1、force vector (cont)2.8 force vector directed along a linev force f acting along the chain can be presented as a cartesian vector by- establish x, y, z axes- form a position vector r along length of chainv unit vector, u = r/r that defines the direction of both the chain and the forcev we get f = fu

2、example 2.13the man pulls on the cord with a force of 350n. represent this force acting on the support a, as a cartesian vector and determine its direction. solutionend points of the cord are a (0m, 0m, 7.5m) and b (3m, -2m, 1.5m)r = (3m 0m)i + (-2m 0m)j + (1.5m 7.5m)k = 3i 2j 6kmmagnitude = length

3、of cord abunit vector, u = r /r = 3/7i - 2/7j - 6/7kmmmmr7623222solutionforce f has a magnitude of 350n, direction specified by u.f = fu = 350n(3/7i - 2/7j - 6/7k) = 150i - 100j - 300k n = cos-1(3/7) = 64.6 = cos-1(-2/7) = 107 = cos-1(-6/7) = 1492.9 dot productv dot product of vectors a and b is wri

4、tten as ab (read a dot b)v define the magnitudes of a and b and the angle between their tailsab = ab cos where 0 180v referred to as scalar product of vectors as result is a scalar2.9 dot productvlaws of operation1. commutative lawab = ba2. multiplication by a scalara(ab) = (aa)b = a(ab) = (ab)a3. d

5、istribution lawa(b + d) = (ab) + (ad) 2.9 dot productvcartesian vector formulation- dot product of cartesian unit vectorsii = (1)(1)cos0 = 1ij = (1)(1)cos90 = 0- similarlyii = 1jj = 1kk = 1 ij = 0ik = 1jk = 1 2.9 dot productvcartesian vector formulationdot product of 2 vectors a and bab = axbx + ayb

6、y + azbzvapplicationsthe angle formed between two vectors or intersecting lines. = cos-1 (ab)/(ab) 0 180the components of a vector parallel and perpendicular to a line.aa = a cos = auexample 2.17the frame is subjected to a horizontal force f = 300j n. determine the components of this force parallel

7、and perpendicular to the member ab.solutionsincethus nkjijufffkjikjirrubabbbb1 .257)429. 0)(0()857. 0)(300()286. 0)(0(429. 0857. 0286. 0300.cos429. 0857. 0286. 0362362222solutionsince result is a positive scalar, fab has the same sense of direction as ub. express in cartesian formperpendicular compo

8、nentnkjikjijfffnkjikjinuffabababab110805 .73)1102205 .73(3001102205 .73429. 0857. 0286. 01 .257solutionmagnitude can be determined from f or from pythagorean theorem,nnnfffab1551 .2573002222quiz1. which one of the following is a scalar quantity? a) force b) position c) mass d) velocity2. for vector

9、addition, you have to use _ law. a) newtons second b) the arithmetic c) pascals d) the parallelogram3. 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 a

10、t 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. 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) j30 xyf = 80 n6. determine the magnitude of

11、the resultant (f1 + f2) force in n when f1= 10i + 20j n and f2= 20i + 20j n . a) 30 n b) 40 n c) 50 n d) 60 n e) 70 n7. vector algebra, as we are going to use it, is based on a _ coordinate system. a) euclidean b) left-handed c) greek d) right-handede) egyptian8. the symbols , , and designate the _

12、of a 3-d cartesian vector. a) unit vectors b) coordinate direction angles c) greek societies d) x, y and z components 9. 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 d

13、irection of vector a.10. 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 k12. 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

14、 u13. 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 = - rqpc) rpq = 1/rqp d) rpq = 2 rqp14. 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) dimen

15、sionless c) meter d) newton metere) the expression is algebraically illegal.15. 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 bya) 3 i + 3 j + 3 k m b) 3 i 3 j 3 k mc) 5 i + 7 j + 9 k m d) 3 i + 3 j + 3 k me) 4 i + 5 j + 6 k m16. for

16、ce vector, f, directed along a line pq is given bya) (f/ f) rpq b) rpq/rpq c) f(rpq/rpq) d) f(rpq/rpq)17. the dot product of two vectors p and q is defined as a) p q cos b) p q sin c) p q tan d) p q sec 18. the dot product of two vectors results in a _ quantity. a) scalar b) vector c) complex d) zer

17、o19. 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) perpendiculard) cannot be determined. 20. if a dot product of two non-zero vectors equals -1, then the vecto

18、rs must be _ to each other.a) parallel (pointing in the same direction)b) parallel (pointing in the opposite direction)c) perpendiculard) cannot be determined.21. the dot product can be used to find all of the following except _ .a) sum of two vectorsb) angle between two vectorsc) component of a vec

19、tor parallel to another lined) component of a vector perpendicular to another line22. find the dot product of the two vectors p and q. p = 5 i + 2 j + 3 k m q = -2 i + 5 j + 4 k m a) -12 m b) 12 m c) 12 m2 d) -12 m2 e) 10 m2 next chapter.equilibrium of a particlechapter objectivesvconcept of the fre

20、e-body diagram for a particlevsolve particle equilibrium problems using the equations of equilibriumchapter outline1. condition for the equilibrium of a particle 2. the free-body diagram3. coplanar systems4. three-dimensional force systems3.1 condition for the equilibrium of a particlevparticle at e

