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


 Garry Gardner
 2 years ago
 Views:
Transcription
1 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 of today s engineering sciences. A) Newtonian Mechanics B) Relativistic Mechanics C) Greek Mechanics C) Euclidean Mechanics 3. For a statics problem your calculations show the final answer as N. What will you write as your final answer? A) N B) kn C) 12 kn D) 12.3 kn E) 123 kn 4. In three step IPE approach to problem solving, what does P stand for? A) Position B) Plan C) Problem D) Practical E) Possible
2 Chapter 2: Statics 1. The dot product of two vectors P and Q is defined as A) P Q cos θ B) P Q sin θ C) P Q tan θ D) P Q sec θ 2. The dot product of two vectors results in a quantity. A ) Scalar B) Vector C) Complex D) Zero 3. If a dot product of two non zero vectors is 0, then the two vectors must be to each other. A) Parallel (pointing in the same direction) B) Parallel (pointing in the opposite direction) C) Perpendicular D) Cannot be determined. 4. If a dot product of two non zero vectors equals  1, then the vectors must be to each other. A) Parallel (pointing in the same direction) B) Parallel (pointing in the opposite direction) C) Perpendicular D) Cannot be determined. 5. The dot product can be used to find all of the following except. A) sum of two vectors B) angle between two vectors C) component of a vector parallel to another line D) component of a vector perpendicular to another line 6. Find the dot product of the two vectors P and Q. P = {5 i + 2 j + 3 k} m and Q = { 2 i + 5 j + 4 k} m A)  12 m B) 12 m C) 12 m 2 D)  12 m 2 E) 10 m 2
3 Chapter 2 (Cont.): Statics 1. Which one of the following is a scalar quantity? A) Force B) Position C) Mass D) Velocity 2. For vector addition, you have to use law. A) Newton s Second B) the arithmetic C) Pascal s D) the parallelogram 3. Can you resolve a 2 D vector along two directions, which are not at 90 to each other? A) Yes, but not uniquely. B) No. C) Yes, uniquely. 4. Can you resolve a 2 D vector along three directions (say at 0, 60, and 120 )? A) Yes, but not uniquely. B) No. C) Yes, uniquely. 5. Resolve F along x and y axes and write it in vector form, for θ=30 o, F = { } N A) 80 cos (30 ) i 80 sin (30 ) j B) 80 sin (30 ) i + 80 cos (30 ) j C) 80 sin (30 ) i 80 cos (30 ) j D) 80 cos (30 ) i + 80 sin (30 ) j y θ x F=80 N 6. Determine the magnitude of the resultant (F1 + F2) force in N, when F1 = { 10 i + 20 j } N and F2 = { 20 i + 20 j } N. A) 30 N B) 40 N C) 50 N D) 60 N E) 70 N 7. Vector algebra, as we are going to use it, is based on a coordinate system. A) Euclidean B) Left handed C) Greek D) Right handed E) Egyptian
4 8. The symbols α, β, and γ designate the of a 3 D Cartesian vector. A) Unit vectors B) Coordinate direction angles C) Greek societies D) X, Y and Z components 9. If you know only ua, you can determine the of A uniquely. A) magnitude B) angles (α, β and γ) C) components (AX, AY, & AZ) D) All of the above. 10. For a force vector, the following parameters are randomly generated. The magnitude is 0.9 N, α= 30 º, β= 70 º, γ = 100 º. What is wrong with this 3 D vector? A) Magnitude is too small. B) Angles are too large. C) All three angles are arbitrarily picked. D) All three angles are between 0 º to 180 º. 11. What is not true about an unit vector, ua? A) It is dimensionless. B) Its magnitude is one. C) It always points in the direction of positive X axis. D) It always points in the direction of vector A. 12. If F = {10 i + 10 j + 10 k} N and G = {20 i + 20 j + 20 k } N, then F + G = { } N A) 10 i + 10 j + 10 k B) 30 i + 20 j + 30 k C) 10 i 10 j 10 k D) 30 i + 30 j + 30 k 13. A position vector, rpq, is obtained by A) Coordinates of Q minus coordinates of P B) Coordinates of P minus coordinates of Q C) Coordinates of Q minus coordinates of the origin D) Coordinates of the origin minus coordinates of P 14. A force of magnitude F, directed along a unit vector U, is given by F =. A) F (U) B) U / F C) F / U D) F + U E) F U
5 15. P and Q are two points in a 3 D space. How are the position vectors rpq and rqp related? A) rpq = rqp B) rpq =  rqp C) rpq = 1/rQP D) rpq = 2 rqp 16. If F and r are force vector and position vectors, respectively, in SI units, what are the units of the expression (r * (F / F))? A) Newton B) Dimensionless C) Meter D) Newton  Meter E) The expression is algebraically illegal. 17. Two points in 3 D space have coordinates of P (1, 2, 3) and Q (4, 5, 6) meters. The position vector rqp is given by A) {3 i + 3 j + 3 k} m B) { 3 i 3 j 3 k} m C) {5 i + 7 j + 9 k} m D) { 3 i + 3 j + 3 k} m E) {4 i + 5 j + 6 k} m 18. Force vector, F, directed along a line PQ is given by A) (F/ F) rpq B) rpq/rpq C) F(rPQ/rPQ) D) F(rPQ/rPQ)
