails 8 km south, then 15 km 30o E of N, and then 12 km 25o N of W. Use polygon method to determine the magnitude and direction of her resultant displacement. 6. Solve for the required using the given vectors A = 3i – 8j + 6k B = 4 j + 9k – 5i C = 7k – 2j a. A ∙ B b. A × B c. (A × B) × C d. (B × C) ∙ A 41  MODULE2: KINEMATICS MODULE 2 UNIT 1: MOTION ON A STRAIGHT LINE ENGAGE Based on your learning from module 1, how can you apply vectors and scalar quantities in the study of motion? EXPLORE Read Module 2 Unit 1: Motion on a straight line (pp 42 – 45 ) EXPLAIN MOTION Change in position of an object in a given time interval A continuous change in position KINEMATICS A branch of mechanics concerned with the motion of objects regardless of the cause of motion PARTICLE “Idealized model” representing a moving body; no rotation or change in shape occurs during motion COORDINATES (Rectangular) The mathematical method of locating points in a space or plane will also be used for locating bodies as they move MOTION ALONG A STRAIGHT LINE (RECTILINEAR MOTION) We start our study with the simplest type of motion a body can undergo. This is called rectilinear motion or motion along a straight line. For the analysis, we will be considering the line of motion as a coordinate axis, i.e. the x-axis if the line of motion is horizontal or inclined or the y-axis if the line of motion is vertical. 42 BASIC CONCEPTS: 1. Position, x o indicate the location of the body at any time as it moves o the distance from a given reference point along the path at any time o usually given as a function of time like X = (10 m/s2) t2 or X = (5 m/s) t + 10 m, etc o The form of the equation depends on how the body moves along the line. In cases where the line of motion is the y-axis, position is denoted by (y). 2. Displacement(x) o the change in position of a body during a certain length of time or time interval Displacement is different from distance traveled in the sense that displacement is a vector quantity directed from the initial to the final position. Distance is the length of a path followed by a particle. However, in one directional rectilinear motion, the magnitude of the displacement is the same as the distance traveled. 3. time instant vs time interval timeinstant(t) ▪ the time at a certain point, i.e. at the time 3 secs after starting or time 5 secs before stops, etc timeinterval (t) ▪ a length of time, i.e. during the first 5 secs, or during the time from t1 = 4 secs to t2 = 7 secs, etc. 43  4. Speed vs velocity Speed – defines how fast an object moves Velocity (V) – the rate of change in position of an object; the speed of a body including its direction of motion a. Averagevelocity (Vave) ▪ velocity of a body taken during a time interval, or between two points along its path 𝐝𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭 ∆𝐱 𝐕𝐚𝐯𝐞 = = 𝐭𝐢𝐦𝐞𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 ∆𝐭 b. Instantaneous velocity (V) ▪ velocity of an object at a specific time, or specific point along its path ∆𝐱 𝐝𝐱 𝐕 = 𝐥𝐢𝐦 = ∆𝐭 →𝟎 ∆𝐭 𝐝𝐭 Where x is the position of the given as a function of time *velocity is the derivative of displacement 5. Acceleration – the rate of change in velocity of an object; a body is said to be accelerating when the velocity is changing a. Average acceleration (aave) ▪ acceleration of a body taken during a time interval, or between two points along its path (displacement) 𝐜𝐡𝐚𝐧𝐠𝐞𝐢𝐧𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 ∆𝐕 𝐚𝐚𝐯𝐞 = = 𝐭𝐢𝐦𝐞𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 ∆𝐭 b. Instantaneous acceleration (a) ▪ acceleration of an object at a specific time, or specific point along its path ∆𝐕 𝐝𝐕 𝐝𝟐 𝐱 𝐚 = 𝐥𝐢𝐦 = = ∆𝐭 →𝟎 ∆𝐭 𝐝𝐭 𝐝𝐭 𝟐 *acceleration is the derivative of velocityAND the 2nd derivative of the displacement UNIFORMLY ACCELERATED RECTILINEAR MOTION (UARM) A very common and simple type of one-dimensional motion where the acceleration is constant. In such a case, at any instant within the interval, and the velocity changes at the same rate throughout the motion. 