1 Hydrostatic Forces on Submerged Plane Surfaces Hydrostatic forces mean forces exerted by fluid at rest. - A plate exposed to a liquid, such as a gate valve in a dam, the wall of a liquid storage tank, or the hull of a ship at rest, is subjected to fluid pressure distributed over its surface. - Hydrostatic forces form a system of parallel forces, - We need to determine the magnitude of the force; - and its point of application, Centre of Pressure Note: When analyzing hydrostatic forces on submerged surfaces; the atmospheric pressure can be subtracted for simplicity when it acts on both sides of the structure. Consider the plane surface shown in the figure below. The total area is made up of many elemental areas. The force on each elemental area is always normal to the surface but, in general, each force is of different magnitude as the pressure usually varies.
2 Forces on Horizontal Submerged surface: For a horizontal plane submerged in a liquid (or a plane experiencing uniform pressure over its surface), the pressure, p, will be equal at all points of the surface. Thus the resultant force will be given by P = γh : Uniform on the entire plane. Resultant force, F R = P. A = γ. A. h (A: the bottom area of container) Forces on Inclined Submerged surface: This plane surface is totally submerged in a liquid of density ρ and inclined at an angle of θ to the horizontal. Taking pressure as zero at the surface and measuring down from the surface, the pressure on an element γa, submerged distance z, is given by P = ρgz And therefore, the force on the element is F = γzδa
3 The resultant force can be found by summing all of these forces: F R = γ zδa Where, zδa known as 1 st Moment of Area which is equal to AZ. zδa = AZ = 1 st Moment of Area about the line of free surface. Where A is the area of the plane and Z is the depth (distance from the free surface) to the centroid, G. Steps to solve the exerted bodies problems: 1- First, we have the main equation to find the resultant forces: - Location of the force: F R = γ. A. h c 3- h c, for the vertical surface is the distance from the surface to the centre, for example if the depth.5, then the hc = h c, for inclined surface, will be h c = distance from the surface of the fluid to the up edge of the submerged surface + length of the surface/ sinθ 5- The Area term in the resultant force equation can be calculated directly, depend on the shape of submerged surface, for example, if the gate is square, the area will be (Width * Length) as so on. 6- Now we find the distance where the force F acts from the hinge: h c y c =, for inclined surface. sinθ y c = h c, for vertical surface. 7- I x,c, can be calculated based on the shape of the submerged body wich can be shown for common bodies:
4 Shape Dimensions Area I x,c, nd moment of area Rectangle bd bd 3 1 Triangular bh bh 3 36 Circle πr πr 4 4 Semi-Circle πr 0.11R 4 Example 1: A swimming pool is 18m long and 7m wide. Determine the magnitude and the location of the resultant force of the water on the vertical end of the pool where the depth is.5m.
5 The force is The location of force: F R = γ. A. h c F R = = 14.6 kn y c = h c, for vertical surface y c =.5 = 1.5 m The second moment of area can be calculated from the table for the rectangular: I x,c = bd3 1 I x,c = 7.53 = m y R = = 1.66 m Example : A square 3m*3m gate is located in 45 sloping side of a dam. Some measurements indicate that the resultant force of the water on the gate is 500 kn. 1- Determine the pressure at the bottom of the gate. - Show where this force acts. 1- The force is F R = γ. A. h c = h c h c = 5.66 m P = γ (h c + 3 sin45) P = 65.9 kn - The location of force:
