Radius of sun : r s = 6.960*10 8 m. Distance between sun & earth : 1.496*10 11 m Goal: Given: Assume : Draw : Soln:

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Radius of sun : r s = 6.960*10 8 m. Distance between sun & earth : 1.496*10 11 m Goal: Given: Assume : Draw : Soln:"

Transcription

1 ME2100 Homework 1. Determine the gravitational force exerted by a. The moon on the earth, using the following data. Make sure that you show your work Mass of moon : m m = 7.35*10 22 kg Mass of earth : m e = 5.976*10 24 kg Radius of moon : r m = 1.738*10 6 m Radius of earth : r e = 6.731*10 6 m Distance between moon & earth : 3.844*10 8 m b. The sun on the earth, using the following data. Make sure that you show your work Mass of sun : m s = 1.990*10 30 kg Mass of earth : m e = 5.976*10 24 kg Radius of sun : r s = 6.960*10 8 m Radius of earth : r e = 6.731*10 6 m Distance between sun & earth : 1.496*10 11 m Goal: Given: Assume : Draw : Soln:

2 2. Determine the force of gravity acting on a satellite when it is in orbit 20.2 * 10 6 m above the surface of the earth. Its weight when on the surface of the earth is 8450 N. Use the data given in previous problem as needed. Goal: Given: Assume : Draw : Soln:

3 3. At what distance, in kilometers, from the surface of the earth on a line from center to center would the gravitational force of the earth on a body be exactly balanced by the gravitational force of the moon on the body? Use the data in problem as needed. (As a challenge first work out the problem using only symbols and then plug in the numbers at the end) Goal: Given: Assume : Draw : Soln:

4 4. Two forces are applied at point B of beam AB. Determine the magnitude and direction of the resultant using trigonometry. Challenge : use variables like P,Q for the forces and θ,φ for the angles and derive the solution for a whole class of problems. NOTE : the resultant will not be along line BC. GOAL : Given : P = 2kN, Q = 3kN, θ = 40,φ = 60 Find : R the resultant Solution : Draw the parallelogram with sides along P and Q. P C Hint : Use Sine and cosine law with parallelogram law of addition Q

5 5. The post is to be pulled out of the ground using two ropes A and B. Rope A is subjected to a force (tension) of 600 lb and is T lb directed at a known angle φ o from the vertical. If the resultant force acting on the post is to be always 1200 lb, vertically upward, B determine the force T in rope B and the corresponding angle θ o. Challenge : plot the value of T and θ as a function of φ. For students having difficulty use φ = 30 o θ φ A 600 lb

6 6. Using the figure shown a. Determine the magnitude and its direction of the resultant F r = F 1 + F 2 measured from the positive u- axis. b. Determine the components along the u and v axes of F 1 c. Determine the components along the u and v axes of F 2 F 1 = 150N

7 7. Three cables are attached to a tree as shown in figure. a. Represent the vectors AB, AC, and AD. b. Find the unit vector along u AB, u AC, and u AD c. Find the direction angles of each of the above vectors.

8 8. Three cables are attached to a block as shown in figure. a. Find the unit vector along u OA, u OB, and u OC b. Represent the vectors F 1, F 2, and F 3,. c. Find the direction angles of OA, OB, and OC

9 9. A cable is attached to B to the right angle pipe OAB in figure. The tension in the cable is 750 lb and a moment/torque of 10 lb.ft is applied at B as shown along line of action AB. a. Represent the tension in the cable as a vector F BC b. Find the unit vector u BC c. Find the direction angles of the vector F BC d. Represent the moment as a vector. M=10 lb.ft

10 10. Two forces F 1 and F 2 are applied to an eyelet as shown in figure. Determine the resultant F R = F 1 + F 2, and write in vector notation. Find the unit vector along the resultant and then use it to obtain the direction angles of the resultant.

11 11. The flex-headed ratchet wrench is subjected to a force P lb, applied perpendicular to the handle as shown. Find the moment or torque this imparts along the vertical axis of the bolt at A assuming θ as a given angle. θ

12 12. Determine a. Moment M 1 of F 1 about moment center A b. Moment M 2 of F 2 about moment center A c. Sum of M 1 and M 2 which we will call M 3 d. If F 1 and F 2 are parallel and opposite to each other will your answer in part c. Why OR why not? e. Find the scalar component of M 3 about the axis of the shaft AB. f. Would your answer in part e be different if M 1 and M 2 had been calculated for a moment center at B? (first reason your answer and than confirm by computing)

13 13. A 2 kn force acts on one end of the curved rod. Section AB of the rod lies in the xy plane and section BC lies in the zy plane. Determine the moment about moment center A and about moment center B Now find the magnitude about the line AB and then about line BC. Reason your answers. (For hints see on page 182)

14 14. For the beam pinned at A and acted by T BC, F B and M C, replace the loads by equivalent loading at A. Present your answers as vectors and also sketch the equivalent force and moment.

15 15. Find equivalent loading at A and represent as a vector and also as a diagram.