6 Chapter 3. Statics 1. Particle P is in equilibrium with five (5) forces acting on it in 3 D space. How many scalar equations of equilibrium can be written for point P? A) 2 B) 3 C) 4 D) 5 E) 6 2. In 3 D, when a particle is in equilibrium, which of the following equations apply? A) (Σ Fx) i + (Σ Fy) j + (Σ Fz) k = 0 B) Σ F = 0 C) Σ Fx = Σ Fy = Σ Fz = 0 D) All of the above. E) None of the above. 3. In 3 D, when you know the direction of a force but not its magnitude, how many unknowns corresponding to that force remain? A) One B) Two C) Three D) Four 4. If a particle has 3 D forces acting on it and is in static equilibrium, the components of the resultant force (Σ Fx, Σ Fy, and Σ Fz ). A) have to sum to zero, e.g.,  5 i + 3 j + 2 k B) have to equal zero, e.g., 0 i + 0 j + 0 k C) have to be positive, e.g., 5 i + 5 j + 5 k D) have to be negative, e.g.,  5 i  5 j  5 k 5. In 3 D, when you don t know the direction or the magnitude of a force, how many unknowns do you have corresponding to that force? A) One B) Two C) Three D) Four 6. When a particle is in equilibrium, the sum of forces acting on it equals. (Choose the most appropriate answer) A) A constant B) A positive number C) Zero D) A negative number E) An integer 7. For a frictionless pulley and cable, tensions in the cable (T1 and T2) are related as. A) T1 > T2 B) T1 = T2 C) T1 < T2 D) T1 = T2 sin θ
7 8. Assuming you know the geometry of the ropes, you cannot determine the forces in the cables in which system below? A, B, or C? A) B) C) 9. Why? A) The weight is too heavy. B) The cables are too thin. C) There are more unknowns than equations. D) There are too few cables for a 1000 lb weight. 10. The correct answer is: (D) 11. The correct answer is: (B)
8 Chapter 4. Statics 1. When determining the moment of a force about a specified axis, the axis must be along. A) the x axis B) the y axis C) the z axis D) any line in 3 D space E) any line in the x y plane 2. The triple scalar product u ( r F ) results in A) a scalar quantity ( + or  ). B) a vector quantity. C) zero. D) a unit vector. E) an imaginary number. 3. The vector operation (P Q) R equals A) P (Q R). B) R (P Q). C) (P R) (Q R). D) (P R) (Q R ). 4. For finding the moment of the force F about the x axis, the position vector in the triple scalar product should be. A) rac B) rba C) rab D) rbc 5. If r = {1 i + 2 j} m and F = {10 i + 20 j + 30 k} N, then the moment of F about the y axis is N m. A) 10 B)  30 C)  40 D) None of the above. 6. In statics, a couple is defined as separated by a perpendicular distance. A) two forces in the same direction B) two forces of equal magnitude C) two forces of equal magnitude acting in the same direction D) two forces of equal magnitude acting in opposite directions 7. The moment of a couple is called a vector. A) Free B) Spin C) Romantic D) Sliding
9 8. The correct answer is: (B) 9. If three couples act on a body, the overall result is that A) The net force is not equal to 0. B) The net force and net moment are equal to 0. C) The net moment equals 0 but the net force is not necessarily equal to 0. D) The net force equals 0 but the net moment is not necessarily equal to The correct answer is: (B) 11. You can determine the couple moment as M = r F. If F = {  20 k} lb, then r is A) rbc B) rab C) rcb D) rba The correct answer is: (D)
10 12. The correct answer is: (C) 13. The line of action of the distributed load s equivalent force passes through the of the distributed load. A) Centroid B) Mid point C) Left edge D) Right edge 14. What is the location of FR, i.e., the distance d? A) 2 m B) 3 m C) 4 m D) 5 m E) 6 m 15. If F1 = 1 N, x1 = 1 m, F2 = 2 N and x2 = 2 m, what is the location of FR, i.e., the distance x. A) 1 m B) 1.33 m C) 1.5 m D) 1.67 m E) 2 m 16. FR = A) 12 N B) 100 N C) 600 N D) 1200 N 17. x =. A) 3 m B) 4 m C) 6 m D) 8 m
11 18. The correct answer is: (B) 19. The moment of force F about point O is defined as MO =. A) r x F B) F x r C) r F D) r * F 20. If M = r F, then what will be the value of M r? A) 0 B) 1 C) r 2 F D) None of the above. 21. The correct answer is: (D) 22. If r = { 5 j } m and F = { 10 k } N, the moment r x F equals { } N m. A) 50 i B) 50 j C) 50 i D) 50 j E) Using the CCW direction as positive, the net moment of the two forces about point P is A) 10 N m B) 20 N m C)  20 N m D) 40 N m E)  40 N m
12 24. A general system of forces and couple moments acting on a rigid body can be reduced to a. A) single force B) single moment C) single force and two moments D) single force and a single moment 25. The original force and couple system and an equivalent force couple system have the same effect on a body. A) internal B) external C) internal and external D) microscopic 26. Consider two couples acting on a body. The simplest possible equivalent system at any arbitrary point on the body will have A) One force and one couple moment. B) One force. C) One couple moment. D) Two couple moments. Z S 27. The forces on the pole can be reduced to a single force and a single moment at point. A) P B) Q C) R D) S E) Any of these points. 28. For this force system, the equivalent system at P is. A) FRP = 40 lb (along +x dir.) and MRP = +60 ft lb B) FRP = 0 lb and MRP = +30 ft lb C) FRP = 30 lb (along +y dir.) and MRP =  30 ft lb D) FRP = 40 lb (along +x dir.) and MRP = +30 ft lb X R Q P Y 29. Consider three couples acting on a body. Equivalent systems will be at different points on the body. A) Different when located B) The same even when located C) Zero when located D) None of the above.
