Presentation on theme: "STATICS OF PARTICLES Forces are vector quantities; they add according to the parallelogram law. The magnitude and direction of the resultant R of two forces."— Presentation transcript:
1 STATICS OF PARTICLESForces are vector quantities; they add according to theparallelogram law. The magnitude and direction of theresultant R of two forces P and Q can be determined eithergraphically or by trigonometry.RPAQ
2 Any given force acting on a particle can be resolved into two or more components, i.e.., it can be replaced by two or more forces which have the same effect on the particle.A force F can be resolvedinto two components Pand Q by drawing aparallelogram which hasF for its diagonal; thecomponents P and Qare then represented bythe two adjacent sidesof the parallelogramand can be determinedeither graphically or bytrigonometry.QFAP
3 F = Fx i + Fy j Fx = F cos q Fy = F sin q Fy tan q = Fx F = Fx + Fy 2 A force F is said to have been resolved into two rectangularcomponents if its components are directed along the coordinateaxes. Introducing the unit vectors i and j along the x and y axes,F = Fx i + Fy jyFx = F cos q Fy = F sin qtan q =FyFxFy = Fy jFjF = Fx + Fy2qxiFx = Fx i
4 Rx = S Rx Ry = S Ry Ry tan q = R = Rx + Ry 2 Rx When three or more coplanar forces act on a particle, therectangular components of their resultant R can be obtainedby adding algebraically the corresponding components of thegiven forces.Rx = S Rx Ry = S RyThe magnitude and direction of R can be determined fromtan q =RyRxR = Rx + Ry2
5 Fx = F cos qx Fy = F cos qy Fz = F cos qz Fy Fy qy F F qx Fx Fx Fz Fz BBFyFyqyAFAFDDOOqxFxFxxxFzFzEECCzzyA force F in three-dimensional space can be resolved into componentsBFyFFx = F cos qx Fy = F cos qyADOFz = F cos qzFxxqzEFzCz
6 F = F (cosqx i + cosqy j + cosqz k ) l (Magnitude = 1)The cosines ofqx , qy , and qzare known as thedirection cosines of the force F. Using the unit vectors i , j, and k, we writeFy jF = F lcos qy jFx icos qz kxFz kF = Fx i + Fy j + Fz kcos qx izorF = F (cosqx i + cosqy j + cosqz k )
7 l = cosqx i + cosqy j cos2qx + cos2qy F = Fx + Fy + Fz 2 cosqx = Fx F l (Magnitude = 1)cos qy jl = cosqx i + cosqy j+ cosqz kFy jcos qz kF = F lSince the magnitudeof l is unity, we haveFx ixcos2qx + cos2qy+ cos2qz = 1Fz kzcos qx iIn addition,F = Fx + Fy + Fz2cosqx =FxFcosqy =FyFcosqz =FzF
8 MN = dx i + dy j + dz k MN 1 l = = ( dx i + dy j + dz k ) d y A force vector Fin three-dimensionsis defined by itsmagnitude F andtwo points M andN along its line ofaction. The vectorMN joining pointsM and N isN (x2, y2, z2)Fdy = y2 - y1ldz = z2 - z1< 0dx = x2 - x1M (x1, y1, z1)xzMN = dx i + dy j + dz kThe unit vector l along the line of action of the force isl = = ( dx i + dy j + dz k )MN1d
9 F = F l = ( dx i + dy j + dz k ) F d Fdx d Fdy d Fdz d Fx = Fy = Fz = d = dx + dy + dz222N (x2, y2, z2)dy = y2 - y1A force F isdefined as theproduct of F andl. Therefore,dz = z2 - z1< 0dx = x2 - x1M (x1, y1, z1)xzF = F l = ( dx i + dy j + dz k )FdFrom this it follows thatFdxdFdydFdzdFx =Fy =Fz =
10 When two or more forces act on a particle in three-dimensions, the rectangular components of their resultant R is obtained by adding the corresponding components of the given forces.Rx = S FxRy = S FyRz = S FzThe particle is in equilibrium when the resultant of all forces acting on it is zero.
11 S Fx = 0 S Fy = 0 S Fz = 0 S Fx = 0 S Fy = 0 To solve a problem involving a particle in equilibrium, draw a free-body diagram showing all the forces acting on the particle. The conditions which must be satisfied for particle equilibrium areS Fx = S Fy = S Fz = 0In two-dimensions , only two of these equations are neededS Fx = S Fy = 0
12 V = P x Q V = PQ sin q Q x P = - (P x Q) V = P x Q The vector product of twovectors is defined asQV = P x QqPThe vector product of P andQ forms a vector which isperpendicular to both P and Q, of magnitudeV = PQ sin qThis vector is directed in such a way that a person located at thetip of V observes as counterclockwise the rotation through qwhich brings vector P in line with vector Q. The three vectors P,Q, and V - taken in that order - form a right-hand triad. Itfollows thatQ x P = - (P x Q)
13 P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k It follows from the definition of the vectorproduct of two vectors that the vectorproducts of unit vectors i, j, and k arekii x i = j x j = k x k = 0i x j = k , j x k = i , k x i = j , i x k = - j , j x i = - k , k x j = - iThe rectangular components of the vector product V of twovectors P and Q are determined as follows: GivenP = Px i + Py j + Pz k Q = Qx i + Qy j + Qz kThe determinant containing each component of P and Q isexpanded to define the vector V, as well as its scalarcomponents
14 P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k V = P x Q == Vx i + Vy j + Vz kwhereVx = Py Qz - Pz QyVy = Pz Qx - Px QzVz = Px Qy - Py Qx
15 MO = r x F MO = rF sin q = Fd Mo The moment of force F about point O is definedas the vector productMO = r x FFwhere r is the positionvector drawn from point Oto the point of applicationof the force F. The anglebetween the lines of actionof r and F is q.OrqdAThe magnitude of the moment of F about O can be expressed asMO = rF sin q = Fdwhere d is the perpendicular distance from O to the line of actionof F.