16 16. The wing of the jet aircraft is subjected to a thrust of T= 8,000N from its engine, and the moment of 6000 N*m due to the rotation and the resultant lift force L= 45,000N (This is normally a distributed force, but has been reduced to a concentrated force as shown). If the mass of the wing is 2,100 Kg and the mass center is at G, determine the components of reaction where the wing is fixed to the fuselage at A. GOAL: Given : Assume : Soln : Draw Free Body diagram The body of the aircraft provides a fixed support for the wing (meaning 6 components of reaction, 2 each in x,y,z co-ordinates) Write equations of equilibrium six equations and six unknowns M = 6kN-m

17 17. The pipe assembly supports the vertical loads shown. Determine the components of reaction at the ball and socket joint A and the tension in the supporting cables BC and BD. GOAL : Given : 100N/m Assume: Soln : Draw FBD. Reduce distributed load to conc. load and find its line of action (i.e. 3m*100N/m=300N force acting at the mid-point Already shown for you in FBD) A ball and socket joint does NOT allow translation in all directions, but lets free rotation in each direction. 300N 1.5m

18 18. Determine the reactions at the fixed wall A. The 150 N force is parallel to the z axis, the 200N force is parallel to the y axis and the 300N force is parallel to the x axis. The moment of a couple has a magnitude of 100 N-m and has direction angles of θ x =67.4ºand θ z =39.8º. GOAL : 100N-m 300N Given : Assume: Soln :

19 19. A 200-N force is applied to the handle of the hoist in the direction shown. The bearing at A is a thrust bearing, and the bearing at B is a journal bearing. If the hoist is in equilibrium, what forces act on the shaft at A? What forces acts on the shaft at B? What is the maximum mass m in kilogram that can be lifted? GOAL : Given : Find : m Soln : Draw FBD. A journal bearing does NOT permit translation and rotation in two axes. A thrust bearing does NOT permit translation in three directions and does NOT permit rotation in two axes. After showing the above we will have 10 unknowns which includes the mass. Now make the assumption that the bearings are perfectly aligned this will take care of 4 moment reactions, two each at the two bearings and leave you with six unknowns.

20 20. A shaft is loaded through a pulley and a lever that are fixed to the shown shown in figure. Friction between the belt and the pulley prevents the belt from slipping. The support at A is a journal bearing, and the support at B is thrust bearing. Determine a. the force P required for equilibrium b. the loads acting on the shaft at supports at A and at B. c. find the direction angles of the reaction at A and B d. It is known that the bearing at A fails if the force on it exceeds 1000N and bearing at B fails if force on it exceeds 400 N. Discuss the failure of this mechanism based on your answers in part b.

21 21. Determine the location of the centroid of the beam s crosssectional area. Neglect the size of the corner welds at A and B for the calculation. GOAL: Given : Assume:: Draw: Solution:

22 22. The gravity wall is made of concrete. Determine the location of the center of gravity G for the wall. GOAL: Given : Assume:: Draw: Solution :

23 23. Determine the location of the centroid C of the area. GOAL: Given : Assume:: Draw: Solution :

24 24. Determine the location of the center of gravity of the three-wheeler. The location of the center of gravity of each component and its weight are tabulated in the figure. If the three-wheeler is symmetrical with respect to the x-y plane, determine the normal reactions each of its wheels exerts on the ground. GOAL: Given : Assume:: Draw: Solution :

25 25. Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects a. member AB, measured from A. b. member BC, measured from C. GOAL: Given : Assume:: Draw: Solution :

26 26. The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position a on beam such that the resultant force and couple moment acting on the beam are zero. GOAL: Given : Assume:: Draw: Solution :

27 27. Replace the distributed loading by an equivalent resultant force, and specify its location on the beam, measured from the pin at C. GOAL: Given : Assume:: Draw: Solution :

28 28. The truss is made from five members, each having a length of 4m and a mass of 7kg/m. If the mass of the gusset plates at the joints and the thickness of the members can be neglected, determine the distance d to where the hoisting cable must be attached, so that the truss does not tip (not rotate) when it is lifted. GOAL: Given : Assume:: Draw: Solution :

29 29. The wall crane supports a load of 1000lb as shown. Determine the horizontal and vertical components of reaction at the pins A and D. Also what is the force in the cable at the winch? The jib ABC has a weight of 100lb and member BD has a weight of 40lb. Each member is uniform and has a center of gravity at its center. Outline : First consider the pulley at E as shown below and find the tension in the cable. Use this tension and find the reactions at C as shown in the diagram below. Equivalent system of forces Similarly reduce the pulley at B into an equivalent system of forces. Now consider the structure with pulleys removed and equivalent forces shown for their point of attachment. (Remember Newton s third law). Do not forget the weight of the members. Now draw the FBD of structure which will now consist of ABC and BD. Write equilibrium equations Next break it into members ABC and BD. Again do not forget the weight of the members 4 3 P=1000lb T E T T 3 P=1000lb 4 Pulley E R CX T T 3 4 Pulley C R CY T R B1 Pulley B T R B2

30 30. Determine the reactions at the supports of the frame shown. The pin attached to member BCD, passes through a smooth slot in member AB. (Hint : Frame has two members AB and BD. Draw their FBD and apply equilibrium conditions. You will need to reduce distributed load to concentrated load) Outline Determine support reactions (FBD of whole structure, Eqm equations) Breakup structure and draw FBD, Eqm equations. Do not forget that pin attached to BCD behaves like a roller reactions are normal to surface of contact. Do not forget external moment at A when drawing FBD of AB or FBD of structure. 100 lb-ft

31 31. The two-bar mechanism consists of a lever arm AB and smooth link CD, which has a fixed collar at its end C and a roller at the other end D. a) Determine the force P needed to hold the lever in the position θ. The spring has a stiffness k and unstretched length 2L. Assume that the roller contacts either the top or bottom portion of the horizontal guide at D. b) Using any software of your choice (Matlab, Mathcad), draw a graph of θ vs P. (θ = 30 to 60). From the graph determine the maximum P required and at what angle does this occur. For part b, assume L = 1m, k = 100N/m. Challenge : Assume that P is acting at an angle β from the horizontal measured in clockwise direction instead of as shown in figure.

32 32. Determine the force in each member of the truss. First solve using P 1 and P 2 as the forces and the given length. Then substitute values a) P 1 = 240lb, P 2 = 100lb. state if the members are in tension OR compression. b) Determine the largest permissible load P 2 if P 1 = 0lb. No member should exceed 500lb in tension and 350 lb in compression.