44 BASIC KINEMATIC EQUATIONS for UARM: **To simplify the equations, it will be assumed here that at time t = 0, the position xi = 0,Thus the time interval Δt will be the same as time instant t (Δt = t) because ti = 0 and tf = t. The displacement Δx will also become same as position x (Δx = x) because xi = 0 and xf = x at time instant t. Derivation: ∆X X Take Vf from Eq 3.a and substitute in Eq 2 Vave = = → Eq a, ∆t t 2x But − Vi = Vi + at t Vi + Vf 2x Vave = → Eq b = 2Vi + at → 2x = 2Vi t + at 2 2 t Equating Eq a and Eq b 𝟏 𝐗 = 𝐕𝐢 𝐭 + 𝐚𝐭 𝟐 → 𝐄𝐪 𝟑 𝐕𝐢 + 𝐕𝐟 𝟐 𝐗= 𝐭 → Eq 1 𝟐 Take Vi from Eq 1.a and substitute in Eq 2 Eq 1 can be written as 𝟏 𝐗 = 𝐕𝐟 𝐭 − 𝐚𝐭 𝟐 → 𝐄𝐪 𝟒 2x 𝟐 = Vi + Vf → Eq 1. a t Take t from Eq 1.a and substitute in Eq 2 From acceleration: 2X t= 𝐕𝐟 = 𝐕𝐢 + 𝐚𝐭 → Eq 2 Vi + Vf 2X 2X Vf = Vi + a ( ) → Vf − Vi = a ( ) Vi + Vf Vi + Vf (Vf − Vi )(Vi + Vf ) = 2aX 𝐕𝐟𝟐 = 𝐕𝐢𝟐 + 𝟐𝐚𝐗 → 𝐞𝐪 𝟓 FREE FALLING MOTION A freely falling body is one that moves vertically in air under the action of its weight alone. Air resistance is neglected and there is no external additional factor that is causing the body to move except gravity. The specific equations for free fall are the following (based from Basic equations of UARM: From UARM → Free Fall Sign convention: 1. 𝐗 = 𝐕𝐢 + 𝐕𝐟 𝐭 1. 𝐘 = 𝐕𝐢 + 𝐕𝐟 𝐭 Y is positive (+) if measured above the 𝟐 𝟐 starting point. It is negative (-) if measured below the starting point. 2. 𝐕𝐟 = 𝐕𝐢 + 𝐚𝐭 2. 𝐕𝐟 = 𝐕𝐢 + 𝐠𝐭 Vi and Vfare positive (+) if directed upward, 𝟏 𝟏 negative (-) if directed downward. 3. 𝐗 = 𝐕𝐢 𝐭 + 𝟐 𝐚𝐭 𝟐 3. 𝐘 = 𝐕𝐢 𝐭 + 𝐠𝐭 𝟐 𝟐 Time t is always positive (+). 𝟏 𝟏 4. 𝐗 = 𝐕𝐟 𝐭 − 𝟐 𝐚𝐭 𝟐 4. 𝐘 = 𝐕𝐟 𝐭 − 𝟐 𝐠𝐭 𝟐 g is always negative (-): (g=-9.8 m/s2; -980 cm/s2; -32 ft/s2) 5. 𝐕𝐟𝟐 = 𝐕𝐢𝟐 + 𝟐𝐚𝐗 5. 𝐕𝐟𝟐 = 𝐕𝐢𝟐 + 𝟐𝐠𝐘 45 ELABORATE HORIZONTAL MOTION m 1. The motion of a particle along the x –axis is described by the equation x = (6 s ) t + m (0.5 3 ) t 3 where x is in meters, and t is in seconds. Determine the following: s a. Position of the particle 3 seconds after 2. Average acceleration during the starting second 3 seconds b. Displacement from 2 sec to 5 sec 3. Acceleration at t = 5 secs c. Average velocity during first 5 seconds 4. Time the particle moves at 20m/s d. Velocity at t= 4secs 5. The acceleration and velocity at start of motion SOLUTION: m m Given: X = (6 s ) t + (0.5 s3 ) t 3 a) The position of the particle 3 secs after starting To solve for the position of the body at any time t, simply substitute the value of t in the given equation. m m m m Therefore at t = 3 secs ; X = (6 ) t + (0.5 3 ) t 3 = (6 ) (3s) + (0.5 3 ) (3s)3 s s s s 𝐗 = 𝟑𝟏. 𝟓 𝐦 b) The displacement during the time interval from 2 sec to 5 secs Displacement is the change in the position of the body. That is ∆X = X 2 − X1. But before ∆X can be solved, the values of X1 and X 2 should be solved first from the equation of X. Let t1 = 2 sec, and x1 = distance travelled at time 2 seconds m m m m X1 = (6 s ) t + (0.5 s3 ) t 3 = (6 s ) (2s) + (0.5 s3 ) (2s)3 𝐗 𝟏 = 𝟏𝟔 𝐦 Let t2 = 5 sec, and x2 = distance travelled at time 5 seconds m m m m X 2 = (6 ) t + (0.5 3 ) t 3 = (6 ) (5s) + (0.5 3 ) (5s)3 s s s s 𝐗 𝟐 = 𝟗𝟐. 𝟓 𝐦 Therefore: the displacement, ∆X = X 2 − X1 = 92.5 m − 16 m ∆𝐗 = 𝟕𝟔. 𝟓 𝐦 (occurs between t1 and t2) 46 c) The average velocity during the first 5 secs Average velocity is the velocity of the body taken during a time interval. That is 𝐃𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭 ∆𝐗 𝐕𝐚𝐯𝐞 = = 𝐭𝐢𝐦𝐞 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 ∆𝐭 The term during the first 5 secs means that the body starts from t 1 = 0 s and ends at t2 = 5 s. Solve first for the values of X1 and X 2 from the equation of X. m m m m At t1 = 0 sec; X1 = (6 s ) t + (0.5 s3 ) t 3 = (6 s ) (0s) + (0.5 s3 ) (0s)3 𝐗 𝟏 = 𝟎𝐦 m m m m At t2 = 5 sec; X 2 = (6 s ) t + (0.5 s3 ) t 3 = (6 s ) (5s) + (0.5 s3 ) (5s)3 𝐗 𝟐 = 𝟗𝟐. 𝟓 𝐦 𝐃𝐢𝐬𝐩𝐥𝐚𝐜𝐞𝐦𝐞𝐧𝐭 ∆𝐗 𝐗 𝟐 −𝐗 𝟏 𝟗𝟐.𝟓 𝐦−𝟎 𝐦 Therefore: 𝐕𝐚𝐯𝐞 = 𝐭𝐢𝐦𝐞 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 = ∆𝐭 = 𝐭 𝟐 −𝐭 𝟏 = 𝟓 𝐬−𝟎 𝐬 𝐕𝐚𝐯𝐞 = 𝟏𝟖. 𝟓 𝐦/𝐬 d) The velocity at time t = 4 secs Instantaneous velocity is the velocity of the body at a specific point in time. To get its value, first determine the equation of instantaneous velocity by getting the derivative of X with respect to time, and then simply substitute the value of time in the derived equation. m m 𝐝𝐗 𝐝 [(6 )t+(0.5 3)t3 ] 𝐦 𝐦 𝐕= 𝐝𝐭 = s 𝐝𝐭 s =𝟔 𝐬 + (𝟏. 𝟓 𝐬𝟑 ) 𝐭 𝟐 m m m m Therefore at t = 4 sec; V=6 s + (1.5 s3 ) t 2 = 6 s + (1.5 s3 ) (4 s)2 𝐕 = 𝟑𝟎 𝐦/𝐬 e) The average acceleration during the second 3 seconds Average acceleration is the acceleration of the body taken during a time interval. 