I .The concept of work3, The work of the resultant forceIf there are n forces acting on a particle, the netforce is R- ZE , When the particle moves from M, to M,the work done by the resultant force R isMM2 R.dr(m (E +F, +..+F).drWi2 = -JM,MMM2M2 F. .drA?dr + M? F, dr +..JM,JMUM=W +W, +...+W, =ZWSo, the work done by the resultant force in the convergentforce system at any distance is egual to the algebraic sumof the work done by the components at the same distance
3、The work of the resultant force If there are forces acting on a particle, the net force is , When the particle moves from to , the work done by the resultant force is M1 M2 R n i M M n M M M M n M M M M W W W W F dr F dr F dr W R dr F F F dr = + + + = = + + + = = + ++ 1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 1 2 1 ( ) So, the work done by the resultant force in the convergent force system at any distance is equal to the algebraic sum of the work done by the components at the same distance. Ⅰ.The concept of work n R Fi =
II. The work of a common force1, The work of gravityLet the mass of the particle bem, it'sgoing to move from M, to M, under theZMVforce of gravity. Establish the coordinateM2Zas shown in the figure, X = O,Y = O,Z = -mgy22mgOByWi2= fmXdx+ Ydy +ZdzJMxWe have Wi2 = (2 (-mg)dz = mg(z1 - z2)For a system of particles, the work done by its gravity isW12 = Zm,g(zi1 - zi2) =(Zm,zi1 -Zm,zi2)g= (Mzcl - Mzc2)g = Mg(zc1 - zc2)Thus, the work of gravity depends only on the startingand ending position of the center of gravity, and hasnothing to do with the path the center of gravity travels
Ⅱ. The work of a common force 1、The work of gravity x y z M M2 M1 O 1 z 2 z mg Let the mass of the particle be ，it's going to move from to under the force of gravity. Establish the coordinate as shown in the figure, m M1 M2 X = 0,Y = 0,Z = −mg By = + + 2 1 1 2 M M W Xdx Ydy Zdz We have ( ) ( ) 1 2 1 2 2 1 W m g dz m g z z z z = − = − For a system of particles, the work done by its gravity is ( ) ( ) ( ) ( ) 1 2 1 2 12 1 2 1 2 C C C C i i i i i i i Mz Mz g Mg z z W m g z z m z m z g = − = − = − = − Thus, the work of gravity depends only on the starting and ending position of the center of gravity, and has nothing to do with the path the center of gravity travels
I. The work of a common force2, The work of the elastic forceF&Let the particle M move alongMMfrthe trajectory under the action ofKS0t2elastic forces, the original length ofMloOthe spring is lo, elasticity coefficientis k, in the elastic range, the elasticforce Fis F=-k(r-l.)rM2From Wiz = JM F.dr we have Wi2 = JM -k(r-lo)-drMMr2Wi2 = ["2 - k(r -lo)dr2=k[(ri -l。)2-(rz -l。)2]ork(82 -82)Wi2 =22
2、The work of the elastic force O M1 M2 M F 0 l 1 r 2 r r 0 r 1 2 Let the particle M move along the trajectory under the action of elastic forces, the original length of the spring is , elasticity coefficient is , in the elastic range, the elastic force is 0 l k F 0 0 F k(r l )r = − − From = 2 1 12 M M W F dr we have = − − 2 1 1 2 0 0 ( ) M M W k r l r dr 2 2 0 2 1 0 2 1 2 0 0 ( ) ( ) 2 1 ( ) 2 1 ( ) 2 1 2 1 k r l r l W k r l dr k r l r r r r = − − − = − − = − − or ( ) 2 1 2 2 2 W1 2 = k 1 − Ⅱ. The work of a common force
II. The work of a common force3, The work applied on a rigid body rotating in a fixed axisZThe force Facting on a fixed axis rotating rigidbody is shown in the figure. The work done by the百rigid body as it moves from positionPrto position PLis Wi2 =|M F,ds = [" F,-O,MdpJM00M.(F)= F ·OMOf whichMSoWi2 = [" M.(F)dp0When M.(F)is constant, W2 = M,(F)·(P2 -P,)If applied to a rigid body it is a couple of forces, torque vectoris m, then the work done by the couple isWi2 = [" m.dpWhen m, is constant,W2 = m,(P2 -P)
3、The work applied on a rigid body rotating in a fixed axis z O1 M O F F Fb Fn The force acting on a fixed axis rotating rigid body is shown in the figure. The work done by the rigid body as it moves from position to position is F 1 2 W F ds F O Md M M 12 1 2 1 2 1 = = So W M z (F)d 2 1 12 = When is constant, M (F) z ( ) ( ) 12 2 −1 W = M F z Of which Mz (F) = F O1 M If applied to a rigid body it is a couple of forces, torque vector is m , then the work done by the couple is W mz d = 2 1 12 When is constant, mz ( ) W12 = mz 2 −1 Ⅱ. The work of a common force
I. The work of a common forceZ4, The work of frictionmgAs shown in the figure, when the objectslides on a fixed surface, the work of slidingNXfriction is Wi2 = [-Fds =-{[ fNdsSWhenfN is constant, Wz =-fNsWhere s is the arc length of the particle. It follows that the workdone by friction depends on the path traveled by the particle.As shown in the figure, when the rigid body rollspurely on a fixed plane, the frictionalforce and theFnormalreaction force act on the instantaneous777C*/77777center. Since the velocity of the instantaneous centerNis equal to zero, there is no displacement in theinstantaneous center, so the frictionalforce and thenormalreaction forcedono work
4、The work of friction x y z O mg N F s M1 M2 As shown in the figure, when the object slides on a fixed surface, the work of sliding friction is = − = − 2 1 2 1 12 s s s s W Fds f Nds When is constant, f N W = − f Ns 12 Where is the arc length of the particle. It follows that the work done by friction depends on the path traveled by the particle. s N F C As shown in the figure, when the rigid body rolls purely on a fixed plane, the frictional force and the normal reaction force act on the instantaneous center. Since the velocity of the instantaneous center is equal to zero, there is no displacement in the instantaneous center, so the frictional force and the normal reaction force do no work. Ⅱ. The work of a common force