33 33. Determine the force in each member of the truss in terms of the load P and state if the members are in tension OR compression. Challenge Use the result to solve P a. Members AB and BC can each support a maximum compressive force of 800lb, and members AD, DC, and BD can support a maximum tensile force of 1500lb. If a = 10ft, determine the greatest load P the truss can support. b. Members AB and BC can each support a maximum compressive force of 800lb, and members AD, DC, and BD can support a maximum tensile force of 2000lb. If a = 6ft, determine the greatest load P the truss can support.

34 34. Determine the force in a) members GF, GD and CD b) members BG, BC, and HG using the method of sections. c) using method of joints now solve for forces in AB, AH, BH, CG, DF, FE, DE. In all cases state if the members are in tension OR compression.

35 35. Draw the axial force, shear force and bending moment for the structure shown. OUTLINE : First obtain the support reactions R Ax, R Ay and R By by drawing R By Fx = 0 R Ax = 0N r M A = i 2000 j i = Fy = i 4000 j + + 8i R By j 2000 j i 2000 j i the FBD of whole structure and using equilibrium equations j i 4000 j + Now recognize that we have seven regions in the beam so you will have to draw seven FBD and write 7*3=21 equilibrium equations to obtain axial force, shear force and bending moments.

36 36. Draw the axial force, shear force and bending moment diagrams. (Even though not asked you will have to find the support reactions before proceeding) OUTLINE : First obtain the support reactions R Ax, R Ay and R By by drawing the FBD of whole structure and using equilibrium equations. Remember to reduce distributed load to concentrated load which will be area under curve area of R By Fx = 0 R Ax = 30 lb r M A = 0 6i 225 j + 9i R By j 200 k = Fy = 0 triangle. 225lb acting at 6ft from A. Now realize that the beam has two regions. So you will write 3*2 = 6 equations to obtain the axial force, shear force and bending moment. Be sure to consider the area of triangle as we did in ICE lb

37 37. Draw the shear force and bending moment OUTLINE : Find support reactions. Identify regions we did this in class for this problem. Write eqm. equations for each section. (There are five regions in this beam most often people will make a mistake of not identifying point E where a concentrated moment is applied). E

38 700lb 38. The beam consists of two segments pinconnected at B. Draw the axial force, shear force and bending moment diagrams for the E D 300lb beam. Assume that the moment is applied at point E for the analysis. (HINT : We have solved 300lb*ft a similar problem in class) CLUE : once you A find the reaction in pin B, you can actually look at 8ft 4ft the two members independently for the axial force, shear force and bending moment diagrams. 200lb/ft B 6ft C

Mechanics 1. Revision Notes

Mechanics 1. Revision Notes Mechanics 1 Revision Notes July 2012 MECHANICS 1... 2 1. Mathematical Models in Mechanics... 2 Assumptions and approximations often used to simplify the mathematics involved:... 2 2. Vectors in Mechanics....

More information

Chapter 28 Applied Mechanics

Chapter 28 Applied Mechanics Chapter 28 Applied Mechanics UNIT 28-1 Forces Weight (mass) and density have many applications in industry and construction. A few examples are Defining quantities of materials, such as bags of mortar

More information

Fy = P sin 50 + F cos = 0 Solving the two simultaneous equations for P and F,

Fy = P sin 50 + F cos = 0 Solving the two simultaneous equations for P and F, ENGR0135 - Statics and Mechanics of Materials 1 (161) Homework # Solution Set 1. Summing forces in the y-direction allows one to determine the magnitude of F : Fy 1000 sin 60 800 sin 37 F sin 45 0 F 543.8689

More information

ENGR-1100 Introduction to Engineering Analysis. Lecture 13

ENGR-1100 Introduction to Engineering Analysis. Lecture 13 ENGR-1100 Introduction to Engineering Analysis Lecture 13 EQUILIBRIUM OF A RIGID BODY & FREE-BODY DIAGRAMS Today s Objectives: Students will be able to: a) Identify support reactions, and, b) Draw a free-body

More information

Announcements. Moment of a Force

Announcements. Moment of a Force Announcements Test observations Units Significant figures Position vectors Moment of a Force Today s Objectives Understand and define Moment Determine moments of a force in 2-D and 3-D cases Moment of

More information

IMPORTANT NOTE ABOUT WEBASSIGN:

IMPORTANT NOTE ABOUT WEBASSIGN: Week 8 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

More information

Chapter 1: Statics. A) Newtonian Mechanics B) Relativistic Mechanics

Chapter 1: Statics. A) Newtonian Mechanics B) Relativistic Mechanics Chapter 1: Statics 1. The subject of mechanics deals with what happens to a body when is / are applied to it. A) magnetic field B) heat C ) forces D) neutrons E) lasers 2. still remains the basis of most

More information

Ideal Cable. Linear Spring - 1. Cables, Springs and Pulleys

Ideal Cable. Linear Spring - 1. Cables, Springs and Pulleys Cables, Springs and Pulleys ME 202 Ideal Cable Neglect weight (massless) Neglect bending stiffness Force parallel to cable Force only tensile (cable taut) Neglect stretching (inextensible) 1 2 Sketch a

More information

SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE MEMBERS

SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE MEMBERS SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE MEMBERS Today s Objectives: Students will be able to: a) Define a simple truss. b) Determine the forces in members of a simple truss. c) Identify zero-force

More information

4.2 Free Body Diagrams

4.2 Free Body Diagrams CE297-FA09-Ch4 Page 1 Friday, September 18, 2009 12:11 AM Chapter 4: Equilibrium of Rigid Bodies A (rigid) body is said to in equilibrium if the vector sum of ALL forces and all their moments taken about