13 Chapter 5: Statics 1. If a support prevents translation of a body, then the support exerts a on the body. A) Couple moment B) Force C) Both A and B. D) None of the above 2. Internal forces are shown on the free body diagram of a whole body. A) Always B) Often C) Rarely D) Never 3. The beam and the cable (with a frictionless pulley at D) support an 80 kg load at C. In a FBD of only the beam, there are how many unknowns? A) 2 forces and 1 couple moment B) 3 forces and 1 couple moment C) 3 forces D) 4 forces 4. If the directions of the force and the couple moments are both reversed, what will happen to the beam? A) The beam will lift from A. B) The beam will lift at B. C) The beam will be restrained. D) The beam will break. 5. Internal forces are not shown on a free body diagram because the internal forces are. (Choose the most appropriate answer.) A) Equal to zero B) Equal and opposite and they do not affect the calculations C) Negligibly small D) Not important 6. How many unknown support reactions are there in this problem? A) 2 forces and 2 couple moments B) 1 force and 2 couple moments C) 3 forces D) 3 forces and 1 couple moment
14 7. The three scalar equations FX = FY = MO = 0, are equations of equilibrium in two dimensions. A) Incorrect B) The only correct C) The most commonly used D) Not sufficient 8. A rigid body is subjected to forces as shown. This body can be considered as a member. A) Single force B) Two force C) Three force D) Six force 9. For this beam, how many support reactions are there and is the problem statically determinate? A) (2, Yes) B) (2, No) C) (3, Yes) D) (3, No) 10. The beam AB is loaded and supported as shown: a) how many support reactions are there on the beam, b) is this problem statically determinate, and c) is the structure stable? A) (4, Yes, No) B) (4, No, Yes) C) (5, Yes, No) D) (5, No, Yes) 11. Which equation of equilibrium allows you to determine FB right away? A) FX = 0 B) FY = 0 C) MA = 0 D) Any one of the above. 12. A beam is supported by a pin joint and a roller. How many support reactions are there and is the structure stable for all types of loadings? A) (3, Yes) B) (3, No) C) (4, Yes) D) (4, No) 13. If a support prevents rotation of a body about an axis, then the support exerts a on the body about that axis. A) Couple moment B) Force C) Both A and B. D) None of the above.
15 14. When doing a 3 D problem analysis, you have scalar equations of equilibrium. A) 3 B) 4 C) 5 D) The rod AB is supported using two cables at B and a ball and socket joint at A. How many unknown support reactions exist in this problem? A) 5 force and 1 moment reaction B) 5 force reactions C) 3 force and 3 moment reactions D) 4 force and 2 moment reactions 16. If an additional couple moment in the vertical direction is applied to rod AB at point C, then what will happen to the rod? A) The rod remains in equilibrium as the cables provide the necessary support reactions. B) The rod remains in equilibrium as the ball and socket joint will provide the necessary resistive reactions. C) The rod becomes unstable as the cables cannot support compressive forces. D) The rod becomes unstable since a moment about AB cannot be restricted. 17. A plate is supported by a ball and socket joint at A, a roller joint at B, and a cable at C. How many unknown support reactions are there in this problem? A) 4 forces and 2 moments B) 6 forces C) 5 forces D) 4 forces and 1 moment 18. What will be the easiest way to determine the force reaction BZ? A) Scalar equation FZ = 0 B) Vector equation MA = 0 C) Scalar equation MZ = 0 D) Scalar equation MY = 0
16 Chapter 6: Statics 1. In the method of sections, generally a cut passes through no more than members in which the forces are unknown. A) 1 B) 2 C) 3 D) 4 2. If a simple truss member carries a tensile force of T along its length, then the internal force in the member is. A) Tensile with magnitude of T/2 B) Compressive with magnitude of T/2 C) Compressive with magnitude of T D) Tensile with magnitude of T 3. Can you determine the force in member ED by making the cut at section a a? Explain your answer. A) No, there are 4 unknowns. B) Yes, using Σ MD = 0. C) Yes, using Σ ME = 0. D) Yes, using Σ MB = If you know FED, how will you determine FEB? A) By taking section b b and using Σ ME = 0 B) By taking section b b, and using Σ FX = 0 and Σ FY = 0 C) By taking section a a and using Σ MB = 0 D) By taking section a a and using Σ MD = 0 5. As shown, a cut is made through members GH, BG and BC to determine the forces in them. Which section will you choose for analysis and why? A) Right, fewer calculations. B) Left, fewer calculations. C) Either right or left, same amount of work. D) None of the above, too many unknowns. 6. When determining the force in member HG in the previous question, which one equation of equilibrium is best to use? A) Σ MH = 0 B) Σ MG = 0 C) Σ MB = 0 D) Σ MC = 0
17 7. Frames and machines are different as compared to trusses since they have. A) Only two force members B) Only multiforce members C) At least one multiforce member D) At least one two force member 8. Forces common to any two contacting members act with on the other member. A) Equal magnitudes but opposite sense B) Equal magnitudes and the same sense C) Different magnitudes but opposite sense D) Different magnitudes but the same sense 9. The figures show a frame and its FBDs. If an additional couple moment is applied at C, then how will you change the FBD of member BC at B? A) No change, still just one force (FAB) at B. B) Will have two forces, BX and BY, at B. C) Will have two forces and a moment at B. D) Will add one moment at B. 10. The figures show a frame and its FBDs. If an additional force is applied at D, then how will you change the FBD of member BC at B? A) No change, still just one force (FAB) at B. B) Will have two forces, BX and BY, at B. C) Will have two forces and a moment at B. D) Will add one moment at B. 11. When determining the reactions at joints A, B, and C, what is the minimum number of unknowns for solving this problem? A) 3 B) 4 C) 5 D) For the above problem, imagine that you have drawn a FBD of member AB. What will be the easiest way to write an equation involving unknowns at B? A) MC = 0 B) MB = 0 C) MA = 0 D) FX = 0