𝐂𝐡𝐚𝐧𝐠𝐞 𝐢𝐧 𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 ∆𝐕 That is 𝐚𝐚𝐯𝐞 = 𝐭𝐢𝐦𝐞 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 = ∆𝐭 The term during the second 3 secs means that the body starts from t1 = 3 s and ends at t2 = 6 s. Solve first for the values of V1 and V2 from the equation of V. m m m m At t1 = 3 s; V1 = 6 s + (1.5 s3 ) t 2 = 6 s + (1.5 s3 ) (3 s)2 = 19.5 m/s m m m m At t2 = 6 s; V2 = 6 + (1.5 3 ) t 2 = 6 + (1.5 3 ) (6 s)2 = 60 m/s s s s s m m Change in velocity ∆V V2 −V1 60 −19.5 Therefore: aave = time interval = ∆t = t2 −t1 = s 6 s−3 s s 𝟐 𝐚𝐚𝐯𝐞 = 𝟏𝟑. 𝟓 𝐦/𝐬 47 f) What is the acceleration at time t = 5 secs? Instantaneous acceleration is the acceleration at a specific point in time. To get its value, first determine the equation of instantaneous acceleration by getting the derivative of V with respect to time, and then simply substitute the value of time in the derived equation. 𝐦 𝐦 𝐝𝐕 𝐝 [𝟔 𝐬 + (𝟏. 𝟓 𝐬𝟑) 𝐭 𝟐 ] 𝐦 𝐚= = = (𝟑 𝟑 ) 𝐭 𝐝𝐭 𝐝𝐭 𝐬 𝐦 𝐦 Therefore, at t = 5 secs; 𝐚 = (𝟑 𝟑) 𝐭 = (𝟑 𝟑 ) (𝟓 𝐬) 𝐬 𝐬 𝟐 𝐚 = 𝟏𝟓 𝐦/𝐬 g) At what time was it moving at 20 m/s? Given is an instantaneous velocity. Use the equation of instantaneous velocity to solve for the value of time. 𝐦 𝐦 𝐕 = 𝟔 𝐬 + (𝟏. 𝟓 𝐬𝟑) 𝐭 𝟐 m m m 20 = 6 + (1.5 3 ) t 2 → = 𝟑. 𝟎𝟓𝟓 𝐬 s s s h) What are the acceleration and velocity when it started? The term “when it started” means time t = 0. To solve for the value of acceleration and velocity at t = 0, use the equation of instantaneous acceleration and instantaneous velocity, respectively. m 𝐦 At t = 0 s; a = (3 s3 ) t = 𝟎 𝐬𝟐 m m 𝐦 At t = 0 s; V=6 s + (1.5 s3 ) t 2 = 𝟔 𝐬 48 UARM: 1. A car starting from rest moves with constant acceleration. after 10 sec, the velocity is now 30 m/s. Compute for (a) acceleration, (b) distance travelled after 10 sec, (c) time it took to travel 100meters, and (d) its velocity after travelling 50 meters. SOLUTION: First step of the solution is to identify the given values. Draw a simple diagram representing the given. Given: Solution: a) the acceleration To solve for the acceleration, choose an equation wherein the given values can be substituted directly. Given are the Vi, Vf and t. Therefore, use equation (2). m m Vf = Vi + at → 30 = 0 + a (10 s) → 𝒂 = 𝟑 𝒎/𝒔𝟐 s s b) the distance traveled after 10 secs To solve for the distance traveled given again Vi ,Vf and t, choose equation (1) m m Vi + Vf 0 + 30 s s X= t= (10 s) → 𝑿 = 𝟏𝟓𝟎 𝒎/𝒔 2 2 c) the time it took to travel 100 meters This time the value of acceleration can be used to solve for the time the car took to travel 100 m since the car is moving at constant acceleration. Given now are Vi , X and 𝒂. Therefore, use equation (3) to solve for t. 1 X = Vi t + at 2 2 m 1 m 100m = (0 ) t + (3 2 ) t 2 → 𝒕 = 𝟖. 𝟏𝟔𝟓 𝒔 s 2 s d) its velocity after traveling 50 meters. Use equation (5) to solve for Vf after traveling 50 m since the given are X, Vi and 𝒂. 𝒎 𝒎 𝑽𝟐𝒇 = 𝑽𝟐𝒊 + 𝟐𝒂𝑿 = ( 𝟎 )𝟐 + 𝟐 (𝟑 𝟐 ) (𝟓𝟎 𝒎) → 𝑽𝒇 = 𝟏𝟕. 𝟑𝟐𝟏 𝒎/𝒔 𝒔 𝒔 49 2. A subway train starts from rest at a station and accelerates at a rate of 1.6 m/s 2 for 14 s. It runs at constant speed for 70 s and slows down at a rate of 3.5 m/s 2 until it stops at the next station. Find the total distance covered. GIVEN: REQUIRED: Total distance traveled by the subway train, XT SOLUTION: The total distance traveled by the train is the sum of the distances it has traveled in the different situations it has encountered. Let X1 = distance traveled at 𝒂 =1.6 m/s2 X2 = distance traveled at constant velocity X3 = distance traveled at 𝒂 = -3.5 m/s2 Therefore, 𝑋𝑇 = 𝑋1 + 𝑋2 + 𝑋3 Solve for 𝑿𝟏 : Given 𝑎, t and Vi , use equation (3) to solve for X1. 𝟏 𝑚 1 𝑚 𝑿𝟏 = 𝑽𝒊 𝒕 + 𝒂𝒕𝟐 = (0 ) (14𝑠) + (1.6 2 ) (14𝑠)2 → 𝑿𝟏 = 𝟏𝟓𝟔. 𝟖 𝒎 𝟐 𝑠 2 𝑠 Solve for 𝑿𝟐 : Before 𝑋2 can be solved, the value of the constant velocity(let it be V1) should be solved first. Given 𝒂 =1.6 m/s2 ,t = 14 s and Vi = 0 m/s, use equation (2) to solve for V1. 𝑚 𝒎 𝑽𝟏 = 𝑽𝒊 + 𝒂𝒕 = 0 + (1.6 2 ) (14 𝑠) = 𝟐𝟐. 𝟒 𝑠 𝒔 At constant velocity, the only equation that is applied is 𝑿 = 𝑽𝒕 𝑚 𝑋2 = 𝑉1 𝑡 = (22.4 ) (70 𝑠) → 𝑿𝟐 = 𝟏𝟓𝟔𝟖 𝒎 𝑠 𝑚 Solve for 𝑿𝟑 : Given 𝒂 = -3.5 m/s2 , 𝑉1 = 22.4 𝑠 and Vf = 0, use equation (5) to solve for 𝑿𝟑. 𝑉𝑓2 = 𝑉12 + 2𝑎𝑋3 0𝑚 2 22.4𝑚 2 𝑚 ( ) =( ) + 2 (−3.5 2 ) (𝑋3 ) → 𝑿𝟑 = 𝟕𝟏. 𝟔𝟖 𝒎 𝑠 𝑠 𝑠 Solving for 𝑋𝑇 : 𝑋𝑇 = 𝑋1 + 𝑋2 + 𝑋3 = 156.8 m + 1568 m + 71.68 m →𝑿𝑻 = 𝟏𝟕𝟗𝟔. 