More information

Announcements. Dry Friction

Announcements. Dry Friction Announcements Dry Friction Today s Objectives Understand the characteristics of dry friction Draw a FBD including friction Solve problems involving friction Class Activities Applications Characteristics

More information

Two-Force Members, Three-Force Members, Distributed Loads

Two-Force Members, Three-Force Members, Distributed Loads Two-Force Members, Three-Force Members, Distributed Loads Two-Force Members - Examples ME 202 2 Two-Force Members Only two forces act on the body. The line of action (LOA) of forces at both A and B must

More information

Physics, Chapter 3: The Equilibrium of a Particle

Physics, Chapter 3: The Equilibrium of a Particle University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 1-1-1958 Physics, Chapter 3: The Equilibrium of a Particle

More information

Chapter 11 Equilibrium

Chapter 11 Equilibrium 11.1 The First Condition of Equilibrium The first condition of equilibrium deals with the forces that cause possible translations of a body. The simplest way to define the translational equilibrium of

More information

EQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS

EQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS EQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS Today s Objectives: Students will be able to: 1. Analyze the planar kinetics In-Class Activities: of a rigid body undergoing rotational motion. Check Homework

More information

MECE 3400 Summer 2016 Homework #1 Forces and Moments as Vectors

MECE 3400 Summer 2016 Homework #1 Forces and Moments as Vectors ASSIGNED 1) Knowing that α = 40, determine the resultant of the three forces shown: 2) Two cables, AC and BC, are tied together at C and pulled by a force P, as shown. Knowing that P = 500 N, α = 60, and

More information

CE 201 (STATICS) DR. SHAMSHAD AHMAD CIVIL ENGINEERING ENGINEERING MECHANICS-STATICS

CE 201 (STATICS) DR. SHAMSHAD AHMAD CIVIL ENGINEERING ENGINEERING MECHANICS-STATICS COURSE: CE 201 (STATICS) LECTURE NO.: 28 to 30 FACULTY: DR. SHAMSHAD AHMAD DEPARTMENT: CIVIL ENGINEERING UNIVERSITY: KING FAHD UNIVERSITY OF PETROLEUM & MINERALS, DHAHRAN, SAUDI ARABIA TEXT BOOK: ENGINEERING

More information

Announcements. 2-D Vector Addition

Announcements. 2-D Vector Addition Announcements 2-D Vector Addition Today s Objectives Understand the difference between scalars and vectors Resolve a 2-D vector into components Perform vector operations Class Activities Applications Scalar

More information

SOLUTION 5 1. The Significance of Each Force: W is the effect of gravity (weight) on the dumpster. A x. A y. are the pin A reactions on the dumpster.

SOLUTION 5 1. The Significance of Each Force: W is the effect of gravity (weight) on the dumpster. A x. A y. are the pin A reactions on the dumpster. 5 1. Draw the free-body diagram of the dumpster D of the truck, which has a weight of 5000 lb and a center of gravity at G. It is supported by a pin at and a pin-connected hydraulic cylinder C (short link).

More information

5 Solutions /23/09 5:11 PM Page 377

5 Solutions /23/09 5:11 PM Page 377 5 Solutions 44918 1/23/09 5:11 PM Page 377 5 71. The rod assembly is used to support the 250-lb cylinder. Determine the components of reaction at the ball-andsocket joint, the smooth journal bearing E,

More information

CHAPTER 6 Space Trusses

CHAPTER 6 Space Trusses CHAPTER 6 Space Trusses INTRODUCTION A space truss consists of members joined together at their ends to form a stable threedimensional structures A stable simple space truss can be built from the basic

More information

Structural Analysis: Space Truss

Structural Analysis: Space Truss Structural Analysis: Space Truss Space Truss - 6 bars joined at their ends to form the edges of a tetrahedron as the basic non-collapsible unit - 3 additional concurrent bars whose ends are attached to

More information

10 Space Truss and Space Frame Analysis

10 Space Truss and Space Frame Analysis 10 Space Truss and Space Frame Analysis 10.1 Introduction One dimensional models can be very accurate and very cost effective in the proper applications. For example, a hollow tube may require many thousands

More information

Copyright 2011 Casa Software Ltd. www.casaxps.com. Centre of Mass

Copyright 2011 Casa Software Ltd. www.casaxps.com. Centre of Mass Centre of Mass A central theme in mathematical modelling is that of reducing complex problems to simpler, and hopefully, equivalent problems for which mathematical analysis is possible. The concept of

More information

Physics 201 Homework 8

Physics 201 Homework 8 Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 N-m is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kg-m 2. What is the

More information

Stress and Deformation Analysis. Representing Stresses on a Stress Element. Representing Stresses on a Stress Element con t

Stress and Deformation Analysis. Representing Stresses on a Stress Element. Representing Stresses on a Stress Element con t Stress and Deformation Analysis Material in this lecture was taken from chapter 3 of Representing Stresses on a Stress Element One main goals of stress analysis is to determine the point within a load-carrying

More information

Homework 4. problems: 5.61, 5.67, 6.63, 13.21

Homework 4. problems: 5.61, 5.67, 6.63, 13.21 Homework 4 problems: 5.6, 5.67, 6.6,. Problem 5.6 An object of mass M is held in place by an applied force F. and a pulley system as shown in the figure. he pulleys are massless and frictionless. Find

More information

Method of Sections for Truss Analysis

Method of Sections for Truss Analysis Method of Sections for Truss Analysis Joint Configurations (special cases to recognize for faster solutions) Case 1) Two Bodies Connected F AB has to be equal and opposite to F BC Case 2) Three Bodies

More information

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 1 NON-CONCURRENT COPLANAR FORCE SYSTEMS 1. Be able to determine the effects

