PHYSICS SEM1 PDF

8/12/2024 2. STATICS August 12, 2024 Static If a Force is applied to a body it will cause that body to move in the direction of the applied force, a force has both magnitude (size) and d...

8/12/2024 2. STATICS August 12, 2024 Static If a Force is applied to a body it will cause that body to move in the direction of the applied force, a force has both magnitude (size) and direction. Some forces require contact between the two objects: - e.g. the force of friction between car tires and the road as the car corners. Some forces do not require contact: - e.g. the force between two magnets. Statics is used to describe study of bodies at rest when forces are balanced. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 2 1  8/12/2024 LEARNING OUTCOMES On completion of this topic you should be able to: Describe about statics. 1. Forces, moments and couples, representation as vectors. 2. Centre of gravity. 3. Elements of theories. 4. Nature of properties. 5. Pressure and buoyancy in liquids (Barometer). Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 3 2.1 FORCES Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 4 2  8/12/2024 Force  Force – anything that tends to cause motion, change of motion, stop motion or prevent motion.  Work is the product of a force applied to an object times the distance the object moves.  Force has a unit of Newtons (N). kams  One Newton is defined as the force which gives a mass of 1 kg an acceleration (or deceleration) of 1 m/s2, i.e. 1 N = 1 kg m/s2. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 5 free body diagram 9 = 0 Vconst , > - R ↑ Forces FapphosFapp , W  Normally more than eight one force acts on an 0 object. F -(ty Fr) = + - F , eaction F = my-  An object resting on a table is pulled down by its weight W and pushed back upwards by a force R due to the table supporting it. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 6 3  8/12/2024 Newton’s Law of Motions First Law of Motion A body at rest will remain at rest unless given an external force, or a body which is moving will keep on moving unless given an external force. (Inertia) Second Law of Motion A force proportional to the rate of change of its velocity is produced whenever a body ( or mass ) is accelerated. F = ma Third Law of Motion For every action, there is an equal and opposite direction. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 7 2.1 SCALAR & VECTOR QUANTITY (Cont.) Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 8 4  8/12/2024 direction SCALAR QUANTITY VECTOR QUANTITY Quantity (by a single number) Quantity (by a number / magnitude and a direction) Number with units (+ve, -ve, 0) Magnitude of vector: |F| = F always +ve Example: length, time, Example: Force, momentum, temperature, mass, density, velocity, displacement, volume acceleration Acceptable symbol for vector is F Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 9 Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 10 5  8/12/2024 Vector Addition If a particles undergoes a displacement A, followed by a second displacement B. The final result is the same as if the particle had started at the same initial point and undergone a single displacement C. We call the displacement C as Vector Sum or Resultant. B A C Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 11 F2 = 20N Vector Forencement (sesaran Eon De > EF F Fnet resultant = = - - Two or more forces may act upon the same point so producing a resultant force. If the forces act in the same straight line the resultant is found by simple subtraction or addition. If the forces are do not act in a straight line then they can be added together using the ‘parallelogram law’. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 12 6  Fy N destroy/close I = cos 8/12/2024 For sin = open · - 200s 30 2 > &X Fx = Fre + Fy" = F2 20 sin 30 Ex Fuse = Fy = E2 (F cos e)" + (Fsine)" F : , Fy = F sin 8 #" Cost + Fr sin'o = F2 Example F (cos'0 + Sino) = " F2 (1) : E2 A UniKL Miat student walks 12 km east one day and 5km east # the following day. Find the resultant vector for the journey of the student? First day 12 km Second day 5 km Fu = 40 cs g 40 sinzo 17 km to the east Fy = 681N = 13. A UniKL Miat student walks 12km east one day and 5km west p - the next day. Find the resultant vector for the journey of the student? PO east, Ex 25 cos 7 km to the = = = 19. 15) N < -25 sin 40 Fy = 13 F2 30 : = - 16 0 TN. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak ↓ EQUILIBRANT Force FinFFsinsonS 1. A single force that can hold the original system of forces in soh equilibrium is known as the EQUILIBRANT. call : toa 2. It is equal in magnitude to the resultant but it is opposite in tano sense. B A = A C B Equilibrant Force R= 2. 324 + 35.98 48 366N =. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 14 tane- 8 : 48. 070 7 - 8/12/2024 p Vectors in 2 Dimension form. (axis – x and axis – y) A vector in two dimensions may be resolved into two component vectors acting along any two mutually perpendicular directions. +y A = Ax + Ay Ax = Acos θ Ay = Asin θ Ay A Magnitude, |A| = √(Ax2 + Ay2) θ Direction, tan θ = Ay / Ax +x Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 15 Component vector along x and y axis depend on the angle, θ Bx – Negative Ax – Positive By - Positive B A Ay - Positive Cx – Negative Dx – Positive C D Cy - Negative Dy - Negative Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 16 8  & 8/12/2024 · C O & Obtain the Resultant Force of the Following Vectors? E |B| = 180 N / lis S θ = 25o |A| = 150 N I in 180 - #Y θ= 20o vectors S 7 Fr =O FAST Fx = 150 cus 20 140 95. F2 = y Fy = 150 sin 20 76 : 07 51 30 Ex Ey 180 - = = cos 25. = = 163. # ↓ Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 17 tane [ - - 51 = 15. 52 + 180 = 195. 5 from Obtain the Resultant Force of the Following Vectors? #x = 60cos35 X axis |A| = 60 N 80 cos 30 = 49 15 Fx - - = θ = 35o Fy = 60sin35 69 28 to - =. θ = 30o 34 41 =. 6908 - x = + 49 15 - Fy = 80 sinso |B| = 80 N = - 20-13 - 48 + 34 4) = 40. Fy = + = - 5 - 59 18 Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak =41) : 04 74 % un + In 132) 15 59. Fr +. 20. S = 71. 64N I - 405. 22 - 31 24. 9 = 20. 89N  8/12/2024 Obtain the Resultant Force of the Following Vectors? |C| = 60 N |A| = 160 N rectors X Y θ = 40o θ= 35o ~ - θ= 30o 160 10535 160 sin 35 Fi = 131. 06 = 91 77. so cos 40 ↑ 60 sin 40 |B| = 80 N F2 46 + 38 6 - = 1 - 80 CUS 38 80 sin3u + 1 - - = 69 28 40 =. = - Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 19 - : 91 74. · & Obtain the Resultant Force of the Following Vectors? Fy IION = Fx = 0 Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 20 10  8/12/2024 E brian Obtain the Resultant Force of the Following Vectors? sin sbb bukak fi A plane flies from base camp to lake A, a distance of 280 km at a vectors X direction of 200 north of east. After dropping off supplies, the plane flies to lake B B, which is 190 km and 300 west of north from lake A. DI 280c0s28 280 sin28 Graphically determine the distance and direction from lake B to. E 1 base camp. 19)) 263. = 95 77 C. = D2 - 1goiso 19000s - 30 - - S I - BBC" 1902 + 280" (199(280) (sin80] 20 = - + - T · _ =......... BC BBC = 249.22 km oset # -ulab) Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 21 up - 2.1 MOMENTS AND COUPLES (Cont.) Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 22 11  = Travis F49 8/12/2024 rotational E X-axis ↓ F =ma -Iran/s Ir Moment of a Force  A force can also be used to produce rotation, as = pm occurs when opening a door or tightening a nut with rotation a spanner. · - distance  This turning effect of the force is known as “the at matter ! moment of the force”. #  It depends on the magnitude of the force and a 9 5. = 0. distance called the lever arm. This is the · perpendicular distance from the force to the axis of rotation. · ↑ 23 F Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak F29 < Moment corace t = IL T Moment (Nm) = Magnitude of the force (N) x Perpendicular distance (d) Applying the force in such a way that its line of action passes through the pivot will not produce a turning effect. In SI units, Newton metres = Newton x metres Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 24 12  8/12/2024 &countertwise Scockwise Line of action of the Applied force force Pivot Pivot Line of action of the force Applied force Pivot If the force causes the lever to move in a clockwise direction, the moment is said to be a clockwise moment, and vice versa. If the force is inclined, the turning effect is reduced i.e. moment is reduced because the perpendicular distance is reduced. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 25 Example clockwise In the diagram above a force of 5 N is applied at a distance of 3 m from the fulcrum, therefore: Moment =5Nx3m = 15 N m Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 26 13  8/12/2024 Moments and Equilibrium  Equilibrium is where all the forces and all the moments acting on a body cancel each other and the net effect on the body is zero.  