𝟒𝟖 𝒎 50 FREE FALL 1. A stone is thrown vertically downward with a velocity of 5 m/s from a window 50 meters above ground level. (a) How much time will it take to travel to the ground? (b) With what velocity will it strike the ground? (c) How far did it fall in 2? Given: Solution: a) Time for the stone to reach the ground Given are Vi , Y and g, hence, use equation (3) to solve for the value of t, but don’t forget to use proper sign convention. 𝟏 𝒀 = 𝑽𝒊 𝒕 + 𝒈𝒕𝟐 𝟐 𝑚 1 𝑚 2 −50 𝑚 = (−5 ) 𝑡 + (−9.8 2 ) 𝑡 𝑠 2 𝑠 𝑚 2 𝑚 (4.9 2 ) 𝑡 + (5 ) 𝑡 − 50 𝑚 = 0 𝑠 𝑠 Use quadratic formula to solve for the values of t: −𝟓 ± √(𝟓)𝟐 − 𝟒(𝟒. 𝟗)(−𝟓𝟎) 𝒕= → 𝒕 = 𝟐. 𝟕𝟐𝟒𝟕 𝒔 𝟐(𝟒. 𝟗) b) Velocity with which the stone strikes the ground Use equation (5) to solve for Vf. 𝒎 𝒎 𝑽𝟐𝒇 = 𝑽𝟐𝒊 + 𝟐𝒈𝒀 = (−𝟓 )𝟐 + 𝟐 (−𝟗. 𝟖 𝟐 ) (−𝟓𝟎 𝒎) 𝒔 𝒔 𝑽𝒇 = 𝟑𝟏. 𝟕𝟎𝟏𝟕 𝒎/𝒔 2. A ball is thrown vertically upward from the roof of a building. It just misses the roof on its way down and passes a point 30 m below its starting point 5 s after it leaves the thrower’s hand. Air resistance may be ignored. (a) What is the initial speed of the ball? (b) How high does it rise above its starting point? (c) What is its velocity 5 s after it leaves the thrower’s hand? Given: Solution: a) Initial speed of the ball Use equation (3) to solve for Vi , since Y, t and g are given. 𝟏 𝒀 = 𝑽𝒊 𝒕 + 𝒈𝒕𝟐 𝟐 1 𝑚 −30 𝑚 = 𝑉𝑖 (5 𝑠) + (−9.8 2 )(5 𝑠)2 2 𝑠 𝑽𝒊 = 𝟏𝟖. 𝟓 𝒎/𝒔 51  b) Height, Y, the ball reached above its starting point When the ball reaches its highest point its velocity becomes zero. Knowing the values of Vi, Vf and g, use equation (5) to solve for the value of Y. 𝑽𝟐𝒇 = 𝑽𝟐𝒊 + 𝟐𝒈𝒀 𝑚 2 𝑚 2 𝑚 (0 ) = (18.5 ) + 2 (−9.8 2 ) 𝑌 𝑠 𝑠 𝑠 𝒀 = 𝟔𝟗. 𝟖𝟒𝟕 𝒎 c) Velocity of the ball 5 s after it leaves the thrower’s hand Use equation (2) to solve for the velocity, Vf, of the ball 5 s after it leaves the thrower’s hand, since the given are Vi, t and g. 𝒎 𝒎 𝑽𝒇 = 𝑽𝒊 + 𝒈𝒕 = 𝟏𝟖. 𝟓 + (−𝟗. 𝟖 𝟐 ) (𝟓𝒔) 𝒔 𝒔 𝑽𝒇 = −𝟑𝟎. 𝟓 𝒎/𝒔 ( - ) negative means velocity is downward 3. A stone is dropped from a building 50 m high. At what time will the stone reach the ground? What is the stone’s velocity upon hitting the ground? What is the stones velocity at 1 sec? what is the stones position at t=2 secs and t= 3 secs? 4. At what velocity must an object be thrown vertically up to reach its maximum height in 2.5 seconds? How high is the peak from the starting point? EVALUATE HORIZONTAL MOTION m 1. The motion of a particle along the x –axis is described by the equation v = (3 s2 ) t + m (0.5 s2 ) t 2 where x is in meters, and t is in seconds. Determine the following: a. Position of the particle 3 seconds after e. Average acceleration during the starting second 3 seconds b. Displacement from 2 sec to 5 sec f. Acceleration at t = 5 secs c. Average velocity during first 5 seconds g. Time the particle moves at 20m/s d. Velocity at t= 4secs h. The acceleration and velocity at start of motion UARM: 1. A car is moving at uniform acceleration. after traveling for 2 seconds, it is found 12 meters from where it started. If the car’s velocity at 5 seconds is 10 m/s, determine (a) the car’s acceleration, (b) its displacement at t = 5 seconds measured from its starting point and (c) its velocity at the start. 52 2. A ball is rolled on a flat surface with an initial velocity of 2 m/s. what is the ball’s acceleration if it was found 15 meters away from its starting position after 5 seconds? What is the ball’s velocity at a distance 20m? how long will the ball travel to achieve a velocity of 20 m/s? FREE FALL: 1. If the stone in example problem 3 (elaborate) was initially thrown downwards at 4 m/s, how much time will it take to reach the ground? At what velocity will it hit the ground? At what time will it reach a velocity of 12m/s? Where is the stone at v=6 m/s? 2. Resolve evaluate problem number 1 (free fall), if the stone was thrown upwards instead. Use the same given values. How high will the stone reach? KINEMATICS UNIT 2: MOTION IN TWO DIMENSIONS ENGAGE From Module 2 Unit 2, what concepts can be applied in two- dimensional motion? What factors would affect the motion of objects in two dimensions? EXPLORE Read Module 2 Unit 2: Motion in Two Dimensions (pp 53 – 55) EXPLAIN MOTION IN TWO DIMENSIONS Motion in two dimensions can be modelled as two independent motions in each of the two perpendicular directions associated with the x and y axes. That is, any influence in the y direction does not affect the motion in the x direction and vice versa. PROJECTILE MOTION o A curvilinear motion which moves