18 13. One of the assumptions used when analyzing a simple truss is that the members are joined together by. A) Welding B) Bolting C) Riveting D) Smooth pins E) Super glue 14. When using the method of joints, typically equations of equilibrium are applied at every joint. A) Two B) Three C) Four D) Six 15. Truss ABC is changed by decreasing its height from H to 0.9 H. Width W and load P are kept the same. Which one of the following statements is true for the revised truss as compared to the original truss? A) Force in all its members have decreased. B) Force in all its members have increased. C) Force in all its members have remained the same. D) None of the above. 16. For this truss, determine the number of zero force members. A) 0 B) 1 C) 2 D) 3 E) Using this FBD, you find that FBC = 500 N. Member BC must be in. A) Tension B) Compression C) Cannot be determined 18. For the same magnitude of force to be carried, truss members in compression are generally made as compared to members in tension. A) Thicker B) Thinner C) The same size
19 Chapter 7: Internal Forces 1. In a multiforce member, the member is generally subjected to an internal. A) Normal force B) Shear force C) Bending moment D) All of the above. 2. In mechanics, the force component V acting tangent to, or along the face of, the section is called the. A) Axial force B) Shear force C) Normal force D) Bending moment 3. A column is loaded with a vertical 100 N force. At which sections are the internal loads the same? A) P, Q, and R B) P and Q C) Q and R D) None of the above. 4. A column is loaded with a horizontal 100 N force. At which section are the internal loads largest? A) P B) Q C) R D) S 5. Determine the magnitude of the internal loads (normal, shear, and bending moment) at point C. A) (100 N, 80 N, 80 N m) B) (100 N, 80 N, 40 N m) C) (80 N, 100 N, 40 N m) D) (80 N, 100 N, 0 N m ) 6. A column is loaded with a horizontal 100 N force. At which section are the internal loads the lowest? A) P B) Q C) R D) S
20 Chapter 8: Friction 1. A friction force always acts to the contact surface. A) Normal B) At 45 C) Parallel D) At the angle of static friction 2. If a block is stationary, then the friction force acting on it is. A) µs N B) = µs N C) µs N D) = µk N 3. A 100 lb box with a wide base is pulled by a force P and µs = 0.4. Which force orientation requires the least force to begin sliding? A) P(A) B) P(B) C) P(C) D) Can not be determined 4. A ladder is positioned as shown. Please indicate the direction of the friction force on the ladder at B. A) B) C) D) 5. A 10 lb block is in equilibrium. What is the magnitude of the friction force between this block and the surface? A) 0 lb B) 1 lb C) 2 lb D) 3 lb 6. The ladder AB is positioned as shown. What is the direction of the friction force on the ladder at B. A) B) C) D)
21 Chapter 9: Centroid, Center of Gravity, and Center of Mass 1. The is the point defining the geometric center of an object. A) Center of gravity B) Center of mass C) Centroid D) None of the above 2. To study problems concerned with the motion of matter under the influence of forces, i.e., dynamics, it is necessary to locate a point called. A) Center of gravity B) Center of mass C) Centroid D) None of the above 3. The steel plate with known weight and non uniform thickness and density is supported as shown. Of the three parameters (CG, CM, and centroid), which one is needed for determining the support reactions? Are all three parameters located at the same point? A) (center of gravity, no) B) (center of gravity, yes) C) (centroid, yes) D) (centroid, no) 4. When determining the centroid of the area above, which type of differential area element requires the least computational work? A) Vertical B) Horizontal C) Polar D) Any one of the above. 5. If a vertical rectangular strip is chosen as the differential element, then all the variables, including the integral limit, should be in terms of. A) x B) y C) z D) Any of the above. 6. If a vertical rectangular strip is chosen, then what are the values of x and y? A) (x, y) B) (x / 2, y / 2) C) (x, 0) D) (x, y / 2)
22 7. A composite body in this section refers to a body made of. A) Carbon fibers and an epoxy matrix B) Steel and concrete C) A collection of simple shaped parts or holes D) A collection of complex shaped parts or holes 8. The composite method for determining the location of the center of gravity of a composite body requires. A) Integration B) Differentiation C) Simple arithmetic D) All of the above. 9. Based on the typical centroid information, what is the minimum number of pieces you will have to consider for determining the centroid of the area shown at the right? A) 1 B) 2 C) 3 D) A storage box is tilted up to clean the rug underneath the box. It is tilted up by pulling the handle C, with edge A remaining on the ground. What is the maximum angle of tilt (measured between bottom AB and the ground) possible before the box tips over? A) 30 B) 45 C) 60 D) A rectangular area has semicircular and triangular cuts as shown. For determining the centroid, what is the minimum number of pieces that you can use? A) Two B) Three C) Four D) Five 12. For determining the centroid of the area, two square segments are considered; square ABCD and square DEFG. What are the coordinates (x, y ) of the centroid of square DEFG? A) (1, 1) m B) (1.25, 1.25) m C) (0.5, 0.5 ) m D) (1.5, 1.5) m
