同济大学理论力学孙杰习题解答4(练习册P17-P26).ppt

理论力学,习题解答,静力学 4,练习册,运动学 1,P17 习题：1,Fcosb = FScosq + FNsinq,FNcosq + Fsinb = P + FSsinq,FN + Fsin(b -q )= Pcosq,FS + Psinq = Fcos(b -q ),Fcos( b -q -jm )= Psin(jm+q ),b -q -jm = 0,b = jm + q,当：,时,Fmin = P sin(jm+q ),Fy = 0,Fx = 0,即,= 常数,几何法,Fmin,而 FS = fSFN,= FNtanjm,受力图,Fy = 0,FNcosq + FSsinq = Q,而 | FS | fSFN,Fx = 0,FNsinq FScosq = F,且 | FS | fSFN,(1),(2),(3),令 Fmin 0,q arctan fS,设：B有下滑的趋势,sinq fS cosq,P18 习题：3,受力图,已知 fS,P20 习题：4,l,MA = 0,FNAB - MA = 0,MC - FNEB + FSEC = 0,MC = 0,FS = fSFN,设：MC较小，,套筒B有向上滑动趋势；,临界平衡。,= 49.6 Nm,设：MC较大，,FS反向；,临界平衡。,MC - FNEB - FSEC = 0,= 70.4 Nm,49.6 Nm MC 70.4 Nm,平衡时：,x = ( l + AM )cosj,y = ( l - AM )sinj,= 100cos4t,= 20sin4t,(sinj)2 + (cosj)2 = 1,= - 1600 cos4t,= - 320 sin4t,当 j = 4t = 0；,ax = - 16 m/s2,ay = 0,a = 16 m/s2,椭圆,P21 习题：2,直角坐标法,坐标,轨迹方程,(固定),(向左),即,j = 4t,-,= - c,匀变速,= - 23.87 m/s2,= 0.209 s,= 25 m/s2,= 34.57 m/s2,v = 20 - ct,P22 习题：3,自然法,匀变速,(公式),B点：,t,t,v = 20 - ct,P22 习题：3,vA = 20 ct1,vB = 20 ct2,ct1 = 10 m/s,ct2 = 15 m/s,sAB = sB sA,则：,D t = t2 t1,而：,= - c,( 常数 ),= 25 m/s2,= 34.57 m/s2,O1定轴转动,j = 15p t,= 15p rad/s,= 0,AB平行移动,vA = rw,aAt = ra,aAn = rw 2,= 3p,= 9.42 m/s,vM =,= 0,aMt =,= 45p2,= 444 m/s2,aMn =,= aM,P23 习题：1,刚体基本运动,杆的位置,n,w5 = w4,= p rad/s,n1= 30 r/min,= 16,= 0.785 cm/s,则：,vB = r5w5,P24 习题：3,i 14,齿轮箱,rad/s,A,B,C,D,E,O,P,r1,vD = rw,vE = Rw,= 5 p cm/s,瞬心 P：,EP + DP = 2r1,则：,vO = vD,= 5.236 cm/s,= wO,P25 习题：1,瞬心法,图(P ),O,A,B,C,D,E,vA = ( r1 + r2 )w,= 50 cm/s,= 10 rad/s,vB = 0,= 70.7 cm/s,= vE,vD = 2 r1w1,= 100 cm/s,瞬心 B ：,P26 习题：3,刚体运动,瞬心法,方向,vA = Rw,D,PAB,vB = Lsin30wAB,瞬心 D ：,P26 习题：4,v 方向关系,w ：,刚体,瞬心法,(某瞬时),vA = rw,vBA = 2vA,= 2rw,= w,= 3 rad/s,= 5.2 rad/s,60,60,P25 习题：2,基点法,BA,r,vBA = 2rw