under the sole effect of gravity. Air resistance is neglected just like in free falling motion. The difference is that for a projectile, the initial velocity called the velocity of projection is not vertical. It is directed either horizontally, at an angle below the horizontal, or at anangle above the horizontal. o the path is that of an inverted parabola, symmetric about a vertical line that passes thru the vertex (highest point) 53  TRAJECTORY o curved path followed by a projectile and is always in the form of an inverted Parabola HORIZONTAL COMPONENTS VERTICAL COMPONENTS Displacement R = range Y Velocity Vx = constant (Vix = Vfx) Vy (under free fall) acceleration ax=0 ay = g = -9.81 m/s2 Equations for Projectile Motion: Vertical: Sign convention: 1. 𝐘 = 𝐕𝐢𝐲 + 𝐕𝐟𝐲 𝐭 Y is positive (+) if measured above the starting point. It is negative 𝟐 (-) if measured below the starting point. 2. 𝐕𝐟𝐲 = 𝐕𝐢𝐲 + 𝐠𝐭 Viy and Vfyare positive (+) if directed upward, negative (-) if 𝟏 directed downward. 3. 𝐘 = 𝐕𝐢𝐲 𝐭 + 𝐠𝐭 𝟐 𝟐 Time t is always positive (+). 𝟏 4. 𝐘 = 𝐕𝐟𝐲 𝐭 − 𝐠𝐭 𝟐 g is always negative (-). (g=-9.8 m/s2; -980 cm/s2; -32 ft/s2) 𝟐 5. 𝐕𝐟𝐲 𝟐 𝟐 = 𝐕𝐢𝐲 + 𝟐𝐠𝐘 Horizontal: 6. 𝐑 = 𝐗 = 𝐕𝐱 𝐭 54 x and y components of Initial Velocity 𝐯𝐱 = 𝐯𝐢𝐱 = 𝐯𝐢 𝐜𝐨𝐬 𝛉 𝐯𝐢𝐲 = 𝐯𝐢 𝐬𝐢𝐧 𝛉 𝒗𝒊𝒚 𝒗𝒇𝒙 = 𝒗𝒙 2 2 𝒗𝒊 𝑉𝑓 = √𝑉𝑓𝑥 + 𝑉𝑓𝑦 𝛉 𝑉𝑓𝑦 𝜃 = 𝑇𝑎𝑛−1 θ 𝑉𝑓𝑥 𝒗𝒇𝒚 𝒗𝒇 𝒗𝒊𝒙 CASES: 1. Projectile launched HORIZONTALLY θ at initial is zero vi = vix = vx = vfx Y is negative 2. Projectile launched at an angle BELOWthe horizontal vi = vix = vx = vfx Viy is negative Y is negative 3. Projectile launched at an angle ABOVEthe horizontal vi = vix = vx = vfx Viy is positive V approaches zero as it reaches Ymax 55 ELABORATE 1. A ball is thrown horizontally with a velocity of 30 m/s from a window 40 m above ground level. Solve for (a) the time it takes to travel to the ground, (b) the velocity 2 secs after being thrown, (c) the horizontal displacement or range. Given: SOLUTION: a) The time it takes to travel to the ground Use equation (3) to solve for the time the ball takes to travel to the ground since 𝑉𝑖 , Y and g are given. But 𝑉𝑖𝑦 is zero since 𝑉𝑖 is horizontal. 𝟏 𝒀 = 𝑽𝒊𝒚 𝒕 + 𝒈𝒕𝟐 𝟐 1 𝑚 −40 𝑚 = 0 + (−9.8 2 )𝑡 2 2 𝑠 𝒕 = 𝟐. 𝟖𝟓𝟕𝟏 𝒔 b) The velocity 2 secs after being thrown To solve for the velocity after 2 s, use the equation 𝑉 = √𝑉𝑥2 + 𝑉𝑦2 ; 𝑉𝑖𝑥 = 𝑉𝑖 = 𝑉𝑥 = 𝟑𝟎 𝒎/𝒔 To solve for 𝑉𝑦 at t = 2 s use equation (2) since Vi , t and g are given. But 𝑉𝑖𝑦 is again zero since 𝑉𝑖 is horizontal. 𝑚 𝒎 𝑉𝑦 = 𝑉𝑖𝑦 + 𝑔𝑡 = 0 + (−9.8 2 ) (2 𝑠)2 = −𝟑𝟗. 𝟐 ↓ 𝑠 𝒔 𝑚 𝒎 Therefore: 𝑉 = √𝑉𝑥2 + 𝑉𝑦2 = √(30 𝑠 )2 + (−39.2 𝑚/𝑠)2 → 𝑽 = 𝟒𝟗. 𝟑𝟔𝟐𝟑 𝒔 2. An airplane was diving at an angle of 30o below the horizontal when it dropped a bomb from a height of 1000 meters. The bomb hits the ground 5 sec later. (a) What was the velocity of the airplane? (b) With what velocity did the bomb hit the ground? Given: SOLUTION: a) The velocity of the airplane The velocity of the plane V is the same as the initial velocity of the bomb Vi which is dropped from the plane with an initial velocity that is 30o below the horizontal. To solve for Vi of the plane use equation (3) since Y, t and g are given. Use the correct sign convention. 𝟏 1 𝑚 𝒀 = 𝑽𝒊𝒚 𝒕 + 𝒈𝒕𝟐 → −1000 𝑚 = (−𝑉𝑖 𝑠𝑖𝑛30𝑜 )(5 𝑠) + (−9.8 2 )(5 𝑠)2 → 𝑽𝒊 = 𝟑𝟓𝟏 𝒎/𝒔 𝟐 2 𝑠 56  b) With what velocity did the bag of sand hit the ground 𝑽𝒇 = √𝑽𝟐𝒇𝒙 + 𝑽𝟐𝒇𝒚 𝑚 𝒎 But 𝑉𝑖𝑥 = 𝑉𝑓𝑥 = 𝑉𝑥 = 351 𝑠 𝑥 𝑐𝑜𝑠 30𝑜 → 𝑽𝒇𝒙 = 𝟑𝟎𝟑. 𝟗𝟕𝟒𝟗 𝒔 Use equation (2) to solve for Vfy. 𝑚 𝑚 𝑉𝑓𝑦 = 𝑉𝑖𝑦 + 𝑔𝑡 = (−351 𝑥 𝑠𝑖𝑛30𝑜 ) + (−9.8 2 ) (5 𝑠) 𝑠 𝑠 𝑽𝒇𝒚 = − 𝟐𝟐𝟒. 𝟓 𝒎/𝒔 ↓ 𝑚 2 𝑉𝑓 = √(303.9749 ) + (− 224.5 𝑚/𝑠)2 → 𝑽𝒇 = 𝟑𝟕𝟕. 𝟖𝟗 𝒎/𝒔 𝑠 3. A cannonball is fired with a velocity of 40 m/s 30o above the horizontal from the top of a cliff 50 meters high. (a) How much time will it take to travel to ground level? (b) What is the highest point reached measured from ground level? (c) What is the range of the cannonball? (d) With what velocity did the cannonball strike the ground? Given: SOLUTION: a) Time t the cannonball will take to travel to ground level Use equation (3) to solve for t since Y, Viy and g are 𝟏 given :𝒀 = 𝑽𝒊𝒚 𝒕 + 𝟐 𝒈𝒕𝟐 𝑚 1 𝑚 − 50 𝑚 = (40 𝑥 𝑠𝑖𝑛30𝑜 ) 𝑡 + (−9.8 2 ) 𝑡 2 𝑠 2 𝑠 2 4.9 𝑡 − 20 𝑡 − 50 = 0 Use quadratic formula to solve for the value of t: 𝒕 = 𝟓. 𝟖𝟑𝟏𝟒 𝒔 b) the highest point reached measured from ground level The height reached by the cannonball measured from the ground level, Yt, is the sum of 50 m and height Y as shown from the diagram. To solve for the value of Y, use equation (5). The velocity y-component Vy of the cannonball at the highest point is zero. 