23 Chapter 10: Area Moment of Inertia 1. The definition of the Moment of Inertia for an area involves an integral of the form A) x da. B) x 2 da. C) x 2 dm. D) m da. 2. Select the SI units for the Moment of Inertia for an area. A) m 3 B) m 4 C) kg m 2 D) kg m 3 3. A pipe is subjected to a bending moment as shown. Which property of the pipe will result in lower stress (assuming a constant cross sectional area)? A) Smaller Ix B) Smaller Iy C) Larger Ix D) Larger Iy 4. In the figure to the right, what is the differential moment of inertia of the element with respect to the y axis (diy)? A) x 2 y dx B) (1/12) x 3 dy C) y 2 x dy D) (1/3) y dy 5. When determining the MoI of the element in the figure, diy equals A) x 2 dy B) x 2 dx C) (1/3) y 3 dx D) x 2.5 dx 6. Similarly, dix equals A) (1/3) x 1.5 dx B) y 2 da C) (1 /12) x 3 dy D) (1/3) x 3 dx
24 7. The parallel axis theorem for an area is applied between A) An axis passing through its centroid and any corresponding parallel axis. B) Any two parallel axis. C) Two horizontal axes only. D) Two vertical axes only. 8. The moment of inertia of a composite area equals the of the MoI of all of its parts. A) Vector sum B) Algebraic sum (addition or subtraction) C) Addition D) Product 9. For the area A, we know the centroid s (C) location, area, distances between the four parallel axes, and the MoI about axis 1. We can determine the MoI about axis 2 by applying the parallel axis theorem. A) Between axes 1 and 3 and then between the axes 3 and 2. B) Directly between the axes 1 and 2. C) Between axes 1 and 4 and then axes 4 and 2. D) None of the above. 10. For the same case, consider the MoI about each of the four axes. About which axis will the MoI be the smallest number? A) Axis 1 B) Axis 2 C) Axis 3 D) Axis 4 E) Can not tell. 11. For the given area, the moment of inertia about axis 1 is 200 cm 4. What is the MoI about axis 3 (the centroidal axis)? A) 90 cm 4 B) 110 cm 4 C) 60 cm 4 D) 40 cm The moment of inertia of the rectangle about the x axis equals A) 8 cm 4. B) 56 cm 4. C) 24 cm 4. D) 26 cm 4.
25 Chapter 10 (cont.): Mass Moment of Inertia 1. The formula definition of the mass moment of inertia about an axis is. A) r dm B) r 2 dm C) m dr D) m 2 dr 2. The parallel axis theorem can be applied to determine. A) Only the MoI B) Only the MMI C) Both the MoI and MMI D) None of the above. Note: MoI is the moment of inertia of an area and MMI is the mass moment inertia of a body 3. Consider a particle of mass 1 kg located at point P, whose coordinates are given in meters. Determine the MMI of that particle about the z axis. A) 9 kg m 2 B) 16 kg m 2 C) 25 kg m 2 D) 36 kg m 2 4. Consider a rectangular frame made of four slender bars with four axes (zp, zq, zr and zs) perpendicular to the screen and passing through the points P, Q, R, and S respectively. About which of the four axes will the MMI of the frame be the largest? A) zp B) zq C) zr D) zs E) Not possible to determine 5. A particle of mass 2 kg is located 1 m down the y axis. What are the MMI of the particle about the x, y, and z axes, respectively? A) (2, 0, 2) B) (0, 2, 2) C) (0, 2, 2) D) (2, 2, 0) 6. Consider a rectangular frame made of four slender bars and four axes (zp, zq, zr and zs) perpendicular to the screen and passing through points P, Q, R, and S, respectively. About which of the four axes will the MMI of the frame be the lowest? A) zp B) zq C) zr D) zs E) Not possible to determine.
4.2 Free Body Diagrams
CE297FA09Ch4 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 informationAnnouncements. 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 2D and 3D cases Moment of
More informationChapter 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 informationSIMPLE TRUSSES, THE METHOD OF JOINTS, & ZEROFORCE MEMBERS
SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZEROFORCE 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 zeroforce
More informationENGR1100 Introduction to Engineering Analysis. Lecture 13
ENGR1100 Introduction to Engineering Analysis Lecture 13 EQUILIBRIUM OF A RIGID BODY & FREEBODY DIAGRAMS Today s Objectives: Students will be able to: a) Identify support reactions, and, b) Draw a freebody
More informationCopyright 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 informationChapter 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 informationMechanics 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 informationIntroduction to Statics. .PDF Edition Version 1.0. Notebook
Introduction to Statics.PDF Edition Version 1.0 Notebook Helen Margaret Lester Plants Late Professor Emerita Wallace Starr Venable Emeritus Associate Professor West Virginia University, Morgantown, West
More informationStatically Indeterminate Structure. : More unknowns than equations: Statically Indeterminate
Statically Indeterminate Structure : More unknowns than equations: Statically Indeterminate 1 Plane Truss :: Determinacy No. of unknown reactions = 3 No. of equilibrium equations = 3 : Statically Determinate
More informationAnnouncements. 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 informationStructural 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 informationChapter 17 Planar Kinetics of a Rigid Body: Force and Acceleration
Chapter 17 Planar Kinetics of a Rigid Body: Force and Acceleration 17.1 Moment of Inertia I 2 2 = r dm, 單位 : kg m 或 slug m ft 2 M = Iα resistance to angular acceleration dm = ρdv I = ρ V r 2 dv 172 MOMENT
More informationVELOCITY, 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 informationA vector is a directed line segment used to represent a vector quantity.