6 y c = h c sinθ y c = 5.66 sin45 = 8 m The second moment of area can be calculated from the table for the rectangular: I x,c = bd3 1 I x,c = 3 33 = 6.75 m4 1 y R = = m Example 3: the vertical cross section of a 7m long closed storage tank is shown below. The tank contains ethyl alcohol and air pressure is 40 kpa. Determine the magnitude of the resultant fluid force acting on one end of the tank. Break the figure to three parts as shown below: 1- For part 1: - For part : F 1 = γ. A. h c F R = P A F R = 40 = 160 kn
7 F R = P air A + γ ethyl h c A F = ( 4 ) 4 = 444 kn 3- For part 3: F R = P air A + γ ethyl h c A Area of part 3: F 3 = ( 3 4) 1 4 = 43 kn A = 1 bottom height, h c = 3 height Now, the resultant force can be evaluated by: F R = F 1 + F + F 3 = = 847 kn Example 4: the rectangular gate CD shown in the below figure is 1.8m wide and m long. Assuming the material of the gate to be homogeneous and neglecting friction at the hinge C, determine the resultant and the location of this force of the gate necessary to keep it shut until the water level rises to m above the hinge. To start solving this, we need to find the angle first: θ = tan 1 ( 4 3 ) = F R = γ. A. h c F R = ( + sin (53.13) = kn Now, the location can be found by: Where,
8 y c = h c sin53.13 = + sin53.13 = 3.5m sin53.13 And from table the second moment of area can be calculated by: I x,c = bd3 1 y R = = = 1. m = 3.595m Example 5: A 60-cm square gate (a = 0.6) has its top edge 1 m below the water surface. It is on a 45 o angle and its bottom edge is hinged as shown in figure below. What force P is needed to just open the gate? the force can be calculated by: F R = γ. A. h c F R = ( Now we find the distance d where the force F acts from the hinge: Where, y c = h c sin45 = = 17.7m sin45 sin45 And from table the second moment of area can be calculated by: y R = The force P can be calculated: I x,c = bd3 1 = = m = 17.7m d = 0.3 (y R y c ) = 0.3 ( ) = 0.98m sin (45) = kn
9 P = d F R a = 1.04kN Example 6: An inclined rectangular gate (1.5m wide) contains water on one side. Determine the total resultant force acting on the gate and the location. the force can be calculated by: F R = γ. A. h c F R = ( Now we find the distance d where the force F acts from the hinge: Where, y c = h c sin30 = sin30 sin30 = 5.4m And from table the second moment of area can be calculated by: I x,c = bd3 1 y R = = = 0.16 m = 5.4m sin (30) = kn Example 7: An inclined circular with water on one side is shown in the fig. Determine the total resultant force acting on the gate and the location.
10 Area of the circular gate, A = π 4 d A = m the force can be calculated by: F R = γ. A. h c F R = ( sin(60)) = 17. kn Now we find the distance where the force F acts: Where, y c = h c sinθ = sin60 sin60 =.578m And from table the second moment of area can be calculated by: I x,c = πr4 4 y R = = π = 0.049m =.6m Example 8: Gate AB in the figure below is 1m long and 0.7m wide. Calculate force F on the gate and position X of c.p. (specific gravity of oil is 0.81)
11 Area of the rectangular gate, A = length width A = 0.7 m The density of oil can be calculated by: And h c can be evaluated as below: the force can be calculated by: ρ oil = = 810 kg m 3 h c = sin sin50 = 4.15 m F R = γ. A. h c F R = = kn Now we find the distance where the force F acts: Where, y c = h c sinθ = 4.15 sin50 = 5.4m And from table the second moment of area can be calculated by: I x,c = bd3 1 = = m 4 y R = = 5.435m Example 9: The gate in figure below is 5 m wide, is hinged at point B, and rests against a smooth wall at point A. Compute: a) The force on the gate due to the water pressure.