More information

EQUILIBRIUM AND ELASTICITY

EQUILIBRIUM AND ELASTICITY Chapter 12: EQUILIBRIUM AND ELASTICITY 1 A net torque applied to a rigid object always tends to produce: A linear acceleration B rotational equilibrium C angular acceleration D rotational inertia E none

More information

General tests Algebra

General tests Algebra General tests Algebra Question () : Choose the correct answer : - If = then = a)0 b) 6 c)5 d)4 - The shape which represents Y is a function of is : - - V A B C D o - - y - - V o - - - V o - - - V o - -if

More information

Lesson 4 Rigid Body Statics. Taking into account finite size of rigid bodies

Lesson 4 Rigid Body Statics. Taking into account finite size of rigid bodies Lesson 4 Rigid Body Statics When performing static equilibrium calculations for objects, we always start by assuming the objects are rigid bodies. This assumption means that the object does not change

More information

Mechanics Cycle 2 Chapter 13+ Chapter 13+ Revisit Torque. Revisit Statics

Mechanics Cycle 2 Chapter 13+ Chapter 13+ Revisit Torque. Revisit Statics Chapter 13+ Revisit Torque Revisit: Statics (equilibrium) Torque formula To-Do: Torque due to weight is simple Different forms of the torque formula Cross product Revisit Statics Recall that when nothing

More information

Chapter 4 Dynamics: Newton s Laws of Motion. Copyright 2009 Pearson Education, Inc.

Chapter 4 Dynamics: Newton s Laws of Motion. Copyright 2009 Pearson Education, Inc. Chapter 4 Dynamics: Newton s Laws of Motion Force Units of Chapter 4 Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the Normal

More information

9 ROTATIONAL DYNAMICS

9 ROTATIONAL DYNAMICS CHAPTER 9 ROTATIONAL DYNAMICS CONCEPTUAL QUESTIONS 1. REASONING AND SOLUTION The magnitude of the torque produced by a force F is given by τ = Fl, where l is the lever arm. When a long pipe is slipped

More information

Explaining Motion:Forces

Explaining Motion:Forces Explaining Motion:Forces Chapter Overview (Fall 2002) A. Newton s Laws of Motion B. Free Body Diagrams C. Analyzing the Forces and Resulting Motion D. Fundamental Forces E. Macroscopic Forces F. Application

More information

Engineering Mechanics I. Phongsaen PITAKWATCHARA

Engineering Mechanics I. Phongsaen PITAKWATCHARA 2103-213 Engineering Mechanics I Phongsaen.P@chula.ac.th May 13, 2011 Contents Preface xiv 1 Introduction to Statics 1 1.1 Basic Concepts............................ 2 1.2 Scalars and Vectors..........................

More information

Structural Axial, Shear and Bending Moments

Structural Axial, Shear and Bending Moments Structural Axial, Shear and Bending Moments Positive Internal Forces Acting Recall from mechanics of materials that the internal forces P (generic axial), V (shear) and M (moment) represent resultants

More information

Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 4. Forces and Newton s Laws of Motion. continued Chapter 4 Forces and Newton s Laws of Motion continued Clicker Question 4.3 A mass at rest on a ramp. How does the friction between the mass and the table know how much force will EXACTLY balance the gravity

More information

RELATIVE MOTION ANALYSIS: VELOCITY

RELATIVE MOTION ANALYSIS: VELOCITY RELATIVE MOTION ANALYSIS: VELOCITY Today s Objectives: Students will be able to: 1. Describe the velocity of a rigid body in terms of translation and rotation components. 2. Perform a relative-motion velocity

More information

Rotation: Moment of Inertia and Torque

Rotation: Moment of Inertia and Torque Rotation: Moment of Inertia and Torque Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Through experience we learn

More information

4 Shear Forces and Bending Moments

4 Shear Forces and Bending Moments 4 Shear Forces and ending oments Shear Forces and ending oments 8 lb 16 lb roblem 4.3-1 alculate the shear force and bending moment at a cross section just to the left of the 16-lb load acting on the simple

More information

MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS

MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS This is the second tutorial on bending of beams. You should judge your progress by completing the self assessment exercises.

More information

Chapter 18 Static Equilibrium

Chapter 18 Static Equilibrium Chapter 8 Static Equilibrium 8. Introduction Static Equilibrium... 8. Lever Law... Example 8. Lever Law... 4 8.3 Generalized Lever Law... 5 8.4 Worked Examples... 7 Example 8. Suspended Rod... 7 Example

More information

P4 Stress and Strain Dr. A.B. Zavatsky MT07 Lecture 4 Stresses on Inclined Sections

P4 Stress and Strain Dr. A.B. Zavatsky MT07 Lecture 4 Stresses on Inclined Sections 4 Stress and Strain Dr. A.B. Zavatsky MT07 Lecture 4 Stresses on Inclined Sections Shear stress and shear strain. Equality of shear stresses on perpendicular planes. Hooke s law in shear. Normal and shear

More information

Chapter 4. Forces and Newton s Laws of Motion. continued

Chapter 4. Forces and Newton s Laws of Motion. continued Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting

More information

6 1. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft.

