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- 1. 1.1 MECHANICS Body of Knowledge which Deals with the Study and Prediction of the State of Rest or Motion of Particles and Bodies under the action of Forces
- 2. PARTS OF MECHANICS
- 3. 1.2 STATICS Statics Deals With the Equilibrium of Bodies, That Is Those That Are Either at Rest or Move With a Constant Velocity. Dynamics Is Concerned With the Accelerated Motion of Bodies and Will Be Dealt in the Next Semester.
- 4. ME13A: ENGINEERING STATICS CHAPTER TWO: STATICS OF PARTICLES
- 5. A particle has a mass but a size that can be neglected. When a body is idealised as a particle, the principles of mechanics reduce to a simplified form, since the geometry of the body will not be concerned in the analysis of the problem. 2.1 PARTICLE
- 6. PARTICLE CONTINUED All the forces acting on a body will be assumed to be applied at the same point, that is the forces are assumed concurrent.
- 7. 2.2 FORCE ON A PARTICLE A Force is a Vector quantity and must have Magnitude, Direction and Point of action. F P
- 8. Force on a Particle Contd. Note: Point P is the point of action of force and and are directions. To notify that F is a vector, it is printed in bold as in the text book. Its magnitude is denoted as |F| or simply F.
- 9. Force on a Particle Contd. There can be many forces acting on a particle. The resultant of a system of forces on a particle is the single force which has the same effect as the system of forces. The resultant of two forces can be found using the paralleolegram law.
- 10. 2.2.VECTOR OPERATIONS 2.3.1 EQUAL VECTORS Two vectors are equal if they are equal in magnitude and act in the same direction. pP Q
- 11. Equal Vectors Contd. Forces equal in Magnitude can act in opposite Directions S R
- 12. Q P R 2.3.2 Vector Addition Using the Paralleologram Law, Construct a Parm. with two Forces as Parts. The resultant of the forces is the diagonal.
- 13. Vector Addition Contd. Triangle Rule: Draw the first Vector. Join the tail of the Second to the head of the First and then join the head of the third to the tail of the first force to get the resultant force, R Q P R = Q + P
- 14. Triangle Rule Contd. Also: P Q R = P + Q Q + P = P + Q. This is the cummutative law of vector addition
- 15. Polygon Rule Can be used for the addition of more than two vectors. Two vectors are actually summed and added to the third.
- 16. Polygon Rule contd. P Q S P Q S R R = P + Q + S (P + Q)
- 17. Polygon Rule Contd. P + Q = (P + Q) ………. Triangle Rule i.e. P + Q + S = (P + Q) + S = R The method of drawing the vectors is immaterial . The following method can be used.
- 18. Polygon Rule contd. P Q S P Q S R R = P + Q + S (Q + S)
- 19. Polygon Rule Concluded Q + S = (Q + S) ……. Triangle Rule P + Q + S = P + (Q + S) = R i.e. P + Q + S = (P + Q) + S = P + (Q + S) This is the associative Law of Vector Addition
- 20. 2.3.3. Vector Subtraction P - Q = P + (- Q) P Q P -Q P -Q Q P P - Q Parm. Rule Triangle Rule
- 21. 2.4 Resolution of Forces It has been shown that the resultant of forces acting at the same point (concurrent forces) can be found. In the same way, a given force, F can be resolved into components. There are two major cases.