12 b) The distance between the centre of gate and location of force act. c) The reactions at hinge B. To start solving this, we need to find the angle first: Now, the location can be found by: Where, θ = tan 1 ( 6 ) = h c = 15 3 = 1m F R = γ. A. h c F R = = 5886 kn y c = h c sin37 = 1 sin37 = 19.94m And from table the second moment of area can be calculated by: I x,c = bd3 1 = = m 4 1 y R = = 0.4m Note: the distance between the centre of the gate and centre of pressure can be calculated by: d = y R y c = = 0.46m Summing moments counter clockwise about B gives: P Lsinθ = F(5 d)
13 P 10sin37 = 5886(5 0.46) P = kn With F and P known, the reactions Bx and Bz are found by summing forces on the gate. F x = 0 P = Bx + Fsin37 Bx = 898kN Example 10: A tank holding water has a triangular gate, hinged at the top, in one wall. Find the moment at the hinge required to keep this triangular gate closed. The force can be calculated by: h c = = 1.7m A = bh = 1.5 = 1.5 m F R = γ. A. h c F R = = 5 kn Now we find the distance where the force F acts: Where, y c = h c = 1.7m And from table the second moment of area can be calculated by: I x,c = bh3 36 = 1.53 = m 4 36
14 y R = = m The moment on the hinge from the water is: Moment = 5 ( ) = 14.35kN Submerged Curved surface: The easiest way to determine the resultant hydrostatic force FR acting on a twodimensional curved surface is to determine the horizontal and vertical components FH and FV separately. Horizontal force component on curved surface: F H = F x Horizontal force component on curved surface: F V = F y + W where the summation F y + W is a vector addition (i.e., add magnitudes if both act in the same direction and subtract if they act in opposite directions). Thus, we conclude that: 1- The horizontal component of the hydrostatic force acting on a curved surface is equal (in both magnitude and the line of action) to the hydrostatic force acting on the vertical projection of the curved surface. - The vertical component of the hydrostatic force acting on a curved surface is equal to the hydrostatic force acting on the horizontal projection of the curved surface, plus (minus, if acting in the opposite direction) the weight of the fluid block. The magnitude of the resultant hydrostatic force acting on the curved surface is: F R = F H + F V And the tangent of the angle it makes with the horizontal is α = F V F H
15 Example 1: The water is on the right side of the curved surface AB, which is one quarter of a circle of radius 1.3m. The tank s length is.1m. Find the horizontal and vertical component of the hydrostatic acting on the curved surface. Horizontal force, F H = γ. A. h c F H = ( ) = 84.4 kn Vertical force, F V = γ. volume F V = ( π ) = 94.97kN Example : The gate AB shown is hinged at A and is in the form of quarter-circle wall of radius 1m. If the width of the gate is 30m, calculate the horizontal and vertical force. Horizontal force, F H = γ. A. h c F H = ( 1 ) 1 30 = kn Vertical force, F V = γ. volume F V = ( π ) = kN
16 Example 3: Three gates of negligible weight are used to hold back water in the channel of width b as shown in the figure below. The force of the gate against the block for gate (b) is R. determine in term of R, the force against the blocks for the other two gates. For gate (b) F R = γ water h c A F R = 9810 h h b F R = 4905h b Where, y c = h c y c = h And from table the second moment of area can be calculated by: Thus: y R = I x,c = bd3 1 = bh3 1 bh h h b h = h 6 + h = 4h 6 = h 3 M = 0 R = y R F R R = h h b R = 370 h b eq. 1 For case (a): The weight can be calculated by: W = 9810 volume W = 9810 ( π 4 (h ) b) W = 196. h b
17 This lead to: M H = 0 W ( h 4h 6π ) + F R ( h 3 ) = F B h F B = 385.9h b With equation 1: F B = 1.17R For case (c): The force FB on the curve section passes through the hinge and therefore does not contribute to the moment around H. on the bottom part of the gate: F R = γ water h c A F R = h 4 h b F R = h b Where, y c = h c y c = 3h 4 And from table the second moment of area can be calculated by: And with equation 1: y R = I x,c = bd3 1 = bh3 1 bh 3 1 3h + 3h 4 h b 4 = 0.777h M H = 0 F R 0.777h = F B h F B = 0.875R
18 Example 3: A 4m long curved gate is located in the side of a reservoir containing water as shown in figure below. Determine the magnitude of the horizontal and vertical components of the force of the water on the gate. For equilibrium: Based on the figure: Similarly: Now: F x = 0 F H = F F H = γ water h c A = 9810 (6 + 3 ) 3 4 = 88.9 kn F y = 0 F V = F 1 + W F 1 = (3 4) = kn W = γ water Volume W = 9810 ( π 4 3 4) = 77.4 kn F V = F 1 + W = = kn Buoyancy and Stability It is a common experience that an object feels lighter and weighs less in a liquid than it does in air. This can be demonstrated easily by weighing a heavy object in water by a waterproof spring scale. Also, objects made of wood or other light materials float on water. These and other observations suggest that a fluid exerts an upward force on a