6 1. Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft. 06 Solutions 46060_Part1 5/27/10 3:51 PM Page 329 6 1. Draw the shear and moment diagrams for the shaft. The bearings at and exert only vertical reactions on the shaft. 250 mm 800 mm 24 kn 6 2. Draw the

More information

Vectors and the Inclined Plane

Vectors and the Inclined Plane Vectors and the Inclined Plane Introduction: This experiment is designed to familiarize you with the concept of force as a vector quantity. The inclined plane will be used to demonstrate how one force

More information

ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P

ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P This material is duplicated in the Mechanical Principles module H2 and those

More information

Ground Rules. PC1221 Fundamentals of Physics I. Force. Zero Net Force. Lectures 9 and 10 The Laws of Motion. Dr Tay Seng Chuan

Ground Rules. PC1221 Fundamentals of Physics I. Force. Zero Net Force. Lectures 9 and 10 The Laws of Motion. Dr Tay Seng Chuan PC1221 Fundamentals of Physics I Lectures 9 and 10 he Laws of Motion Dr ay Seng Chuan 1 Ground Rules Switch off your handphone and pager Switch off your laptop computer and keep it No talking while lecture

More information

04-1. Newton s First Law Newton s first law states: Sections Covered in the Text: Chapters 4 and 8 F = ( F 1 ) 2 + ( F 2 ) 2.

04-1. Newton s First Law Newton s first law states: Sections Covered in the Text: Chapters 4 and 8 F = ( F 1 ) 2 + ( F 2 ) 2. Force and Motion Sections Covered in the Text: Chapters 4 and 8 Thus far we have studied some attributes of motion. But the cause of the motion, namely force, we have essentially ignored. It is true that

More information

The Free Body Diagram. The Concurrent System

The Free Body Diagram. The Concurrent System T A B The Free Body Diagram The Concurrent System Free Body Diagrams Essential step in solving Equilibrium problems Complex Structural systems reduced into concise FORCE systems WHAT IS A FREE BODY DIAGRAM?

More information

Newton s Third Law. object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1

Newton s Third Law. object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1 Newton s Third Law! If two objects interact, the force exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1!! Note on notation: is

More information

ME 230 Kinematics and Dynamics

ME 230 Kinematics and Dynamics ME 230 Kinematics and Dynamics Wei-Chih Wang Department of Mechanical Engineering University of Washington Planar kinematics of a rigid body Chapter objectives Chapter 16 To classify the various types

More information

EQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS

EQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS EQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS Today s Objectives: Students will be able to: 1. Analyze the planar kinetics of a rigid body undergoing rotational motion. In-Class Activities: Applications

More information

This week s homework. 2 parts Quiz on Friday, Ch. 4 Today s class: Newton s third law Friction Pulleys tension. PHYS 2: Chap.

This week s homework. 2 parts Quiz on Friday, Ch. 4 Today s class: Newton s third law Friction Pulleys tension. PHYS 2: Chap. This week s homework. 2 parts Quiz on Friday, Ch. 4 Today s class: Newton s third law Friction Pulleys tension PHYS 2: Chap. 19, Pg 2 1 New Topic Phys 1021 Ch 7, p 3 A 2.0 kg wood box slides down a vertical

More information

Physics-1 Recitation-7

Physics-1 Recitation-7 Physics-1 Recitation-7 Rotation of a Rigid Object About a Fixed Axis 1. The angular position of a point on a wheel is described by. a) Determine angular position, angular speed, and angular acceleration

More information

Lecture 15. Torque. Center of Gravity. Rotational Equilibrium. Cutnell+Johnson:

Lecture 15. Torque. Center of Gravity. Rotational Equilibrium. Cutnell+Johnson: Lecture 15 Torque Center of Gravity Rotational Equilibrium Cutnell+Johnson: 9.1-9.3 Last time we saw that describing circular motion and linear motion is very similar. For linear motion, we have position

More information

PROBLEMS. m s TAC. m = 60 kg/m, determine the tension in the two supporting cables and the reaction at D.

PROBLEMS. m s TAC. m = 60 kg/m, determine the tension in the two supporting cables and the reaction at D. PRLEMS 1. he uniform I-beam has a mass of 60 kg per meter of its length. Determine the tension in the two supporting cables and the reaction at D. (3/64) PRLEMS A( 500) m (5 23) m m = 60 kg/m determine

More information

6. Vectors. 1 2009-2016 Scott Surgent (surgent@asu.edu)

6. Vectors. 1 2009-2016 Scott Surgent (surgent@asu.edu) 6. Vectors For purposes of applications in calculus and physics, a vector has both a direction and a magnitude (length), and is usually represented as an arrow. The start of the arrow is the vector s foot,

More information

Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy. An Eccentric Professor

Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy. An Eccentric Professor 1/5 2009/11/14 上午 11:08 Manage this Assignment: Chapter 12 Due: 12:00am on Saturday, July 3, 2010 Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy

More information

Mechanics of Materials. Chapter 4 Shear and Moment In Beams

Mechanics of Materials. Chapter 4 Shear and Moment In Beams Mechanics of Materials Chapter 4 Shear and Moment In Beams 4.1 Introduction The term beam refers to a slender bar that carries transverse loading; that is, the applied force are perpendicular to the bar.

More information

OUTCOME 2 KINEMATICS AND DYNAMICS TUTORIAL 2 PLANE MECHANISMS. You should judge your progress by completing the self assessment exercises.

OUTCOME 2 KINEMATICS AND DYNAMICS TUTORIAL 2 PLANE MECHANISMS. You should judge your progress by completing the self assessment exercises. Unit 60: Dynamics of Machines Unit code: H/601/1411 QCF Level:4 Credit value:15 OUTCOME 2 KINEMATICS AND DYNAMICS TUTORIAL 2 PLANE MECHANISMS 2 Be able to determine the kinetic and dynamic parameters of

More information

Solution Derivations for Capa #11

Solution Derivations for Capa #11 Solution Derivations for Capa #11 1) A horizontal circular platform (M = 128.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A student (m = 68.3 kg) walks slowly from the rim of the platform

More information

Mechanics 2. Revision Notes

Mechanics 2. Revision Notes Mechanics 2 Revision Notes November 2012 Contents 1 Kinematics 3 Constant acceleration in a vertical plane... 3 Variable acceleration... 5 Using vectors... 6 2 Centres of mass 8 Centre of mass of n particles...