- 22. Resolution of Forces: Case 1 (a)When one of the two components, P is known: The second component Q is obtained using the triangle rule. Join the tip of P to the tip of F. The magnitude and direction of Q are determined graphically or by trignometry. F P Q i.e. F = P + Q
- 23. Resolution of Forces: Case 2 (b) When the line of action of each component is known: The force, F can be resolved into two components having lines of action along lines ‘a’ and ‘b’ using the paralleogram law. From the head of F, extend a line parallel to ‘a’ until it intersects ‘b’. Likewise, a line parallel to ‘b’ is drawn from the head of F to the point of intersection with ‘a’. The two components P and Q are then drawn such that they extend from the tail of F to points of intersection. a Q F P b
- 24. Example Determine graphically, the magnitude and direction of the resultant of the two forces using (a) Paralleolegram law and (b) the triangle rule. 900 N 600 N 30o 45o
- 25. Solution Solution: A parm. with sides equal to 900 N and 600 N is drawn to scale as shown. The magnitude and direction of the resultant can be found by drawing to scale. 600 N R 15o 900 N 45o 30o The triangle rule may also be used. Join the forces in a tip to tail fashion and measure the magnitude and direction of the resultant. 600 N R 45o 135o C B 30o 900 N 900N 600N 30o 45o
- 26. Trignometric Solution Using the cosine law: R2 = 9002 + 6002 - 2 x 900 x 600 cos 1350 R = 1390.6 = 1391 N Using the sine law: R B i e B The angle of the resul t sin sin . . sin sin . tan . . 135 600 600 135 1391 17 8 30 17 8 47 8 1 ie. R = 139N 47.8o R 900 N 600N 135o 30o B
- 27. Example Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 30 kN and Q = 20 kN, determine the magnitude and direction of the resultant force exerted on the bracket. P 25o 50o
- 28. Solution Solution: Using Triangle rule: 75o 30 kN 20 kN 105o 25o Q R R2 = 302 + 202 - 2 x 30 x 20 cos 1050 - cosine law R = 40.13 N Using sine rule: 4013 105 20 20 105 4013 288 288 25 38 401 38 1 . sin . . . . . . , . N Sin Sin and Sin Angle R i e R N o o o o o o o
- 29. 2.5 RECTANGULAR COMPONENTS OF FORCE x F j i Fx = Fx i Fy = Fy j y
- 30. RECTANGULAR COMPONENTS OF FORCE CONTD. In many problems, it is desirable to resolve force F into two perpendicular components in the x and y directions. Fx and Fy are called rectangular vector components. In two-dimensions, the cartesian unit vectors i and j are used to designate the directions of x and y axes. Fx = Fx i and Fy = Fy j i.e. F = Fx i + Fy j Fx and Fy are scalar components of F
- 31. RECTANGULAR COMPONENTS OF FORCE CONTD. While the scalars, Fx and Fy may be positive or negative, depending on the sense of Fx and Fy, their absolute values are respectively equal to the magnitudes of the component forces Fx and Fy, Scalar components of F have magnitudes: Fx = F cos and Fy = F sin F is the magnitude of force F.
- 32. Example Determine the resultant of the three forces below. 25o 45o 350 N 800 N 600 N 60o y x
- 33. Solution F x = 350 cos 25o + 800 cos 70o - 600 cos 60o = 317.2 + 273.6 - 300 = 290.8 N F y = 350 sin 25o + 800 sin 70o + 600 sin 60o = 147.9 + 751 + 519.6 = 1419.3 N i.e. F = 290.8 N i + 1419.3 N j Resultant, F F N 2908 1419 3 1449 1419 3 2908 78 4 2 2 1 0 . . tan . . . F = 1449 N 78.4o 25o 45o 350 N 800 N 600 N 60o y
- 34. Example A hoist trolley is subjected to the three forces shown. Knowing that = 40o , determine (a) the magnitude of force, P for which the resultant of the three forces is vertical (b) the corresponding magnitude of the resultant. 1000 N P 2000 N
- 35. Solution 1000 N P 2000 N 40o 40o (a) The resultant being vertical means that the horizontal component is zero. F x = 1000 sin 40o + P - 2000 cos 40o = 0 P = 2000 cos 40o - 1000 sin 40o = 1532.1 - 642.8 = 889.3 = 889 kN (b) Fy = - 2000 sin 40o - 1000 cos 40o = - 1285.6 - 766 = - 2052 N = 2052 N
- 36. 2.6. EQUILIBRIUM OF A PARTICLE A particle is said to be at equilibrium when the resultant of all the forces acting on it is zero. It two forces are involved on a body in equilibrium, then the forces are equal and opposite. .. 150 N 150 N If there are three forces, when resolving, the triangle of forces will close, if they are in equilibrium. F2 F1 F2 F3 F1 F3
- 37. EQUILIBRIUM OF A PARTICLE CONTD. If there are more than three forces, the polygon of forces will be closed if the particle is in equilibrium. F3 F2 F2 F3 F1 F4 F1 F4 The closed polygon provides a graphical expression of the equilibrium of forces. Mathematically: For equilibrium: R = F = 0 i.e. ( Fx i + Fy j) = 0 or (Fx) i + (Fy) j
- 38. EQUILIBRIUM OF A PARTICLE CONCLUDED For equilibrium: Fx = 0 and F y = 0. Note: Considering Newton’s first law of motion, equilibrium can mean that the particle is either at rest or moving in a straight line at constant speed.
- 39. FREE BODY DIAGRAMS: Space diagram represents the sketch of the physical problem. The free body diagram selects the significant particle or points and draws the force system on that particle or point. Steps: 1. Imagine the particle to be isolated or cut free from its surroundings. Draw or sketch its outlined shape.
- 40. Free Body Diagrams Contd. 2. Indicate on this sketch all the forces that act on the particle. These include active forces - tend to set the particle in motion e.g. from cables and weights and reactive forces caused by constraints or supports that prevent motion.