19 body immersed in it. This force that tends to lift the body is called the buoyant force and is denoted by FB. The buoyant force is caused by the increase of pressure in a fluid with depth. Consider, for example, a flat plate of thickness h submerged in a liquid of density ρ f parallel to the free surface, as shown in figure below. The area of the top (and also bottom) surface of the plate is A, and its distance to the free surface is s. The pressures at the top and bottom surfaces of the plate are ρ f. g. s and ρ f. g. (s + h), respectively. Then the hydrostatic force F top = ρ f. g. s. A acts downward on the top surface, and the larger force F top = ρ f. g. (s + h). A acts upward on the bottom surface of the plate. The difference between these two forces is a net upward force, which is the buoyant force. F buoyant = F bottom F top = F top = ρ f. g. (s + h). A ρ f. g. s. A F buoyant = ρ f. g. h. A = ρ f. g. V, where, V = A. h, volume of the plate we conclude that the buoyant force acting on the plate is equal to the weight of the liquid displaced by the plate. Archimedes principle: The buoyant force acting on a body immersed in a fluid is equal to the weight of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume. For floating bodies, the weight of the entire body must be equal to the buoyant force, which is the weight of the fluid whose volume is equal to the volume of the submerged portion of the floating body. That is, F B = w ρ f. g. V submerged = ρ average,body. g. V total V sub V total = ρ average,body ρ f Notes:
20 1- The submerged volume fraction of a floating body is equal to the ratio of the average density of the body to the density of the fluid. Note that when the density ratio is equal to or greater than one, the floating body becomes completely submerged. - A body immersed in a fluid (1) remains at rest at any point in the fluid when its density is equal to the density of the fluid, () sinks to the bottom when its density is greater than the density of the fluid, and (3) rises to the surface of the fluid and floats when the density of the body is less than the density of the fluid. 3- The buoyant force is proportional to the density of the fluid, and thus we might think that the buoyant force exerted by gases such as air is negligible. This is certainly the case in general, but there are significant exceptions. For example, the volume of a person is about 0.1 m 3, and taking the density of air to be 1. kg/m 3, the buoyant force exerted by air on the person is: F buoyant = ρ f. g. h. A = N The weight of an 80kg person is 80 * 9.81 = 788 N. Therefore, ignoring the buoyancy in this case results in an error in weight of just 0.15 percent, which is negligible. Example 1: A crane is used to lower weights into the sea (density =105 kg/m 3 ) for an underwater construction project as shown below. Determine the tension in the rope of the crane due to a rectangular 0.4m * 0.4m * 3m concrete block
21 (density = 300 kg/m 3 ) when it is (a) suspended in the air and (b) completely immersed in water. The volume of the rectangle, V = = 0.48 m 3 F T,air = ρ concret. g. V F T,air = = 10.8 kn And the force when the body immersed in water is: Now the force is: F B = ρ water. g. V F B = = 4.8 kn F water = w F B = = 6 kn Note: that the weight of the concrete block, and thus the tension of the rope, decreases by ( )/10.8 = 55% percent in water. Example : A spherical buoy has a diameter of a 1.5m, weight 8.5KN and is anchored to the seafloor with a cable as shown in figure below. Although, the buoy normally floats on the surface, at certain times the water depth increases so that the buoy is
22 completely immersed as illustrated. For this conditions what is the tension of the cable? The volume of the spherical shape, V = π 6 D3 = π 6 (1.5)3 = 1.77m 3 γ seawater = 10.1 kn m 3, given F B = γ seawater V F B = = kn The tension in the cable can be calculated by: T = F B w = = kn
Force on a Curved Surface due to Hydrostatic Pressure If the surface is curved, the forces on each element of the surface will not be parallel (normal to the surface at each point) and must be combined
CHARACTERISTIC OF FLUIDS A fluid is defined as a substance that deforms continuously when acted on by a shearing stress at any magnitude. In a fluid at rest, normal stress is called pressure. 1 Dimensions,
hapter 13 Archimedes Up-thrust In science, buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed object. The buoyant force is also called Archimedes Up-thrust force. Proof
1 Upthrust and Archimedes Principle Objects immersed in fluids, experience a force which tends to push them towards the surface of the liquid. This force is called upthrust and it depends on the density
Physics 07 HOMEORK ASSIGNMENT #9 Cutnell & Johnson, 7 th edition Chapter : Problems 6, 8, 33, 40, 44 *6 A 58-kg skier is going down a slope oriented 35 above the horizontal. The area of each ski in contact
Chapter 3 Fluid Statics 3.1 Pressure Pressure : The ratio of normal force to area at a point. Pressure often varies from point to point. Pressure is a scalar quantity; it has magnitude only It produces
MULTIPLE-CHOICE PROLEMS:(Two marks per answer) (Circle the Letter eside the Most Correct Answer in the Questions elow.) 1. The absolute viscosity µ of a fluid is primarily a function of: a. Density. b.