21、quilibrium if- at rest- moving at constant a constant velocityvnewtons first law of motionf = 0where f is the vector sum of all the forces acting on the particle 3.1 condition for the equilibrium of a particlevnewtons second law of motionf = mavwhen the force fulfill newtons first law of motion, ma

22、= 0 a = 0therefore, the particle is moving in constant velocity or at rest3.2 the free-body diagramvbest representation of all the unknown forces (f) which acts on a bodyva sketch showing the particle “free” from the surroundings with all the forces acting on itvconsider two common connections in th

23、is subject spring cables and pulleys3.2 the free-body diagramvspring linear elastic spring: change in length is directly proportional to the force acting on it spring constant or stiffness k: defines the elasticity of the spring magnitude of force when spring is elongated or compressed f = ks 3.2 th

24、e free-body diagramvcables and pulley cables (or cords) are assumed negligible weight and cannot stretch tension always acts in the direction of the cable tension force must have a constant magnitude for equilibrium for any angle , the cableis subjected to a constant tension t procedure for drawing

25、a fbd1. draw outlined shape2. show all the forces- active forces: particle in motion- reactive forces: constraints that prevent motion 3. identify each forces- known forces with proper magnitude and direction- letters used to represent magnitude and directionsexample 3.1the sphere has a mass of 6kg

26、and is supported. draw a free-body diagram of the sphere, the cord ce and the knot at c.solutionfbd at spheretwo forces acting, weight and the force on cord ce. weight of 6kg (9.81m/s2) = 58.9ncord cetwo forces acting: sphere and knotnewtons 3rd law: fce is equal but oppositefce and fec pull the cor

27、d in tensionfor equilibrium, fce = fecsolutionfbd at knot3 forces acting: cord cba, cord ce and spring cd important to know that the weight of the sphere does not act directly on the knot but subjected to by the cord ce3.3 coplanar systemsva particle is subjected to coplanar forces in the x-y planev

28、resolve into i and j components for equilibrium fx = 0fy = 0vscalar equations of equilibrium require that the algebraic sum of the x and y components to equal to zeroprocedure for analysis1. free-body diagram- establish the x, y axes- label all the unknown and known forces 2. equations of equilibriu

29、m- apply f = ks to find spring force - when negative result force- indicates its sense is reverse of that shown on the free body diagram- apply the equations of equilibriumfx = 0fy = 0example 3.4determine the required length of the cord ac so that the 8kg lamp is suspended. the undeformed length of

30、the spring ab is lab = 0.4m, and the spring has a stiffness of kab = 300n/m.solution1. draw fbd at point athree forces acting, force by cable ac, force in spring ab and weight of the lamp.if force on cable ab is known, stretch of the spring is found by f = ks. + fx = 0; tab tac cos30 = 0+ fy = 0; ta

31、bsin30 78.5n = 0solving, tac = 157.0kntab = 136.0knsolutiontab = kabsab; 136.0n = 300n/m(sab) sab = 0.453nfor stretched length, lab = lab+ sablab = 0.4m + 0.453m = 0.853mfor horizontal distance bc, 2m = laccos30 + 0.853m lac = 1.32m3.4 three-dimensional force systemsvfor particle equilibriumf = 0vre

32、solving into i, j, k componentsfxi + fyj + fzk = 0vthree scalar equations representing algebraic sums of the x, y, z forces fxi = 0fyj = 0fzk = 0procedure for analysis free-body diagram- establish the z, y, z axes - label all known and unknown force equations of equilibrium- apply fx = 0, fy = 0 and

33、 fz = 0- substitute vectors into f = 0 and set i, j, k components = 0- negative results indicate that the sense of the force is opposite to that shown in the fbd.example 3.7determine the force developed in each cable used to support the 40kn crate.solution1. draw fbd at point ato expose all three un

34、known forces in the cables.2. equations of equilibriumexpressing each forces in cartesian vectors, fb = fb(rb / rb) = -0.318fbi 0.424fbj + 0.848fbk fc = fc (rc / rc) = -0.318fci 0.424fcj + 0.848fck fd = fdi w = -40k )8()4()3(843222kjibfsolutionfor equilibrium, f = 0; fb + fc + fd + w = 0-0.318fbi 0.

35、424fbj + 0.848fbk - 0.318fci 0.424fcj + 0.848fck + fdi - 40k = 0 fx = 0; -0.318fb - 0.318fc + fd = 0 fy = 0; 0.424fb 0.424fc = 0 fz = 0; 0.848fb + 0.848fc - 40 = 0 solving, fb = fc = 23.6kn fd = 15.0knquiz1. when a particle is in equilibrium, the sum of forces acting on it equals _ . (choose the mos

36、t appropriate answer)a) a constant b) a positive number c) zero d) a negative number e) an integer 2. for a frictionless pulley and cable, tensions in the cables are related asa) t1 t2b) t1 = t2c) t1 t2d) t1 = t2 sin t1t23. assuming you know the geometry of the ropes, you cannot determine the forces in the cables in which system above?4. why?a) the weight is too h