𝑽𝒚𝟐 = 𝑽𝒊𝒚𝟐 + 𝟐𝒈𝒀 𝑚 𝑚 0 = (40 𝑥 𝑠𝑖𝑛30)2 + 2 (−9.8 ) 𝑌 𝑠 𝑠 𝒀 = 𝟐𝟎. 𝟒𝟎𝟖𝟐 𝒎 Solving for the total height Yt from the ground level: 𝑌𝑡 = 50 𝑚 + 𝑌 = 50 𝑚 + 20.4082 𝑚 𝒀𝒕 = 𝟕𝟎. 𝟒𝟎𝟖𝟐 𝒎 57  c) The range of the cannonball 𝑚 𝑅 = 𝑉𝑥 𝑡 = (40 𝑥 𝑐𝑜𝑠30𝑜 ) (5.8314 𝑠) → 𝑹 = 𝟐𝟎𝟐. 𝟎𝟎𝟓𝟔 𝒎 𝑠 d) The velocity the cannonball strike the ground ,Vf 𝑽𝒇 = √𝑽𝟐𝒇𝒙 + 𝑽𝟐𝒇𝒚 𝑚 𝑽𝒇𝒙 = 𝑽𝒊𝒙 = 40 𝑥 𝑐𝑜𝑠30𝑜 = 𝟑𝟒. 𝟔𝟒𝟏 𝐦/𝐬 𝑠 𝑚 To solve for 𝑽𝒇𝒚 use equation (5): 𝑽𝟐𝒇𝒚 = 𝑽𝟐𝒊𝒚 + 𝟐𝒈𝒀 = (40 𝑠 𝑥 sin 300 )2 + 𝑚 2 (−9.8 𝑠2 ) (−50 𝑚) 𝑽𝒇𝒚 = −𝟑𝟕. 𝟏𝟒𝟖𝟒 𝒎/𝒔 ↓ 2 2 𝑚 𝑉𝑓 = √𝑉𝑓𝑥 + 𝑉𝑓𝑦 = √(34.641 + (37.1484 𝑚/𝑠)2 → 𝑽𝒇 = 𝟓𝟎. 𝟕𝟗𝟑𝟕 𝒎/𝒔 𝑠)2 4. During a fireworks display, a shell is shot into air at an initial speed of 70 m/s at Angie of 75 above the horizontal. The shell ignites once it reaches its highest point. (a) Calculate the height at which the shell explodes. (b) how much time has passed between the launch and explosion? (c) What is the horizontal displacement of the shell at explosion? EVALUATE 1. In a game war, one team sets a base on a cliff 15m high and 60m away from the opponent’s base. At what velocity must the attack be launched so that the lower base will be hit? The initial launch is at 20 degrees below the horizontal? 2. A baseball is hit at a velocity of 100fps at an angle 20 above the horizontal. Will the ball hit a fence 20ft high and 200ft away from the batter’s plate? How far will the ball land from the fence? 3. A boulder rolls over a 50m high cliff. How fast was the boulder moving if it was found 180 m on the ground? At what time did the boulder hit the ground? At what velocity was it approaching the ground? 4. Graph the trajectory of a particle for every 1 sec, if it is moving at an initial velocity of 25 m/s from a height of 20 m, if (a) it is launched horizontally, (b) it is launched 15 above the horizontal, and (c) it is launched 10 below the horizontal. Mark the highest point it reaches, and its maximum horizontal displacements. 58  MODULE 3: DYNAMICS ENGAGE Enumerate and differentiate the three (3) laws of Newton What happens when a force is subjected to an object? EXPLORE Read Module 3: Dynamics (pp 59 – 64) EXPLAIN KINEMATICS vs DYNAMICS Kinematics is the language for describing motion of an object which includes its position, velocity, and acceleration but there was no consideration of what might influence that motion. Dynamics studies the causes of motion. The two main factors we need to consider to answer question about why the motion of an object will change are the forcesacting on an object and the mass of the object. We begin our study of dynamics by discussing the three basic laws of motion, which deal with forces and masses and were formulated by Isaac Newton. LAWS OF MOTION o Physical laws that are the foundation of classical mechanics o Describes the relation between massive bodies and the interaction of forces o According to Sir Isaac Newton, there are three laws governing motion 1STLAW: LAW OF INERTIA ▪ “Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied” When the vector sum or resultant of the forces acting on a body is equal to zero, the body will remain at rest if it is initially at rest or remains in uniform motion along a straight line if it is initially in motion 59  PARTICLE EQUILIBRIUM ▪ If the acceleration of an object that can be modelled as a particle is zero, the object is said to be in equilibrium. A body in equilibrium is a body that is either at rest or a body that is moving at uniform velocity. Mathematically, the net force acting on the object is zero. ∑𝐅 = 𝟎 That is 𝜮𝑭𝒙 = 𝟎 𝒂𝒏𝒅 𝜮𝑭𝒚 = 𝟎 Inertia and Mass Inertia - The tendency of an object to resist any attempt to change its velocity Mass - is that property of an object that specifies how much resistance an object exhibits to changes in its velocity, a measure of the inertia of an object. 2NDLAW: LAW OF FORCE AND ACCELERATION o “The force exerted on an object is directly dependent on its mass and is inversely dependent on its acceleration” o Any object under the second law has an unbalanced forced which result to an acceleration o The direction of motion is the same as the direction of the force Mathematically represented as : ∑𝐅 𝑭𝒏𝒆𝒕 = 𝐚 𝒐𝒓 𝒂 = 𝐦 𝒎 If an object that can be modeled as a particle experiences an acceleration, there must be a nonzero net force acting on it. 