II. FLUID STATICS From a force analysis on a triangular fluid element at rest, the following three concepts are easily developed: For a continuous, hydrostatic, shear free fluid: 1. Pressure is constant
Chapter 1 INTRODUCTION 1-1 The Fluid. 1-2 Dimensions. 1-3 Units. 1-4 Fluid Properties. 1 1-1 The Fluid: It is the substance that deforms continuously when subjected to a shear stress. Matter Solid Fluid
Pressure Fluid Statics Variation of Pressure with Position in a Fluid Measurement of Pressure Hydrostatic Thrusts on Submerged Surfaces Plane Surfaces Curved Surfaces ddendum First and Second Moment of
Fluid Mechanics Chapter 2: Fluid Statics Lecture 3 Forces on Fluid Elements Fluid Elements - Definition: Fluid element can be defined as an infinitesimal region of the fluid continuum in isolation from
Eric G. Paterson Department of Mechanical and Nuclear Engineering Pennsylvania State University Spring 2005 Reading and Homework Read Chapter 3. Homework Set #2 has been posted. Due date: Friday 21 January.
Chapter 15 Density Often you will hear that fiberglass is used for racecars because it is lighter than steel. This is only true if we build two identical bodies, one made with steel and one with fiberglass.
Chapter / Fluid Statics CHPTER Fluid Statics FE-type Eam Review Problems: Problems - to -9. (C). (D). (C).4 ().5 () The pressure can be calculated using: p = γ h were h is the height of mercury. p= γ h=
1 Hydrostatics 2 Introduction In Fluid Mechanics hydrostatics considers fluids at rest: typically fluid pressure on stationary bodies and surfaces, pressure measurements, buoyancy and flotation, and fluid
Test Midterm 1 F2013 MULTIPLE-CHOICE PROBLEMS :(Two marks per answer) (Circle the Letter Beside the Most Correct nswer in the Questions Below.) 1. The absolute viscosity µ of a fluid is primarily a function
ME 262 BASIC FLUID MECHANICS Assistant Professor Neslihan Semerci Lecture 4 (Buoyancy and Viscosity of water) 16. BUOYANCY Whenever an object is floating in a fluid or when it is completely submerged in
1 Fluid Statics 1.1 Fluid Properties Fluid A fluid is a substance, which deforms when subjected to a force. A fluid can offer no permanent resistance to any force causing change of shape. Fluid flow under
Chapter 2 Hydrostatics uoyancy, Floatation and Stability Zerihun Alemayehu Rm. E119 AAiT Force of buoyancy an upward force exerted by a fluid pressure on fully or partially floating body Gravity Archimedes
Lecture 24: Archimedes Principle and Bernoulli s Law 1 Chapter 15: Fluid Mechanics Dynamics Using Pascal s Law Example 15.1 The hydraulic lift A hydraulic lift consists of a small diameter piston of radius
2.6 Force reacts with planar object in fluid Fluid surface Specific weight (γ) => Object sinks in fluid => C is center of gravity or Centroid => P is center of pressure (always under C) => x axis is cross
ME 262 BSIC FLUID MECHNICS ssistant Professor Neslihan Semerci Lecture 3 (Hydrostatic forces on flat and curved surfaces) 15. HYDROSTTIC PRESSURE 15.1 Hydrostatic pressure force on flat surfaces When a
ME-B41 Lab 1: Hydrostatics In this lab you will do four brief experiments related to the following topics: manometry, buoyancy, forces on submerged planes, and hydraulics (a hydraulic jack). Each experiment
CIVE 401 - HYDRAULIC ENGINEERING PART I Pierre Julien Colorado State University Problems with and are considered moderate and those with are the longest and most difficult. In 2018 solve the problems with
Pressure in stationary and moving fluid Lab-On-Chip: Lecture Lecture plan what is pressure e and how it s distributed in static fluid water pressure in engineering problems buoyancy y and archimedes law;
