Dynamics 4. Work done by special forces: ZAMi 1)Work done by gravity M For a particle the projections of M2 the gravity force acting on it on the three coordinate axes are X=0,y=0,Z=-mg Hence, W=J-mgdz=mg(=1-22) For a system of particles, W=2W=2mgzn-En)=Mg(=ci-zc2) The work done by the gravity on a system is equal to the product of the total weight of the system times the difference between the initial and final height of the center of gravity of the system. It does not depend on the path of each particle
11 4. Work done by special forces: 1) Work done by gravity. For a particle the projections of the gravity force acting on it on the three coordinate axes are Hence, For a system of particles, The work done by the gravity on a system is equal to the product of the total weight of the system times the difference between the initial and final height of the center of gravity of the system. It does not depend on the path of each particle. X =0,Y =0, Z =−mg = − = − 2 1 ( ) 1 2 z z W mgdz mg z z ( ) ( ) i i i1 i2 C1 C2 W =W =m g z −z =Mg z −z
力单 四.常见力的功 M 1.重力的功 M 质点:重力在三轴上的投影: mgh X=0,=0,Z=-mg Wy=「-mgt=mg(=1-=z2) 1 质点系:W=W=m8(=1-=2)=Mg(=1-=c2) 质点系重力的功,等于质点系的重量与其在始末位置重 心的高度差的乘积,而与各质点的路径无关。 12
12 四．常见力的功 1．重力的功 = − = − 2 1 ( ) 1 2 z z W mgdz mg z z 质点系： ( ) ( ) i i i1 i2 C1 C2 W =W =m g z −z =Mg z −z 质点系重力的功，等于质点系的重量与其在始末位置重 心的高度差的乘积，而与各质点的路径无关。 X =0,Y =0, Z =−mg 质点：重力在三轴上的投影：
Dynarnic 2)Work done by an elastic force Assuming the length of the inextensible spring to be lo, the force in the elastic limit is F=k(r-L r where Po=r/r. The factor K is called the stiffness of the spring. It is the force required to extend the spring by a unit length, and its dimension is [K]=N/M, or N/cm W=Fdr=f-k(r-1o )odr M1 元2=d=①d()=d(r2)=h 2r M2 6 =kr(-lo)-(r-1)21 With 6, -r-I, -r) -Io, we get W=(61-82) The work done by an elastic force depends only on the initial and final deformation and does not depend on the path. 13
13 2) Work done by an elastic force Assuming the length of the inextensible spring to be , the force in the elastic limit is , where . The factor K is called the stiffness of the spring. It is the force required to extend the spring by a unit length ,and its dimension is [K]= N/M, or N/cm. 0 l 0 0 F =−k(r−l )r r r/r 0 = = = − − 2 1 2 1 0 0 ( ) m M M M W F dr k r l r dr d r dr r d r r r dr r r r dr = = = ( )= 2 1 ( ) 2 1 2 0 2 0 0 ( ) 2 hence ( ) 2 1 2 1 d r l k W k r l dr r r r r = − − = − − [( ) ( ) ], With , ,we get 2 1 1 0 2 2 0 2 2 0 2 1 0 r l r l r l r l k = − − − = − = − ( ) 2 2 2 2 = 1 − k W The work done by an elastic force depends only on the initial and final deformation and does not depend on the path
学 2.弹性力的功 弹簧原长lo,在弹性极限内F=-k(r-1) k弹簧的刚度系数,表示使弹簧发生单位 变形时所需的力。Nm,Ncm。F=F/r。 w=Fdr=]-k(r-lo Fo dr M M n ro d r==dr= d(rr) )=2n(2)=d EXX 12 \M2 62 2[(-)2-(2-0)2]令6=-062=n2 即W=k(12-02)弹性力的功只与弹簧的起始变形和终了 变形有关,而与质点运动的路径无关
14 2．弹性力的功 弹簧原长 ，在弹性极限内 k—弹簧的刚度系数，表示使弹簧发生单位 变形时所需的力。N/m , N/cm。 。 0 l 0 0 F =−k(r−l )r r r/r 0 = = = − − 2 1 2 1 0 0 ( ) m M M M W F dr k r l r dr d r dr r d r r r dr r r r dr = = = ( )= 2 1 ( ) 2 1 2 0 2 0 0 ( ) 2 ( ) 2 1 2 1 d r l k W k r l dr r r r r = − − =− − 1 1 0 2 2 0 2 2 0 2 1 0 [( ) ( ) ] , 2 r l r l r l r l k = − − − 令 = − = − ( ) 2 2 2 1 2 = − k 即 W 弹性力的功只与弹簧的起始变形和终了 变形有关，而与质点运动的路径无关
Dynamics 3)Work done by the gravitational force: W=Gmm(L-l) The work done by the gravitational force depends only on the initial and final positions r, and r, and does not depend on the path 4)Work done by forces applied to a rotating body Let a force F act on the point m of a rigid body rotating about an axis Z. Determine the work done by the force F during a turn by a finite angle At the locus of the point M the force iSF=F+Fn+ F W =F ds=F rdo=m ( F)do(=p2-1)hence w=m(F)do The work done by a force applied to a rotating p1 body is equal to the work done by the torque If on a body a force couple acts lying in a plane normal to the axis about which the body rotates Fb we get w= mdo If m=const, then W=m(2-p) M Note the definition of the sign of the work
15 3) Work done by the gravitational force: . The work done by the gravitational force depends only on the initial and final positions r1 and r2 and does not depend on the path. 4)Work done by forces applied to a rotating body: Let a force F act on the point M of a rigid body rotating about an axis Z. Determine the work done by the force F during a turn by a finite angle , . ) 1 1 ( 2 1 0 r r W = Gmm − At the locus of the point M the force is F F Fn Fb = + + W =F ds=F rd=mz (F)d ( ) =2 −1 = 2 1 hence ( ) W mz F d The work done by a force applied to a rotating body is equal to the work done by the torque. If on a body a force couple acts lying in a plane normal to the axis about which the body rotates we get = 2 1 W md If m=const, then ( ) W = m 2 −1 Note the definition of the sign of the work