16 Mx = y Fz - z Fy My = zFx - x Fz Mz = x Fy - y Fx Fy jA (x , y, z )y jThe rectangularcomponents of themoment Mo of a force F are determined byexpanding thedeterminant of r x F.Fx irFz kOx ixz kzixFxjyFykzFzMo = r x F == Mx i + My j + MzkwhereMx = y Fz - z Fy My = zFx - x FzMz = x Fy - y Fx
17 rA/B = xA/B i + yA/B j + zA/B k Fy jA (x A, yA, z A)B (x B, yB, z B)In the more general case of the moment about an arbitrary point B of a force F applied at A, we haveFx irFz kOxzixA/BFxjyA/BFykzA/BFzMB = rA/B x F =rA/B = xA/B i + yA/B j + zA/B kwherexA/B = xA- xB yA/B = yA- yB zA/B = zA- zBand
18 MB = (xA- xB )Fy + (yA- yB ) Fx In the case of problems involving only two dimensions, the force F can be assumed to lie in the xy plane. Its moment about point B is perpendicular to that plane. It can be completely defined by the scalarFy jArA/BFx i(yA - yB ) jB(xA - xB ) iOxzMB = MB kMB = (xA- xB )Fy + (yA- yB ) FxThe right-hand rule is useful for defining the direction of themoment as either into or out of the plane (positive or negativek direction).
19 P Q = PQ cos q P Q = Px Qx + Py Qy + Pz Qz The scalar product of two vectors P and Q is denoted as P Q ,and is defined asQqP Q = PQ cos qwhere q is the angle betweenthe two vectorsPThe scalar product of P and Q is expressed in terms of therectangular components of the two vectors asP Q = Px Qx + Py Qy + Pz Qz
20 POL = Px cos qx + Py cos qy + Pz cos qz The projection of a vectorP on an axis OL can be obtained by forming the scalar product of P and the unit vector l along OL.qyAlPqxxOqzzPOL = P lUsing rectangular components,POL = Px cos qx + Py cos qy + Pz cos qz
21 Sx Px Qx Sy Py Qy Sz Pz Qz S (P x Q ) = The mixed triple product of three vectors S, P, and Q isSxPxQxSyPyQySzPzQzS (P x Q ) =The elements of the determinant are the rectangular components of the three vectors.
22 l x x Fx l y y Fy l z z Fz MOL = l MO = l (r x F) = y The moment of a force F about an axis OL is the projection OC on OL of the moment MO of the force F. This can be written as a mixed triple product.MOFClA (x, y, z)rOxzl xxFxl yyFyl zzFzMOL = l MO = l (r x F) =lx, ly , lz = direction cosines of axis OLx, y , z = components of rFx, Fy , Fz = components of F
23 M- FdFTwo forces F and -F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple.The moment of a couple is independent of the point about which it is computed; it is a vector M perpendicular to the plane of the couple and equal in magnitude to the product Fd.
24 Myy- Fy(M = Fd)MyMdFOMxOOxxxzzzMzTwo couples having the same moment M are equivalent (they have the same effect on a given rigid body).
25 FFMOrAAOOAny force F acting at a point A of a rigid body can be replacedby a force-couple system at an arbitrary point O, consistingof the force F applied at O and a couple of moment MO equal tothe moment about point O of the force F in its original position.The force vector F and the couple vector MO are alwaysperpendicular to each other.
26 Any system of forces can be reduced to a force-couple system at a given point O. First, each of the forces of thesystem is replaced by an equivalent force-couple system at O.Then all of the forces are added to obtain a resultant force R,and all of couples are added to obtain a resultant couple vectorMO. In general, the resultant force R and the couple vector MOwill not be perpendicular to each other.RR
27 S F = S F’ S Mo = S Mo’ F3 F1 R A3 A1 r1 r3 r2 O A2 O F2 M As far as rigid bodies are concerned, two systems of forces, F1, F2, F , and F’1, F’2, F’ , are equivalent if, andonly if,S F = S F’andS Mo = S Mo’
28 If the resultant force R and the resultant couple vector MO are perpendicular to each other, the force-couple system at O can be further reduced to a single resultant force.RA3Rr1A1r2OA2OROF2MThis is the case for systems consisting of either(a) concurrent forces,(b) coplanar forces, or(c) parallel forces.If the resultant force and couple are directed along the sameline, the force-couple system is termed a wrench.