Discussion Prepared By: Dr.Khalil M. Al-Astal Eng.Ahmed S. Al-Agha Eng.Ruba M. Awad 2014-2015 Chapter (1) Fluids and their Properties Fluids and their Properties Fluids (Liquids or gases) which a substance
1896 1920 1987 2006 Introduction to Marine Hydrodynamics (NA235) Department of Naval Architecture and Ocean Engineering School of Naval Architecture, Ocean & Civil Engineering Shanghai Jiao Tong University
Nicholas J. Giordano www.cengage.com/physics/giordano Chapter 10 Fluids Fluids A fluid may be either a liquid or a gas Some characteristics of a fluid Flows from one place to another Shape varies according
Chapter 9: Solids and Fluids State of matters: Solid, Liquid, Gas and Plasma. Solids Has definite volume and shape Can be crystalline or amorphous Molecules are held in specific locations by electrical
Civil Engineering Hydraulics Forces on Curved Surfaces There is a theory which states that if ever anybody discovers exactly what the Universe is for and why it is here, it will instantly disappear and
ELEMENTARY HYDRAULICS National Certificate in Technology (Civil Engineering) Chapter 2 Pressure This section will study the forces acting on or generated by fluids at rest. Objectives Introduce the concept
Set No - 1 I B. Tech I Semester Regular Examinations Feb./Mar. - 2014 ENGINEERING MECHANICS (Common to CE, ME, CSE, PCE, IT, Chem E, Aero E, AME, Min E, PE, Metal E) Time: 3 hours Max. Marks: 70 Question
1. A sphere with a radius of 1.7 cm has a volume of: A) 2.1 10 5 m 3 B) 9.1 10 4 m 3 C) 3.6 10 3 m 3 D) 0.11 m 3 E) 21 m 3 2. A 25-N crate slides down a frictionless incline that is 25 above the horizontal.
CEE 4 Aut 005. Homework # Solutions.9. A free-body diagram of the lower hemisphere is shown below. This set of boundaries for the free body is chosen because it isolates the force on the bolts, which is
Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 9 Fluid Statics Part VI Good morning, I welcome you all to this session of Fluid
Announcements Equilibrium of a Rigid Body Today s Objectives Identify support reactions Draw a free body diagram Class Activities Applications Support reactions Free body diagrams Examples Engr221 Chapter
Hydrostatic Pressure distribution in a static fluid and its effects on solid surfaces and on floating and submerged bodies. M. Bahrami ENSC 283 Spring 2009 1 Fluid at rest hydrostatic condition: when a
TUTORIAL SHEET 1 1. The rectangular platform is hinged at A and B and supported by a cable which passes over a frictionless hook at E. Knowing that the tension in the cable is 1349N, determine the moment
MTE STATICS Example Problem P. Beer & Johnston, 004 by Mc Graw-Hill Companies, Inc. The structure shown consists of a beam of rectangular cross section (4in width, 8in height. (a Draw the shear and bending
MTHE 7 Problem Set Solutions. As a reminder, a torus with radii a and b is the surface of revolution of the circle (x b) + z = a in the xz-plane about the z-axis (a and b are positive real numbers, with
TOPIC B: MOMENTUM ANSWERS SPRING 2019 (Full worked answers follow on later pages) Q1. (a) 2.26 m s 2 (b) 5.89 m s 2 Q2. 8.41 m s 2 and 4.20 m s 2 ; 841 N Q3. (a) 1.70 m s 1 (b) 1.86 s Q4. (a) 1 s (b) 1.5
Chapter 6 Some Applications of the Integral Section 6.1 More on Area a. Representative Rectangle b. Vertical Separation c. Example d. Integration with Respect to y e. Example Section 6.2 Volume by Parallel
Date: February-12-16 Time: 2:00:28 PM TEST REPORT Question file: P12-2006 Copyright: Test Date: 21/10/2010 Test Name: EquilibriumPractice Test Form: 0 Test Version: 0 Test Points: 138.00 Test File: EquilibriumPractice
1. For the manometer shown in figure 1, if the absolute pressure at point A is 1.013 10 5 Pa, the absolute pressure at point B is (ρ water =10 3 kg/m 3, ρ Hg =13.56 10 3 kg/m 3, ρ oil = 800kg/m 3 ): (a)