Chapters and 6 Introduction to Vectors A vector quantity has direction and magnitude. There are many examples of vector quantities in the natural world, such as force, velocity, and acceleration. A vector
More informationCE 201 (STATICS) DR. SHAMSHAD AHMAD CIVIL ENGINEERING ENGINEERING MECHANICSSTATICS
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 informationcos 2u  t xy sin 2u (Q.E.D.)
09 Solutions 46060 6/8/10 3:13 PM Page 619 010 Pearson Education, Inc., Upper Saddle River, NJ. ll rights reserved. This material is protected under all copyright laws as they currently 9 1. Prove that
More informationStructural 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 noncollapsible unit  3 additional concurrent bars whose ends are attached to
More informationToday s Objective COMPOSITE BODIES
Today s Objective: Students will be able to determine: a) The location of the center of gravity, b) The location of the center of mass, c) And, the location of the centroid using the method of composite
More information13.4 THE CROSS PRODUCT
710 Chapter Thirteen A FUNDAMENTAL TOOL: VECTORS 62. Use the following steps and the results of Problems 59 60 to show (without trigonometry) that the geometric and algebraic definitions of the dot product
More informationPHY121 #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 informationMidterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m
Midterm Solutions I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel a solid disc) of mass M, radius R, anchored at its center but free to rotate i) Which of
More informationLecture 6. Weight. Tension. Normal Force. Static Friction. Cutnell+Johnson: 4.84.12, second half of section 4.7
Lecture 6 Weight Tension Normal Force Static Friction Cutnell+Johnson: 4.84.12, second half of section 4.7 In this lecture, I m going to discuss four different kinds of forces: weight, tension, the normal
More informationSHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Exam Name SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 1) A lawn roller in the form of a uniform solid cylinder is being pulled horizontally by a horizontal
More informationAnnouncements. 2D Vector Addition
Announcements 2D Vector Addition Today s Objectives Understand the difference between scalars and vectors Resolve a 2D vector into components Perform vector operations Class Activities Applications Scalar
More informationSolid Mechanics. Stress. What you ll learn: Motivation
Solid Mechanics Stress What you ll learn: What is stress? Why stress is important? What are normal and shear stresses? What is strain? Hooke s law (relationship between stress and strain) Stress strain
More informationFy = 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 ydirection allows one to determine the magnitude of F : Fy 1000 sin 60 800 sin 37 F sin 45 0 F 543.8689
More informationSolution 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 informationwww.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x
Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity
More informationEQUILIBRIUM 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 informationChapter 3 THE STATIC ASPECT OF SOLICITATION
Chapter 3 THE STATIC ASPECT OF SOLICITATION 3.1. ACTIONS Construction elements interact between them and with the environment. The consequence of this interaction defines the system of actions that subject
More informationGround 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 informationPH2213 : Examples from Chapter 4 : Newton s Laws of Motion. Key Concepts
PH2213 : Examples from Chapter 4 : Newton s Laws of Motion Key Concepts Newton s First and Second Laws (basically Σ F = m a ) allow us to relate the forces acting on an object (lefthand side) to the motion
More informationTHEORETICAL MECHANICS
PROF. DR. ING. VASILE SZOLGA THEORETICAL MECHANICS LECTURE NOTES AND SAMPLE PROBLEMS PART ONE STATICS OF THE PARTICLE, OF THE RIGID BODY AND OF THE SYSTEMS OF BODIES KINEMATICS OF THE PARTICLE 2010 0 Contents
More informationEQUATIONS 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 InClass Activities: of a rigid body undergoing rotational motion. Check Homework
More informationEngineering Mechanics I. Phongsaen PITAKWATCHARA
2103213 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 informationWhen the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid.
Fluid Statics When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Consider a small wedge of fluid at rest of size Δx, Δz, Δs
More informationCHAPTER 3 NEWTON S LAWS OF MOTION
CHAPTER 3 NEWTON S LAWS OF MOTION NEWTON S LAWS OF MOTION 45 3.1 FORCE Forces are calssified as contact forces or gravitational forces. The forces that result from the physical contact between the objects
More informationVirtual Work. Scissor Lift Platform
Method of Virtual Work  Previous methods (FBD, F, M) are generally employed for a body whose equilibrium position is known or specified  For problems in which bodies are composed of interconnected members
More information3600 s 1 h. 24 h 1 day. 1 day
Week 7 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 informationUnit 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 information10 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 informationPhysics 201 Homework 8
Physics 201 Homework 8 Feb 27, 2013 1. A ceiling fan is turned on and a net torque of 1.8 Nm is applied to the blades. 8.2 rad/s 2 The blades have a total moment of inertia of 0.22 kgm 2. What is the
More informationBedford, 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 informationStress 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 loadcarrying
More informationIMPORTANT 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 informationPhysics, 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 111958 Physics, Chapter 3: The Equilibrium of a Particle
More informationObjective: Equilibrium Applications of Newton s Laws of Motion I
Type: Single Date: Objective: Equilibrium Applications of Newton s Laws of Motion I Homework: Assignment (111) Read (4.14.5, 4.8, 4.11); Do PROB # s (46, 47, 52, 58) Ch. 4 AP Physics B Mr. Mirro Equilibrium,
More informationStatics and Mechanics of Materials
Statics and Mechanics of Materials Chapter 41 Internal force, normal and shearing Stress Outlines Internal Forces  cutting plane Result of mutual attraction (or repulsion) between molecules on both
More informationv v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )
Week 3 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 informationEDEXCEL 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 NONCONCURRENT COPLANAR FORCE SYSTEMS 1. Be able to determine the effects
More informationA Review of Vector Addition
Motion and Forces in Two Dimensions Sec. 7.1 Forces in Two Dimensions 1. A Review of Vector Addition. Forces on an Inclined Plane 3. How to find an Equilibrant Vector 4. Projectile Motion Objectives Determine
More information5. Forces and MotionI. Force is an interaction that causes the acceleration of a body. A vector quantity.