In other words it will not move if it is in a state of rest, and if in motion it will not slow down or accelerate or change direction. P1 P2 S1 S2 Anticlockwise Clockwise tendency tendency  The product P2 x S2 produces a clockwise moment about the pivot and the product P1 x S1 produces an anti-clockwise moment about the pivot.  For equilibrium of rotation, these two moments must be equal. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 27 Example of One Unknown Force 1. In this case the requirement is to balance the arrangement in figure below by determining the unknown force (P). reaction N force ·. In j re Tnet O = 2(2) + 5(4) - 1(2) - P(b) 0 = Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 28 Sp = 22 p = 3. 7N Tne + = 0 reaction = 5 + 2 + 1 + 3. 7 Tc = Tcn force N = 11. 7 14  Static 8/12/2024 / ↓ Inet = O Fnet = O =0 2 =0 9 Answer Taking moments about the pivot for equilibrium of rotation: Sum of clockwise moments = Sum of anti-clockwise moments (P X 6m) + (1N X 2m) = (2N X 2m) + (5N X 4m) 6P + 2 = 4 + 20 6P = 24 – 2 6P = 22 P = 22/6 = 3 2/3 N Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 29 Example of Pivot Location 2. A uniform bar AB in figure below 7m long has forces of; 25N at a point 0.5m from A, 12N at a point 3m from A, and 12N at a point 1m from B applied to it. Find the position of the pivot which will allow the beam to balance, i.e. be in the state of equilibrium. Inet O se = 0 choose A as a ref point, = 0 (2)(6). (49)(x) - - (257(0 5) · - (12)(3) + & 49X X = = 12 0 2.. 46 5 Jaw few w 1- - Tnet = O They = O choose B as a ref point, 30 Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 12 (1) - 49(y) + 12(4) + 25(65) : 0 222 5 49y. = y = 4 54. 15 = 7 - 4 - 54 = 2. 46 x  8/12/2024 Example of Mass & Forces Pivot Location 3. A uniform beam AB, 4m long and 200N weight has forces of 125N and 20N applied respectively to its ends A and B. Find the point about which the beam will balance. I the ↑order Cur - ↳ cn the A as ref point, B as ref point, 32 Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 125(4) : 0 3 40(k) - 200(2) - 20(4 = 0 200(2) - > 40(4-u) + = x 1. 4m 4 - x = 2. 64 u = 1. 4 m Principle of Moments ‘When a body is in equilibrium under the action of a number of forces, the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point.’ a) Type 1 – Beam balances where arms are of equal length. b) Type 2 – Lever arrangement can best be seen in design of a wheelbarrow. c) Type 3 – Large effort moves through small distance to overcome small load, which moves through a large distance. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 34 16  sin = Son 8/12/2024 cos = can Tnet = Ta + [b 2 + Fada + Fi db : + Fidn CW : -Ve = I 3(866) - 10(34 64). + 5(20) : + Ve CCW Exercise = 13. 8 Nmm , CW out  Calculate the resultant moment of a pivot acting on a bell crank lever, refer to diagram below. cus 30 = 0 · = AO = 100 mm OC = 20 mm so d = 36. 6 mm BC = 20 mm Mo 86 60 =. mm te From cos30 : Ve f + 34 64mm ? d =. -ve assume static Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 35 O Tnef = 3(86 6) · - x(34 6. + 5(20) = 0 x = 10 · 4 N Couple  A special case of moments is a couple. A couple consists of two parallel forces that are equal in magnitude, opposite in sense and do not share a line of action.  It does not produce any translation, only rotation. The resultant force of a couple is zero. BUT, the resultant of a couple is not zero; it is a pure moment. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 37 17  8/12/2024 Whet 0: takebergere - Example  In some situations, for example the winding up of a clockwork mechanism the forces that are applied to the winding key are equal in magnitude but opposite in sense. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 38  In this case the resultant force on the pivot is zero and there is only pure rotation present with no tendency for the pivot to move sideways. The value of the resultant moment ( P x d ) produces rotation.  Such arrangement of forces is called a ‘COUPLE’ and the resultant moment of a couple is called a TORQUE. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 39 18  8/12/2024 X100 m - 2 m Example A nut is to be torque loaded to a maximum of 100 Nm. What is the maximum force that may be applied, perpendicular to the end of the spanner, if the spanner is of length S 30 cm? 100 = F (0 3). F = 333. 