𝜮𝑭 = 𝒎𝒂 3RDLAW: LAW OF ACTION AND REACTION “For every action there is an equal and opposite reaction” Between any two bodies (A and B) exerting force on each other, the force exerted by one body (body A) on the other body (body B) is always of the same magnitude but in the opposite direction to the force exerted by body B on body A. 𝑭𝑨𝑩 = − 𝑭𝑩𝑨 60 BASIC CONCEPTS: 1. FORCES, F o Any interaction between objects, which when unopposed will result to a change in motion o Defined as either a push or a pull exerted on a body o Expressed as Newton (N) in the metric system TYPES OF FORCES a. Contact Forces – interactions between objects that involves physical contact e.g. applied forces, frictional forces Figure 3.1When a coiled spring is pulled, as in Figure a, the spring stretches. When a stationary cart is pulled, as in Figure b, the cart moves. When a football is kicked, as in Figure c, it is both deformed and set in motion. These situations are all examples of a class of forces called contact forces. b. Non-Contact Forces – also called field forces; attracts or repels from a distance or through empty space; no physical contact is required e.g. magnetic force, gravitational force Figure 3.2 The gravitational force of attraction between two objects with mass, illustrated in Figure d, the electric force that one electric charge exerts on another, shown in Figure e (such as the attractive electric force between an electron and a proton that form a hydrogen atom), and the force a bar magnet exerts on a piece of iron (Figure f) are some examples of field forces. 2. Mass, m and WEIGHT, W Mass o amount of matter contained in a body o A scalar quantity o Remains constant wherever it is 61  Weight o Or the force due to gravity; the gravitational force exerted on the body by the earth (if the body is on or near the surface of the earth) o Concentrated weight is considered as acting on the center of gravity of a body and always directed (vertically)downward; o Its value varies based on its location o simply defined as the product of the mass of the object and the gravitational acceleration (where the object is placed) 𝐖 = 𝐦𝐠 o Always directed downwards (no matter what its position or location is *NOTE: the weight of a man standing is directed downward, same when the man is sitting or lying down or face down. Figure 3.3 direction of weight is downward no matter what the orientation of the object is The mass of a person is the same whether he is on earth or on mars or on the moon, but his weight would vary depending on his location, that’s why a person is not fat/heavy, he is just in the wrong planet! ☺ Mass of man = m weight on earth = 9.81m weight on moon = 1.61m Weight of man =mg W moon= 120.75 N If m = 75 kg W earth=735.75 N Figure 3.4 illustration of mass and weight on different ‘g’ 3. TENSION, T o A pulling/ stretching force acting from opposite ends applied axially through the strings, cables, rope, etc 62 4. COMPRESSION, C o Balanced inward force on a material or structure o Occurs when a physical force presses inward on an object causing it to become compacted 5. NORMAL FORCE, Nf and FRICTIONAL FORCE, f Whenever two surfaces in contact move or tend to move past each other, the two surfaces will always exert two forces on each other FRICTIONAL FORCE, f o The interaction between surfaces; a force resisting the relative motion between surfaces; always present whenever a body is in contact with a surface (whether the body is moving or not) o Always opposite the direction of motion or impending motion and is always parallel to the surface of contact NORMAL FORCE o Support force exerted upon an object that is in contact with another object o Always perpendicular and directed towards the surface of contact *Note: for some books, Normal Force is denoted as N, but for this module Nf will be used so as not to be confused with the unit N (Newton) 2 Types of Friction (based on the motional relation between surfaces in contact) a. Static friction, fs ▪ Acts when motion just impending or when there is no relative motion between the surfaces in contact b. Kinetic friction, fk ▪ Acts when one or both the surfaces are in motion ▪ Usually lesser compared to static friction 6. COEFFICIENT OF FRICTION, o The ratio between the force required to move one surface over another and the pressure