5. Forces and MotionI 1 Force is an interaction that causes the acceleration of a body. A vector quantity. Newton's First Law: Consider a body on which no net force acts. If the body is at rest, it will
More informationVersion PREVIEW Practice 8 carroll (11108) 1
Version PREVIEW Practice 8 carroll 11108 1 This printout should have 12 questions. Multiplechoice questions may continue on the net column or page find all choices before answering. Inertia of Solids
More informationPin jointed structures are often used because they are simple to design, relatively inexpensive to make, easy to construct, and easy to modify.
4. FORCES in PIN JOINTED STRUCTURES Pin jointed structures are often used because they are simple to design, relatively inexpensive to make, easy to construct, and easy to modify. They can be fixed structures
More information041. 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 informationPhysics 2A, Sec B00: Mechanics  Winter 2011 Instructor: B. Grinstein Final Exam
Physics 2A, Sec B00: Mechanics  Winter 2011 Instructor: B. Grinstein Final Exam INSTRUCTIONS: Use a pencil #2 to fill your scantron. Write your code number and bubble it in under "EXAM NUMBER;" an entry
More informationState of Stress at Point
State of Stress at Point Einstein Notation The basic idea of Einstein notation is that a covector and a vector can form a scalar: This is typically written as an explicit sum: According to this convention,
More informationSection 1.1. Introduction to R n
The Calculus of Functions of Several Variables Section. Introduction to R n Calculus is the study of functional relationships and how related quantities change with each other. In your first exposure to
More informationFigure 1.1 Vector A and Vector F
CHAPTER I VECTOR QUANTITIES Quantities are anything which can be measured, and stated with number. Quantities in physics are divided into two types; scalar and vector quantities. Scalar quantities have
More informationPhysics Notes Class 11 CHAPTER 5 LAWS OF MOTION
1 P a g e Inertia Physics Notes Class 11 CHAPTER 5 LAWS OF MOTION The property of an object by virtue of which it cannot change its state of rest or of uniform motion along a straight line its own, is
More informationEQUATIONS 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. InClass Activities: Applications
More information11. Rotation Translational Motion: Rotational Motion:
11. Rotation Translational Motion: Motion of the center of mass of an object from one position to another. All the motion discussed so far belongs to this category, except uniform circular motion. Rotational
More informationm i: is the mass of each particle
Center of Mass (CM): The center of mass is a point which locates the resultant mass of a system of particles or body. It can be within the object (like a human standing straight) or outside the object
More informationTwoForce Members, ThreeForce Members, Distributed Loads
TwoForce Members, ThreeForce Members, Distributed Loads TwoForce Members  Examples ME 202 2 TwoForce Members Only two forces act on the body. The line of action (LOA) of forces at both A and B must
More information1.3. DOT PRODUCT 19. 6. If θ is the angle (between 0 and π) between two nonzero vectors u and v,
1.3. DOT PRODUCT 19 1.3 Dot Product 1.3.1 Definitions and Properties The dot product is the first way to multiply two vectors. The definition we will give below may appear arbitrary. But it is not. It
More informationMechanics Cycle 2 Chapter 13+ Chapter 13+ Revisit Torque. Revisit Statics
Chapter 13+ Revisit Torque Revisit: Statics (equilibrium) Torque formula ToDo: Torque due to weight is simple Different forms of the torque formula Cross product Revisit Statics Recall that when nothing
More informationTruss Structures. See also pages in the supplemental notes. Truss: Mimic Beam Behavior. Truss Definitions and Details
Truss Structures Truss: Mimic Beam Behavior Truss Definitions and Details 1 2 Framing of a Roof Supported Truss Bridge Truss Details 3 4 See also pages 1215 in the supplemental notes. 1 Common Roof Trusses
More informationSolving Simultaneous Equations and Matrices
Solving Simultaneous Equations and Matrices The following represents a systematic investigation for the steps used to solve two simultaneous linear equations in two unknowns. The motivation for considering
More informationInterconnected Rigid Bodies with Multiforce Members Rigid Noncollapsible
Frames and Machines Interconnected Rigid Bodies with Multiforce Members Rigid Noncollapsible structure constitutes a rigid unit by itself when removed from its supports first find all forces external
More informationRotation: 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 informationPHYSICS 149: Lecture 4
PHYSICS 149: Lecture 4 Chapter 2 2.3 Inertia and Equilibrium: Newton s First Law of Motion 2.4 Vector Addition Using Components 2.5 Newton s Third Law 1 Net Force The net force is the vector sum of all
More informationNewton s Laws of Motion
Newton s Laws of Motion FIZ101E Kazım Yavuz Ekşi My contact details: Name: Kazım Yavuz Ekşi Email: eksi@itu.edu.tr Notice: Only emails from your ITU account are responded. Office hour: Wednesday 10.0012.00
More informationPhysics 107 HOMEWORK ASSIGNMENT #8
Physics 107 HOMEORK ASSIGMET #8 Cutnell & Johnson, 7 th edition Chapter 9: Problems 16, 22, 24, 66, 68 16 A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself
More informationStatics and Mechanics of Materials
Statics and Mechanics of Materials Chapter 4 Stress, Strain and Deformation: Axial Loading Objectives: Learn and understand the concepts of internal forces, stresses, and strains Learn and understand the
More informationPhysics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion
Physics: Principles and Applications, 6e Giancoli Chapter 4 Dynamics: Newton's Laws of Motion Conceptual Questions 1) Which of Newton's laws best explains why motorists should buckleup? A) the first law
More informationChapter 5 Polar Coordinates; Vectors 5.1 Polar coordinates 1. Pole and polar axis
Chapter 5 Polar Coordinates; Vectors 5.1 Polar coordinates 1. Pole and polar axis 2. Polar coordinates A point P in a polar coordinate system is represented by an ordered pair of numbers (r, θ). If r >
More informationProblem 1: Computation of Reactions. Problem 2: Computation of Reactions. Problem 3: Computation of Reactions
Problem 1: Computation of Reactions Problem 2: Computation of Reactions Problem 3: Computation of Reactions Problem 4: Computation of forces and moments Problem 5: Bending Moment and Shear force Problem
More informationChapter Test. Teacher Notes and Answers Forces and the Laws of Motion. Assessment
Assessment Chapter Test A Teacher Notes and Answers Forces and the Laws of Motion CHAPTER TEST A (GENERAL) 1. c 2. d 3. d 4. c 5. c 6. c 7. c 8. b 9. d 10. d 11. c 12. a 13. d 14. d 15. b 16. d 17. c 18.