FLUID MECHANICS First Year Exam Solutions 03 Q Give answers to all of the following questions (5 marks each): (a) A cylinder of m in diameter is made with material of relative density 0.5. It is moored
27 FE3 EQUILIBRIUM Aims OBJECTIVES In this chapter you will learn the concepts and principles needed to understand mechanical equilibrium. You should be able to demonstrate your understanding by analysing
Name 1. Please complete the following short problems. For parts 1A and 1B, we will consider three M88 recovery vehicles pulling an M1 tank back onto the road as shown below. F2 F1 50 M88 #1 50 M88 #2 y
Q1. Figure 1 shows a solid cylindrical steel rod of length =.0 m and diameter D =.0 cm. What will be increase in its length when m = 80 kg block is attached to its bottom end? (Young's modulus of steel
DIMENSIONS AND UNITS A dimension is the measure by which a physical variable is expressed quantitatively. A unit is a particular way of attaching a number to the quantitative dimension. Primary Dimension
TOPIC B: MOMENTUM EXAMPLES SPRING 2019 (Take g = 9.81 m s 2 ). Force-Momentum Q1. (Meriam and Kraige) Calculate the vertical acceleration of the 50 cylinder for each of the two cases illustrated. Neglect
Answers: Review questions 25 ANSERS O REVIE QUESIONS IMPORAN NOE: READ HIS FIRS. here are three different kinds of answer here. he usual form for quantitative questions is a short entry giving the final
Apex Grammar School O & A Level Evening Classes O Level Power Revision Series EVALUATION TEST PAPER REAL EXAMINATION QUESTIONS for Secondary 4 Name: Time Start: Date: Time End: Total Marks : / 40 40 questions
Topic: Fluid Properties 1. If 6 m 3 of oil weighs 47 kn, calculate its specific weight, density, and specific gravity. 2. 10.0 L of an incompressible liquid exert a force of 20 N at the earth s surface.
TOPIC D: ROTATION EXAMPLES SPRING 018 Q1. A car accelerates uniformly from rest to 80 km hr 1 in 6 s. The wheels have a radius of 30 cm. What is the angular acceleration of the wheels? Q. The University
General Physics I Quiz 6 - Ch. 9 - Static Equilibrium July 15, 2009 Name: Make your work clear to the grader. Show formulas used. Give correct units and significant figures. Partial credit is available
Set No 1 I B. Tech I Semester Regular Examinations, Dec 2016 ENGINEERING MECHANICS (Com. to AE, AME, BOT, CHEM, CE, EEE, ME, MTE, MM, PCE, PE) Time: 3 hours Max. Marks: 70 Question Paper Consists of Part-A
Chapter 10 Solids & Liquids Next 6 chapters use all the concepts developed in the first 9 chapters, recasting them into a form ready to apply to specific physical systems. 10.1 Phases of Matter, Mass Density
CE30 Test 1 Solution Key Date: 26 Sept. 2017 COVER PAGE Write your name on each sheet of paper that you hand in. Read all questions very carefully. If the problem statement is not clear, you should ask
Physics 202 Homework 2 Apr 10, 2013 1. An airplane wing is designed so that the speed of the air across the top of the 192 kn wing is 251 m/s when the speed of the air below the wing is 225 m/s. The density
Chapter 7 Applications of Integration 7.1 Area of a Region Between Two Curves 7.2 Volume: The Disk Method 7.3 Volume: The Shell Method 7.4 Arc Length and Surfaces of Revolution 7.5 Work 7.6 Moments, Centers
PHYS 101 Previous Exam Problems CHAPTER 7 Kinetic Energy and Work Kinetic energy Work Work-energy theorem Gravitational work Work of spring forces Power 1. A single force acts on a 5.0-kg object in such
The online of midterm-tests of Fluid Mechanics 1 1) The information on a can of pop indicates that the can contains 460 ml. The mass of a full can of pop is 3.75 lbm while an empty can weights 80.5 lbf.