between surfaces o Depends on the surfaces in contact and are nearly independent of the area of contact 𝐟 𝛍= 𝐍𝐟 a. Coefficient of Static friction, s b. Coefficient of Kinetic friction, k o The ration between the static o The ration between the kinetic friction and normal force friction and normal force 𝐟𝐬 𝐟𝐤 𝛍= 𝛍= 𝐍𝐟 𝐍𝐟 63 7. FREE BODY DIAGRAM, FBD A diagram representing all forces (magnitude and direction) acting on a body in question isolated from the other parts of the system Its construction is an integral part of the analysis of systems using Newton’s laws of motion ** STEPS in illustrating an FBD: choose one object and draw an arrow to represent each force acting on it. Include every force acting on that object. If the problem involves more than one object, a separate free-body diagram is needed for each object. We consider for now, the likely forces that could be acting are gravity and contact forces (one object pushing or pulling another, normal force, friction). NOTE: use notations used in this module unless your instructor/facilitator uses otherwise ELABORATE FBD: 1. Given the system below, draw the FBD of the blocks A A and B, if block B will move downward and the coefficient of friction between block A and the incline µ is 0.3. B System Diagram Solution: FBD of block A FBD of block B T V T V f Nf WB WA Problem Solving Strategy 1. Draw a sketch of the situation. 2. Consider only one object (at a time) and draw a free-body diagram for that object, showing all the forces acting on that object. Include any unknown forces that you have to solve for. If several objects are involved, draw a free-body diagram for each object separately. 3. The Laws of Motion involve vectors, and it is usually important to resolve vectors into components. Choose x and y axes in a way that simplifies the calculation. 4. For each object, apply Laws of Motion to the x and y components separately. 5. Solve the equation or equations for the unknown(s). 64 NEWTON’S FIRST LAW: 1. For the system to be in equilibrium, determine the tension in each cord. 45° C 60° B A 80 kg Solution: Draw the FBD of the 80kg suspended object To solve for the tension in cord A, TA, apply the equation ∑ 𝐹𝑦 = 0 TA ∑ 𝐹𝑦 = 0 ❖ Sign Convention of Forces: 80kg – 𝑇𝐴 − 𝑊 = 0 𝑚 𝑇𝐴 − (80𝑘𝑔) (9.8 )=0 + + W 𝑠2 𝑇𝐴 = 784 𝑁 – To solve for the tension in cord B and C, draw FBD at the intersection of the 3 cords. Apply 𝜮𝑭𝑿 = 𝟎 Tc 𝑇𝐶𝑥 − 𝑇𝐵𝑥 = 0 45° 𝑇𝐶 cos 45𝑜 − 𝑇𝐵 cos 30𝑜 = 0 → 𝐸𝑞1 30° TB Apply 𝜮𝑭𝒚 = 𝟎 𝑇𝐶𝑦 − 𝑇𝐵𝑦 − 𝑇𝐴 = 0 TA 𝑇𝐶 sin 45𝑜 − 𝑇𝐵 𝑠𝑖𝑛30𝑜 − 784 𝑁 = 0 → 𝐸𝑞2 Solving the two equations simultaneously, the values obtained are: 𝐓𝐁 = 𝟐𝟏𝟒𝟏. 𝟗𝟐𝟕𝟖 𝐍 𝐓𝐂 = 𝟐𝟔𝟐𝟑. 𝟑𝟏𝟓𝟏 65 2. A man pushes a block of mass 25 kg so that it slides at constant velocity along a level floor. Calculate the magnitude of the force if the coefficient of kinetic friction between the block and floor is 0.20. V=k Draw FBD of the block: F f W Nf μk=0.20 To solve for the magnitude of force F To solve for N, apply the equation 𝚺𝐅𝐲 = applied by the man on the block use the 𝟎 equation 𝚺𝐅𝐗 = 𝟎 since force F is 𝚺𝐅𝐲 = 𝟎 horizontal. N−W=0 𝚺𝐅𝐗 = 𝟎 m N = W = 25 kg x 9.8 2 = 245 N 𝐅−𝐟=𝟎 s 𝐅 = 𝐟 = 𝛍𝐍 Therefore: 𝐅 = 𝐟 = 𝛍𝐍 = 𝟎. 𝟐𝟎 𝐱 𝟐𝟒𝟓 𝐍 = 𝟒𝟗 𝐍 3. A man pushes a block of mass 20 kg so that it slides at constant velocity up a ramp that is inclined at 11o. Calculate the magnitude of the force parallel to the incline applied by the man if a) the incline is frictionless; b) the coefficient of kinetic friction between the block and incline is 0.25. Solution: a) Magnitude of force F parallel to the incline applied by the man if the incline is frictionless Draw the FBD of the block. Y For a body placed on an inclined plane, the incline itself is the X-axis and the line perpendicular to it is X the Y-axis. To solve for magnitude of force F parallel to the F incline applied by the man use 𝚺𝐅𝐗 = 𝟎because 11o force F is directed along the X-axis. Nf 𝚺𝐅𝐗 = 𝟎 W F − Wsin11o = 0 m F = Wsin11o = (20 kg x 9.8 2 ) x sin11o s 𝐅 = 𝟑𝟕. 𝟑𝟗𝟖𝟔 𝐍 66 b) Magnitude of force F parallel to the incline applied by the man if the coefficient of kinetic friction between the block and incline is 0.25. Draw FBD of the block Y f X F 11o Nf W To solve for magnitude of force F parallel to the To solve for N, apply the equation 𝚺𝐅𝐲 = 𝟎 incline applied by the man use 𝚺𝐅𝐗 = 𝚺𝐅𝐲 = 𝟎 𝟎because force F is directed along the X-axis. N − Wcos11o = 0 𝚺𝐅𝐗 = 𝟎 m o