More information

INSTANTANEOUS CENTER (IC) OF ZERO VELOCITY (Section 16.6) Today s Objectives: Students will be able to: a) Locate the instantaneous center (IC) of

INSTANTANEOUS CENTER (IC) OF ZERO VELOCITY (Section 16.6) Today s Objectives: Students will be able to: a) Locate the instantaneous center (IC) of INSTANTANEOUS CENTER (IC) OF ZERO VELOCITY (Section 16.6) Today s Objectives: Students will be able to: a) Locate the instantaneous center (IC) of zero velocity. b) Use the IC to determine the velocity

More information

EML 5526 FEA Project 1 Alexander, Dylan. Project 1 Finite Element Analysis and Design of a Plane Truss

EML 5526 FEA Project 1 Alexander, Dylan. Project 1 Finite Element Analysis and Design of a Plane Truss Problem Statement: Project 1 Finite Element Analysis and Design of a Plane Truss The plane truss in Figure 1 is analyzed using finite element analysis (FEA) for three load cases: A) Axial load: 10,000

More information

Unit 4: Science and Materials in Construction and the Built Environment. Chapter 14. Understand how Forces act on Structures

Unit 4: Science and Materials in Construction and the Built Environment. Chapter 14. Understand how Forces act on Structures Chapter 14 Understand how Forces act on Structures 14.1 Introduction The analysis of structures considered here will be based on a number of fundamental concepts which follow from simple Newtonian mechanics;

More information

1 of 7 4/13/2010 8:05 PM

1 of 7 4/13/2010 8:05 PM Chapter 33 Homework Due: 8:00am on Wednesday, April 7, 2010 Note: To understand how points are awarded, read your instructor's Grading Policy [Return to Standard Assignment View] Canceling a Magnetic Field

More information

Physics 113 Exam #4 Angular momentum, static equilibrium, universal gravitation, fluid mechanics, oscillatory motion (first part)

Physics 113 Exam #4 Angular momentum, static equilibrium, universal gravitation, fluid mechanics, oscillatory motion (first part) Physics 113 Exam #4 Angular momentum, static equilibrium, universal gravitation, fluid mechanics, oscillatory motion (first part) Answer all questions on this examination. You must show all equations,

More information

Bedford, Fowler: Statics. Chapter 4: System of Forces and Moments, Examples via TK Solver

Bedford, Fowler: Statics. Chapter 4: System of Forces and Moments, Examples via TK Solver System of Forces and Moments Introduction The moment vector of a force vector,, with respect to a point has a magnitude equal to the product of the force magnitude, F, and the perpendicular distance from

More information

Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N)

Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N) Gravitational Field A gravitational field as a region in which an object experiences a force due to gravitational attraction Gravitational Field Strength The gravitational field strength at a point in

More information

SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS - VELOCITY AND ACCELERATION DIAGRAMS

SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS - VELOCITY AND ACCELERATION DIAGRAMS SOLID MECHANICS TUTORIAL MECHANISMS KINEMATICS - VELOCITY AND ACCELERATION DIAGRAMS This work covers elements of the syllabus for the Engineering Council exams C105 Mechanical and Structural Engineering

More information

Forces. -using a consistent system of units, such as the metric system, we can define force as:

Forces. -using a consistent system of units, such as the metric system, we can define force as: Forces Force: -physical property which causes masses to accelerate (change of speed or direction) -a push or pull -vector possessing both a magnitude and a direction and adds according to the Parallelogram

More information

MECHANICAL PRINCIPLES OUTCOME 4 MECHANICAL POWER TRANSMISSION TUTORIAL 1 SIMPLE MACHINES

MECHANICAL PRINCIPLES OUTCOME 4 MECHANICAL POWER TRANSMISSION TUTORIAL 1 SIMPLE MACHINES MECHANICAL PRINCIPLES OUTCOME 4 MECHANICAL POWER TRANSMISSION TUTORIAL 1 SIMPLE MACHINES Simple machines: lifting devices e.g. lever systems, inclined plane, screw jack, pulley blocks, Weston differential

More information

Center of Mass/Momentum

Center of Mass/Momentum Center of Mass/Momentum 1. 2. An L-shaped piece, represented by the shaded area on the figure, is cut from a metal plate of uniform thickness. The point that corresponds to the center of mass of the L-shaped

More information

Designing With Cylinders

Designing With Cylinders The function of a cylinder in a fluid power system is to convert energy in the fluid stream into an equivalent amount of mechanical energy. Its power is delivered in a straight-line, push-pull motion.

More information

Higher Technological Institute Civil Engineering Department. Lectures of. Fluid Mechanics. Dr. Amir M. Mobasher

Higher Technological Institute Civil Engineering Department. Lectures of. Fluid Mechanics. Dr. Amir M. Mobasher Higher Technological Institute Civil Engineering Department Lectures of Fluid Mechanics Dr. Amir M. Mobasher 1/14/2013 Fluid Mechanics Dr. Amir Mobasher Department of Civil Engineering Faculty of Engineering

More information

Moments and Torques. M = F d

Moments and Torques. M = F d Moments and Torques When a force is applied to an object, the object reacts in six possible ways. It can elongate, compress, translate (moves left, right, up, down, etc.), bend, twist or rotate. The study

More information

Chapter 3. Technical Measurement and Vectors

Chapter 3. Technical Measurement and Vectors Chapter 3. Technical Measurement and Vectors Unit Conversions 3-1. soccer field is 100 m long and 60 m across. What are the length and width of the field in feet? 100 cm 1 in. 1 ft (100 m) 328 ft 1 m 2.54