- 41. Free Body Diagrams Contd. 3. Label known forces with their magnitudes and directions. use letters to represent magnitudes and directions of unknown forces. Assume direction of force which may be corrected later.
- 42. Example The crate below has a weight of 50 kg. Draw a free body diagram of the crate, the cord BD and the ring at B. CRATE B ring C A D 45o
- 43. Solution (a) Crate FD ( force of cord acting on crate) 50 kg (wt. of crate) (b) Cord BD FB (force of ring acting on cord) FD (force of crate acting on cord) CRATE C 45o B A D
- 44. Solution Contd. (c) Ring FA (Force of cord BA acting along ring) FC (force of cord BC acting on ring) FB (force of cord BD acting on ring)
- 45. Example
- 46. Solution Contd. F F F BC AC o o AC sin cos . .............( ) 75 75 373 1 Fy = 0 i.e. FBC sin 75o - FAC cos 75o - 1962 = 0 F F F BC AC AC 1962 0 26 0 966 20312 0 27 2 . . . . ......( ) From Equations (1) and (2), 3.73 FAC = 2031.2 + 0.27 FAC FAC = 587 N From (1), FBC = 3.73 x 587 = 2190 N
- 47. RECTANGULAR COMPONENTS OF FORCE (REVISITED) x j i Fx = Fx i Fy = Fy j y F = Fx + Fy F = |Fx| . i + |Fy| . j |F|2 = |Fx|2 + |Fy|2 F | | | | | | F Fx Fy 2 2
- 48. 2.8 Forces in Space Rectangular Components Fy Fx Fz j i k F
- 49. Rectangular Components of a Force in Space F = Fx + Fy + Fz F = |Fx| . i + |Fy| . j + |Fz| . k |F|2 = |Fx|2 + |Fy|2 + |Fz|2 | | | | | | | | F Fx Fy Fz 2 2 2 | | | | cos | | | | cos | | | |cos , cos , Fx F Fy F Fz F Cos Cos and Cos are called direction ines of angles and x y z x y z x y z
- 50. Forces in Space Contd. i.e. F = F ( cos x i + cos y j + cos z k) = F F can therefore be expressed as the product of scalar, F and the unit vector where: = cos x i + cos y j + cos z k. is a unit vector of magnitude 1 and of the same direction as F. is a unit vector along the line of action of F.
- 51. Forces in Space Contd. Also: x = cos x, y = cos y and z = cos z - Scalar vectors i.e. magnitudes. x 2 + y 2 + z 2 = 1 = 2 i.e. cos2 x, + cos2 y + cos2 z = 1 Note: If components, Fx, Fy, and Fz of a Force, F are known, the magnitude of F, F = Fx 2 + Fy 2 + Fz 2 Direction cosines are: cos x = Fx/F , cos y = Fy/F and cos2 z = Fz/F
- 52. Force Defined by Magnitude and two Points on its Line of Action Contd. Unit vector, along the line of action of F = MN/MN MN is the distance, d from M to N. = MN/MN = 1/d ( dx i + dy j + dz k ) Recall that: F = F F = F = F/d ( dx i + dy j + dz k ) F Fd d F Fd d F Fd d d x x d y y d z z d d d d d d d d d d x x y y z z x y z x y z x x y y z z , , , , cos , cos , cos 2 1 2 1 2 1 2 2 2
- 53. 2.8.3 Addition of Concurrent Forces in Space The resultant, R of two or more forces in space is obtained by summing their rectangular components i.e. R = F i.e. Rx i + Ry j + Rz k = ( Fx i + Fy j + Fz k ) = ( Fx) i + ( Fy)j + ( Fz )k R x = Fx, Ry = Fy , Rz = Fz R = Rx 2 + Ry 2 + Rz 2 cos x = Rx/R cos y = Ry/R cos z = Rz/R
- 54. Solution Solution: Position vector of BH = 0.6 m i + 1.2 m j - 1.2 m k Magnitude, BH = 0 6 12 12 18 2 2 2 . . . . m BH BH BH BH BH BH x y z BH BH m i m j m k T T T BH BH N m m i m j m k T N i N j N k F N F N F N | | . ( . . . ) | |. | | | | . . . . ( ) (500 ) (500 ) , , 1 18 0 6 12 12 750 18 0 6 12 12 250 250 500 500
- 55. 2.9 EQUILIBRIUM OF A PARTICLE IN SPACE For equilibrium: Fx = 0, Fy = 0 and Fz = 0. The equations may be used to solve problems dealing with the equilibrium of a particle involving no more than three unknowns.

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