PHYS 101 Previous Exam Problems CHAPTER 5 Force & Motion I Newton s Laws Vertical motion Horizontal motion Mixed forces Contact forces Inclines General problems 1. A 5.0-kg block is lowered with a downward
TOPIC E: OSCILLATIONS EXAMPLES SPRING 2019 Mathematics of Oscillating Systems Q1. Find general solutions for the following differential equations: Undamped Free Vibration Q2. A 4 g mass is suspended by
Types of Forces Pressure Buoyant Force Friction Normal Force Pressure Ratio of Force Per Unit Area p = F A P = N/m 2 = 1 pascal (very small) P= lbs/in 2 = psi = pounds per square inch Example: Snow Shoes
ME3560 Tentative Schedule Spring 2019 Week Number Date Lecture Topics Covered Prior to Lecture Read Section Assignment Prep Problems for Prep Probs. Must be Solved by 1 Monday 1/7/2019 1 Introduction to
2003 B1. (15 points) A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling, as shown above. One end of the rope is held by Student A of mass 70 kg, who is at rest on
Code No:A109210105 R09 SET-1 B.Tech II Year - I Semester Examinations, December 2011 FLUID MECHANICS (CIVIL ENGINEERING) Time: 3 hours Max. Marks: 75 Answer any five questions All questions carry equal
Torque Definition is a measure of how much a force acting on an object causes that object to rotate, symbol is, (Greek letter tau) = r F = rfsin, r = distance from pivot to force, F is the applied force
FLUID MECHANICS Technical English - I 1 th week Fluid Mechanics FLUID STATICS FLUID DYNAMICS Fluid Statics or Hydrostatics is the study of fluids at rest. The main equation required for this is Newton's
Force 10/01/2010 = = Friction Force (Weight) (Tension), coefficient of static and kinetic friction MIDTERM on 10/06/10 7:15 to 9:15 pm Bentley 236 2008 midterm posted for practice. Help sessions Mo, Tu
General Physics I Exam 4 - Chs. 10,11,12 - Fluids, Waves, Sound Nov. 14, 2012 Name Rec. Instr. Rec. Time For full credit, make your work clear to the grader. Show formulas used, essential steps, and results
Wiley Plus Final Assignment (5) Is Due Today: Before 11 pm! Final Exam Review December 9, 009 3 What about vector subtraction? Suppose you are given the vector relation A B C RULE: The resultant vector
Appendix FLUID MECHANICS Approximate physical properties of selected fluids All properties are given at pressure 101. kn/m and temperature 15 C. Liquids Density (kg/m ) Dynamic viscosity (N s/m ) Surface
Fluid Mechanics HOOKE'S LAW If deformation is small, the stress in a body is proportional to the corresponding strain. In the elasticity limit stress and strain Stress/strain = Const. = Modulus of elasticity.
Fluid Mechanics Prof. S. K. Som Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture - 8 Fluid Statics Part V Good morning, I welcome you all to the session of fluid mechanics.
25/01/2017 Lecture 8 Hydrostatic forces (Cont.) Yesterday, we discussed about hydrostatic forces acting on one side of a submerged plane surface. The plane area centroid was discussed and also the concept