More information

PLANE TRUSSES. Definitions

PLANE TRUSSES. Definitions Definitions PLANE TRUSSES A truss is one of the major types of engineering structures which provides a practical and economical solution for many engineering constructions, especially in the design of

More information

Statics and Mechanics of Materials

Statics and Mechanics of Materials Statics and Mechanics of Materials Chapter 4-1 Internal force, normal and shearing Stress Outlines Internal Forces - cutting plane Result of mutual attraction (or repulsion) between molecules on both

More information

1 of 7 10/2/2009 1:13 PM

1 of 7 10/2/2009 1:13 PM 1 of 7 10/2/2009 1:13 PM Chapter 6 Homework Due: 9:00am on Monday, September 28, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View]

More information

PHY121 #8 Midterm I 3.06.2013

PHY121 #8 Midterm I 3.06.2013 PHY11 #8 Midterm I 3.06.013 AP Physics- Newton s Laws AP Exam Multiple Choice Questions #1 #4 1. When the frictionless system shown above is accelerated by an applied force of magnitude F, the tension

More information

What is a force? Identifying forces. What is the connection between force and motion? How are forces related when two objects interact?

What is a force? Identifying forces. What is the connection between force and motion? How are forces related when two objects interact? Chapter 4: Forces What is a force? Identifying forces. What is the connection between force and motion? How are forces related when two objects interact? Application different forces (field forces, contact

More information

Matrices in Statics and Mechanics

Matrices in Statics and Mechanics Matrices in Statics and Mechanics Casey Pearson 3/19/2012 Abstract The goal of this project is to show how linear algebra can be used to solve complex, multi-variable statics problems as well as illustrate

More information

2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration.

2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration. 2.1 Force and Motion Kinematics looks at velocity and acceleration without reference to the cause of the acceleration. Dynamics looks at the cause of acceleration: an unbalanced force. Isaac Newton was

More information

F mg (10.1 kg)(9.80 m/s ) m

F mg (10.1 kg)(9.80 m/s ) m Week 9 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

More information

8.2 Shear and Bending-Moment Diagrams: Equation Form

8.2 Shear and Bending-Moment Diagrams: Equation Form 8.2 Shear and ending-oment Diagrams: Equation Form 8.2 Shear and ending-oment Diagrams: Equation Form Eample 1, page 1 of 6 1. Epress the shear and bending moment as functions of, the distance from the

More information

Objectives 4 CHAPTER 1 THE PRIME MOVERS

Objectives 4 CHAPTER 1 THE PRIME MOVERS Objectives Define force, and describe how forces are measured. Describe what happens when forces on an object are balanced and when they are unbalanced. Explain the meaning of Newton s first law of motion.

More information

P113 University of Rochester NAME S. Manly Fall 2013

P113 University of Rochester NAME S. Manly Fall 2013 Final Exam (December 19, 2013) Please read the problems carefully and answer them in the space provided. Write on the back of the page, if necessary. Show all your work. Partial credit will be given unless

More information

3 Work, Power and Energy

3 Work, Power and Energy 3 Work, Power and Energy At the end of this section you should be able to: a. describe potential energy as energy due to position and derive potential energy as mgh b. describe kinetic energy as energy

More information

Test - A2 Physics. Primary focus Magnetic Fields - Secondary focus electric fields (including circular motion and SHM elements)

Test - A2 Physics. Primary focus Magnetic Fields - Secondary focus electric fields (including circular motion and SHM elements) Test - A2 Physics Primary focus Magnetic Fields - Secondary focus electric fields (including circular motion and SHM elements) Time allocation 40 minutes These questions were ALL taken from the June 2010

More information

Newton s Laws of Motion

Newton s Laws of Motion Newton s Laws of Motion Newton s Laws and the Mousetrap Racecar Simple version of Newton s three laws of motion 1 st Law: objects at rest stay at rest, objects in motion stay in motion 2 nd Law: force

More information

ANALYTICAL METHODS FOR ENGINEERS

ANALYTICAL METHODS FOR ENGINEERS UNIT 1: Unit code: QCF Level: 4 Credit value: 15 ANALYTICAL METHODS FOR ENGINEERS A/601/1401 OUTCOME - TRIGONOMETRIC METHODS TUTORIAL 1 SINUSOIDAL FUNCTION Be able to analyse and model engineering situations

More information

VELOCITY, ACCELERATION, FORCE

VELOCITY, ACCELERATION, FORCE VELOCITY, ACCELERATION, FORCE velocity Velocity v is a vector, with units of meters per second ( m s ). Velocity indicates the rate of change of the object s position ( r ); i.e., velocity tells you how

More information

SOLUTION (2.14new) Known: The geometry and the loads acting on a pinned assembly are given.

SOLUTION (2.14new) Known: The geometry and the loads acting on a pinned assembly are given. SOLUTION (2.14new) Known: The geometry and the loads acting on a pinned assembly are given. Find: Draw a free-body diagram for the assembly and determine the magnitude of the forces acting on each member

More information

PHYS 1111L LAB 2. The Force Table

PHYS 1111L LAB 2. The Force Table In this laboratory we will investigate the vector nature of forces. Specifically, we need to answer this question: What happens when two or more forces are exerted on the same object? For instance, in

More information

Higher Geometry Problems

Higher Geometry Problems Higher Geometry Problems ( Look up Eucidean Geometry on Wikipedia, and write down the English translation given of each of the first four postulates of Euclid. Rewrite each postulate as a clear statement

More information

Shear Force and Moment Diagrams

Shear Force and Moment Diagrams C h a p t e r 9 Shear Force and Moment Diagrams In this chapter, you will learn the following to World Class standards: Making a Shear Force Diagram Simple Shear Force Diagram Practice Problems More Complex

More information