More informationSpinning Stuff Review
Spinning Stuff Review 1. A wheel (radius = 0.20 m) is mounted on a frictionless, horizontal axis. A light cord wrapped around the wheel supports a 0.50kg object, as shown in the figure below. When released
More informationLecture 15. Torque. Center of Gravity. Rotational Equilibrium. Cutnell+Johnson:
Lecture 15 Torque Center of Gravity Rotational Equilibrium Cutnell+Johnson: 9.19.3 Last time we saw that describing circular motion and linear motion is very similar. For linear motion, we have position
More informationCHAPTER 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 informationStrength of Materials Prof: S.K.Bhattacharya Dept of Civil Engineering, IIT, Kharagpur Lecture no 29 Stresses in Beams IV
Strength of Materials Prof: S.K.Bhattacharya Dept of Civil Engineering, IIT, Kharagpur Lecture no 29 Stresses in Beams IV Welcome to the fourth lesson of the sixth module on Stresses in Beams part 4.
More informationLesson 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 informationHigher 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 informationStatics of Structural Supports
Statics of Structural Supports TYPES OF FORCES External Forces actions of other bodies on the structure under consideration. Internal Forces forces and couples exerted on a member or portion of the structure
More informationMECHANICS OF SOLIDS  BEAMS TUTORIAL 1 STRESSES IN BEAMS DUE TO BENDING. On completion of this tutorial you should be able to do the following.
MECHANICS OF SOLIDS  BEAMS TUTOIAL 1 STESSES IN BEAMS DUE TO BENDING This is the first tutorial on bending of beams designed for anyone wishing to study it at a fairly advanced level. You should judge
More informationExample (1): Motion of a block on a frictionless incline plane
Firm knowledge of vector analysis and kinematics is essential to describe the dynamics of physical systems chosen for analysis through ewton s second law. Following problem solving strategy will allow
More information1 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 informationMECHANICS OF MATERIALS Plastic Deformations of Members With a Single Plane of Symmetry
Plastic Deformations of Members With a Single Plane of Smmetr Full plastic deformation of a beam with onl a vertical plane of smmetr. The neutral axis cannot be assumed to pass through the section centroid.
More informationPLANE 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 informationPhysics 101 Prof. Ekey. Chapter 5 Force and motion (Newton, vectors and causing commotion)
Physics 101 Prof. Ekey Chapter 5 Force and motion (Newton, vectors and causing commotion) Goal of chapter 5 is to establish a connection between force and motion This should feel like chapter 1 Questions
More informationPhysics 1A Lecture 10C
Physics 1A Lecture 10C "If you neglect to recharge a battery, it dies. And if you run full speed ahead without stopping for water, you lose momentum to finish the race. Oprah Winfrey Static Equilibrium
More informationSection 16: Neutral Axis and Parallel Axis Theorem 161
Section 16: Neutral Axis and Parallel Axis Theorem 161 Geometry of deformation We will consider the deformation of an ideal, isotropic prismatic beam the cross section is symmetric about yaxis All parts
More informationPHYSICS 111 HOMEWORK SOLUTION #10. April 8, 2013
PHYSICS HOMEWORK SOLUTION #0 April 8, 203 0. Find the net torque on the wheel in the figure below about the axle through O, taking a = 6.0 cm and b = 30.0 cm. A torque that s produced by a force can be
More informationIntroduction to Engineering Analysis  ENGR1100 Course Description and Syllabus Monday / Thursday Sections. Fall '15.
Introduction to Engineering Analysis  ENGR1100 Course Description and Syllabus Monday / Thursday Sections Fall 2015 All course materials are available on the RPI Learning Management System (LMS) website.
More informationP4 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 informationRotational Mechanics  1
Rotational Mechanics  1 The Radian The radian is a unit of angular measure. The radian can be defined as the arc length s along a circle divided by the radius r. s r Comparing degrees and radians 360
More informationWeight 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