33N Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 40 2.2 CENTRE OF GRAVITY Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 41 19  8/12/2024 F = Fattmc Centre of Gravity  Gravity is a force which is always present and is a pulling force in the direction of the center of the earth.  The centre of gravity is the force acts on every body through an imaginary point. A point where all the weight of a body appears to be concentrated. (total weight can be considered to act through that datum position )  In flight, both airplanes and rockets rotate about their centre of gravity. Determining the centre of gravity is very important for any flying object. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 42 Example of Centroid Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 43 20  8/12/2024 Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 44 Example of Centre of Gravity for some Regular Shaped Object Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 45 21  8/12/2024 Stability / Balancing The lower the C of G, the stable an object is. The wider the base, the more stable an object is – C of G towards the base. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 46 The point of O is the C of G of the rod Only at the particular point O, the rod can stay in a horizontal position. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 47 22  8/12/2024  When force applied to C of G, the body will not rotate.  But if the force is applied offset of the C of G, the body will rotate, or torque will produced. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 48 i Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 49 · invest 23 centera  8/12/2024  Similar to aircraft, force applied will be acted through the C of G, resulting in torque.  The aircraft rotate about its C of G. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 50 The Importance of C of G  To ensure the aircraft is safe to fly, the center-of-gravity must fall within specified limits established by the manufacturer.  To ensure the C of G range – C of G limits are specified longitudinal (forward and aft) and/or lateral (left and right) limits within which the aircraft's center of gravity must be located during flight.  To evenly load the aircraft – equipments, passengers, baggage, cargo, fuel, etc.  So that C of G range will not be exceeded – prevent aircraft unstable during flight.  Also affects C of G in flight – fuel usage, passengers’ movement, etc. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 51 24  8/12/2024 2.3 ELEMENTS OF THEORIES Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 52 pressure Stress  If force is exerted on a body, there will be mechanical pressure acting on the body which is called the stress.  A body with having twice the size of other body subjected to a force, it will be stronger and less likely to fail due to applied the applied force.  So, stress is said : Stress = or  = *units : Newton metre -2 , Nm-2  Components will fail due to over-stressed, not over-loaded. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 53 25  8/12/2024 Example Eg. A tennis ball sealed from atmospheric pressure. So, as long as the external forces acting on it does not exceed the internal forces, the ball will maintain its shape. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 54 ASTM Forces applied to the body will cause distortion of the body and change to the material’s cross-sectional area ; F FMGX eg. Tensile Forces will cause elongation. H Compressive Force will cause reduction in dimension. - Most material have elastic properties ( it will to return to its original shape after the force is removed ) - provided forces does not exceed limit of elasticity. Gress There are 5 types of stress in mechanical bodies : i. Tension 1 ii. Compression iii. Torsion iv. Bending v. Shear + = wx areaPrepared By: Wan Nur Shaqella Bte Wan Abdul Razak 55 > - elongation > strain = ov original length 26  8/12/2024 - very high for aircraft material Tensile  The force that tends to pull an object apart  Flexible steel cable used in aircraft control systems is an example of a component that is in designed to withstand tension loads. Compression  The resistance to an external force that tries to push an object together.  Example: aircraft rivets. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 56 Torsion  Torsional stress is applied to a material when it is twisted.  Torsion is actually a combination of both tension and compression  Example: an engine crankshaft. Bending  In flight, the force of lift tries to bend an aircraft's wing upward. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 57 27  8/12/2024 Shear  Combination of tension and compression is the shear stress, which tries to slide an object apart.  Shear stress exists in a clevis bolt when it is used to connect a cable to a stationary part of a structure.  A fork fitting, such as drawn below, is fastened onto one end of the cable, and an eye is fastened to the structure. The fork and eye are held together by a clevis bolt.  When the cable is pulled there is a shearing action that tries to slide the bolt apart. This is a special form of tensile stress inside the bolt caused by the fork pulling in one direction and the eye pulling in the other. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 58 Strain  Stress is a force inside the object caused by an external force.  If the outside force is great enough to cause the object to change its shape or size, the object is not only under stress, but is also strained.  If a length of elastic is pulled, it stretches. If the pull is increases, it stretches more; if the pull is reduced, it contracts. Hooke’s law states that the amount of stretch (elongation) is proportional to the applied force. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 59 28  8/12/2024  The graph above shows how stress varies with strain when a steel wire is stretched until it breaks.  Strain can be defined as the degree of distortion then has to be the actual distortion divided by the original length (in other words, elongation per unit length). Strain, ε = change in dimension / original dimension (No units). Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 60 exam ? Example it e as percentage Tensile strain If a cable of 10 m length is loaded with a 100 kg weight so that it is stretched to 11 m, what is theC strain placed on the percentt cable? = Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 61 29  8/12/2024 Example Compressive strain A 25 cm rod is subjected to a compressive load so that its length changes by 5 mm. How much strain is the rod under when loaded? Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 62 Shear strain Torsion strain  When the applied load causes  Form of shear stress resulting one 'layer' of material to from a twisting action. move relative to the adjacent layers.  If a torque, or twisting action is applied to the bar shown, one end will twist, or deflect relative to the other end.  Twist will be proportional to the applied torque. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 63 30  8/12/2024 modulus = stiffness = gradient of the linear part Young Modulus Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 64 Exercise X = 0 : 11 F = 30kN A rectangular steel bar 15 mm x 10 mm x 300 mm long extends by 0.11 mm under a tensile force of 30 kN. Find the: a) Stress b) Strain For i c) Elastic modulus 300 0000 > 200 Nmn 9) stress 12010 - 4 = = > Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 65 xo == 54xb) strain = 31  8/12/2024 ↓ physicnica 2.4 NATURE OF PROPERTIES Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 66 rubber glass convectt -higher - lower stiffness stiffness E - high elasticity = e voo ~  Strength Properties of Solid – A strong material requires a strong force to break it. The strength of some materials depends on how the force is applied. – For example, concrete is strong when compressed but weak when stretched, i.e. in tension.  Stiffness – A stiff material resists forces which try to change its shape or size. It is not flexible.  Elasticity – When the force distorting a substance is removed, and that substance has a strong tendency to return to its original shape, it is said to be elastic. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 67 32  8/12/2024 the under area graph ↑ Properties of Solid  Toughness – This is the ability of a substance to resist G breakage when deforming or impact forces are applied to it. Hard substances are usually tough, many softer substances are tough e.g. hammer heads. mid  Hardness – A hard substance has a high resistance to indentation, or to any action tending to penetrate itsG surface. In other words, hardness is the ability of a material to resist scratching, indentation or penetration. The harder a material the more difficult it is to scratch it, dent it or cut it.  Brittleness ↑ ductility – Brittle substances break with little or no change of shape. In most applications, especially where sudden impact-type forces are applied, brittleness is undesirable. At room temperature and below, glass, cast iron, and very hard steel are example of brittle materials. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 68 - Properties of Solid  Malleability – Malleable materials can be beaten, rolled, or pressed into shape without fracture e.g. red hot steel. Malleable metal can be shaped into a design by hitting it. It could also lose that design easily by being hit against countertops, cash register drawers, and other hard surfaces.  Ductility – Ductile materials can be stretched into new shapes without pulling them apart, and keep their new shape after stretching force is removed.  Plasticity - plastic cathra – Plasticity is the ability of a material to have its shape permanently - changed without fracturing by stretching, squashing or twisting. In other words, plasticity is described as a material that does not spring back to its original shape when the load is removed. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 69 33  8/12/2024 liquid tan gas ↑ -kelike Properties of Fluid  Viscosity – As the molecules of a liquid move about due to thermal energy, the milk attractive forces between them try to slow the motion down. The · crae e stronger the forces are the more impediments there is to flow. Such resistance to the flow of a liquid is called viscosity. Viscosity is defined, Coney as the amount of force one layer of liquid of unit area will exert on an -geternaladjacent layer. friction -his  Surface Tension – The molecules of a liquid within the body of the liquid are subjected to forces from all directions. The molecules at the surface are subjected to attractive forces from within and to the sides. – However there are no forces from the outer side of the surface to balance the others. This places the surface molecules under a kind of tension. This surface tension tends to cause the surface molecules to move together and make the surface area as small as possible. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 70 ⑳ Surface Tension This suggests that the surface of a liquid behaves as if it is covered with an elastic skin that is trying to shrink. The surface tension can be reduced if the liquid is ‘contaminated’, adding a detergent to the water will cause our needle to sink. In a liquid, the molecules still partially bond together and prevents liquid from spreading nag expanding out. ↓ ↓ ↓ ↓ Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 71 34  8/12/2024 Example of Surface Tension Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 72 2.5 PRESSURE AND BUOYANCY IN LIQUIDS (BAROMETERS) Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 73 35  8/12/2024 Pressure  The equivalent term associated with fluids is pressure: Pressure (P) = Force (F)/ Area (A)  Pressure is the internal reaction or resistance to that external force.  SI system for pressure is 1 Pa = 1 N/m2 Pascal’s Law : “Pressure acts equally and in all directions throughout that fluid.” Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 74 Atmospheric Pressure The atmosphere is the whole mass of air surrounding the earth. The surface of the earth is at the bottom of an atmospheric sea. The standard atmospheric pressure is measured in various units: 𝟏 𝒂𝒕𝒎𝒐𝒔𝒑𝒉𝒆𝒓𝒆 = 𝟕𝟔𝟎𝒎𝒎𝑯𝒈 = 𝟐𝟗. 𝟗𝟐𝒊𝒏𝑯𝒈 = 𝟏𝟒. 𝟕𝒍𝒃/𝒊𝒏𝟐 = 𝟏𝟎𝟏. 𝟑𝒌𝑷𝒂 Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 75 36  8/12/2024 Measurement of Atmospheric Pressure Atmospheric pressure is typically measured in inches of mercury (in.Hg.) by a mercurial barometer. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 76 Barometer Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 77 37  8/12/2024 Gauge Pressure  Gauge Pressure is the reading taken directly from the gauge devices  It is a pressure relative to the ambient pressure.  Gauge pressure is used to measure engine oil pressure, hydraulic pressure and other operational pressures built up by pumps.  This is because atmospheric pressure acts on the fluid as it enters and as it leaves the pump – only the pressure above atmospheric is of interest. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 78 Absolute Pressure Absolute Pressure is the sum of the available atmospheric pressure and the gauge pressure. Absolute Pressure (PSIA) = Gauge Pressure + Atmospheric Pressure Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 79 38  8/12/2024 Calculation Example : Given (Gauge Pressure) = 150 psig (Atmospheric Pressure) = 14.7psi Absolute Pressure = 150 psig + 14.7 psi = 164.7 PSIA 12 August, 2024 Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 80 Buoyancy Archimedes’ Principle states that when an object is submerged in a liquid, the object displaces a volume of liquid equal to its volume and is supported by a force equal to the weight of the liquid displaced. THE BUOYANCY OF A SUBMERGED BODY = WEIGHT OF DISPLACED LIQUID – WEIGHT OF THE BODY 1. The body will float--if the buoyancy is positive 2.The body will sink--if the buoyancy is negative 3.The body will be stuck--if the buoyancy is neutral Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 81 39  8/12/2024 Buoyancy Archimedes Principle  When an object is submerged in a liquid, the object displaced a volume of liquid equal to its volume and is supported by a force equal to the weight of the liquid it displaced.  The buoyant force of an object which is submerged in a fluid is equal to the weight of the fluid displaced by the object.  A net upward vertical force results because pressure increases with depth and the pressure forces acting from below are larger than the pressure forces acting from above. Buoyant Force, FB  gV Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 82 Archimedes Principle “Any object completely or partially submerged in a fluid experiences an upward force equal in magnitude to the weight of the fluid displaced by the object.” Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 83 40  8/12/2024 THIS EXPLAINS WHY BIG NAVAL SHIP CAN FLOAT !!!!!! A steel ship can encompass a great deal of empty space and so have a large volume and a relatively small density. Weight of ship = weight of water displaced Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 84 41  moron 3. KINETICS + motion Force I I dynamic- static add ma F = mation ↓ takee motion August 26, 2024  LEARNING OUTCOMES On completion of this topic you should be able to: Describe about kinetics. 1. Linear movement. 2. Rotational movement. 3. Periodic motion. 4. Simple theory of vibration, harmonics and resonance. 5. Velocity ratio, mechanical advantage and efficiency. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 3.1 LINEAR MOVEMENT Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak 26 August 2024  & Newton’s Law of Motion As we are considering motion, it is convenient to state some important fundamental laws of Newton’s law of motion. pinersia The first law states “Unless there is a resultant external force /rest acting upon it, a body will move with a constant speed in a straight line”. ma - F = The second law of motion states that “The rate of change of motion of a body is proportional to the resultant force” acting on the body and takes place in the direction of the force”. F MG = Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  first law Inertia Force ❖ All bodies seek a state of equilibrium and are reluctant to change their present state of rest or uniform motion. ❖ This reluctance to change its current state is called INERTIA. O ❖ Inertia is dependant on the mass of the body. The larger the mass the greater the inertia, i.e. the more difficult it is to move when at rest or to stop when in motion ❖ You will probably experience this effect during take-off when you are being push back into your seat. inversia large ↑ Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak Fx)Fo Fre + # 0 Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  - F = ma Newton’s Second Law The acceleration (increase in velocity) of a body is directly proportional to the force applied, and inversely proportional to the mass of the body. (F = m a) A=F/M Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  Finally Newton’s third law of motion states that “The force exerted by one body (A) on another body (B) is equal in magnitude, opposite in direction and acts in the same straight line as the force exerted by B upon A” S In other words “To every action there is an equal and opposite reaction”. 1st + inertia 2nd > - F m9 : 3r -equalstude - opposite direction Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  1 Linear Motion testrrement scaler - distance ❑ Is the uniform motion in a straight line, = distance speed - motion under constant acceleration total time (motion under gravity). dis t ❑ Motion is the change of position of a body velocity = with reference to another body. eg. A person sitting in a moving car and passes a building. ❑ The person is considered to be at a state of rest in reference to the car. ❑ The car is considered to be in motion in relation to the building. ❑ The term "uniform motion" is often used to describe the motion of an object that is travelling at a constant speed or velocity in one direction for a period of time. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  an e ↑ Average speed is defined as the distance traveled by a body along the path of its motion divided by the time taken. Thus if the curved path AB is denoted by ‘s’ and the time taken is ‘t’. distance travelled s Average Speed = = m/s time taken t Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  Example 1 During your trip to UniKL MIAT, you traveled a distance of 5 miles and the trip lasted 0.2 hours (12 minutes). Find the average speed of your car? Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  Example 2 An aircraft travelling from one DME marker to another a distance of 88 km (about 50 miles) in 15 minutes. (DME = Distance Measuring Equipment, a ground based radio navigation aid.) Determine the average G speed for the aircraft? In this example the displacement would be the same as the distance covered and so the magnitude of the average velocity would also be 97.7 m/s. however for it to qualify as a vector quantity the direction (from A to B must also stated. Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  Speed, as a scalar quantity possesses magnitude only, velocity however being a vector quantity possesses both magnitude i and direction. ↓ emer The average velocity is defined in term of displacement, rather than the total distance travelled. In figure 1, although the distance travelled by the body is ‘s’ along the curved path from A to B, the displacement is a straight line from A to B. displacement x Average Velocity = = m/s time taken t Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  Example 3 A helicopter leaving point ‘A’ travels due east, a distance of 18km. It then makes a 90o turn and continues its journey due north to point B, a further 40km. The whole journey from A to B taking 15 minutes. distance Determine: X Time a) The average C speed for the whole journey b) The average velocity N AB direction = = Ttis 43. 86km 4 74m/s S 4 = 48 tan o =. 0 8 = 65.77 Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  SOLUTIONS total distance covered average speed = total time taken 58  10 m 3 average speed = 15  60 s average speed = 64.4 m/s Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  or ha sinc tan= ad cosh = total displacement average velocity = total time taken total displacement AB = 402 + 182 = 1600 + 324 = 43.86km 43.86  103 m average velocity = = 48.7 m/s 15  60 s opposite 40 and since TAN  = = adjacent 18 then  = 65.8o Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  Acceleration Acceleration is a vector which is defined as ‘the rate of change of velocity’ or ‘the change of velocity in unit time. change of velocity acceleration (a) = time if u = initial velocity in m/s and v = final velocity in m/s v-u then acceleration = t Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  Example V An Air Force F-15 fighter cruising at 400 mph. the pilot advances the throttles to full afterburner and V accelerates to 1200 mph in 20 seconds. What is the average acceleration in mph? 400 1200 - = 40mph/g - 28 Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  Rearranging the formula above gives us the formula below; (Equation 1) v = u + at Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  For a body accelerating uniformly, the velocity-time diagram would be as shown below: -a Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  v = u + at v -n + 2as s = Er + ult s : ne + bat" The shaded area is that of a trapezium whose area is calculated from the equation. 1 (Equation 2) s = (v + u) t 2 Substituting equation 1 into equation 2 we obtained; 1 2 (Equation 3) s = ut + at 2 v = u + 2as Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  u = 0 v = 59 - Example 4 u = 0 V A large aircraft has a take-off velocity of 59 m/s (about 132 t mph). It starts from rest and accelerates uniformly for 30 seconds before becoming airborne. v = 59 u = 0 ⑭ & - t = 30s Calculate; a) What is the value of the acceleration in m/s2? b) What take-off distance required? 8 0)(20) a) v = n + at b) s = (59 + 59 = 0 + 9(30) 885m 967 m/gh = a = 7 , Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  SOLUTIONS Prepared By: Wan Nur Shaqella Bte Wan Abdul Razak  O↓ ta terminal velocity I = + a - 8 A body falling freely from a great height will initially be accelerated but will gradually lose this acceleration until it falls with constant velocity. This is known as the body’s terminal velocity, and depends on other things the air resistance. H