运动学 ∴n=V+ 即在任一瞬时动点的绝对速度等于其牵连速度与相对速度的 矢量和,这就是点的速度合成定理 说明:va动点的绝对速度 v动点的相对速度; a动点的牵连速度,是动系上一点(牵连点)的速度 D动系作平动时,动系上各点速度都相等。 I)动系作转动时,v必须是该瞬时动系上与 动点相重合点的速度。 26
26 说明： va—动点的绝对速度； vr—动点的相对速度； ve—动点的牵连速度，是动系上一点(牵连点)的速度 I) 动系作平动时，动系上各点速度都相等。 II) 动系作转动时，ve必须是该瞬时动系上与 动点相重合点的速度。 即在任一瞬时动点的绝对速度等于其牵连速度与相对速度的 矢量和，这就是点的速度合成定理。 a e r v =v +v
Kinematics The equation given in the theorem of composition of velocities is a vector equation at an instant, which includes six components such as magnitudes and directions. If any four components are given, the other two unknowns can be solved from this equation 2. Examples of theorem application Example The small vehicle in a bridge cane v moves along its horizontal track with a uniform velocity Vh, and weight A moves vertically at a velocity v relative to the small vehicle Find the absolute velocity of weight A 27
27 The equation given in the theorem of composition of velocities is a vector equation at an instant, which includes six components such as magnitudes and directions. If any four components are given, the other two unknowns can be solved from this equation. 2. Examples of theorem application [Example1] The small vehicle in a bridge cane moves along its horizontal track with a uniform velocity vh，and weight A moves vertically at a velocity vv relative to the small vehicle. Find the absolute velocity of weight A
动学 点的速度合成定理是瞬时矢量式,共包括大小,方向六个 元素,已知任意四个元素,就能求出其他两个, 二.应用举例 例1桥式吊车已知:小 车水平运行,速度为v平, 物块A相对小车垂直上升 的速度为v。求物块的运p 行速度。 O 28
28 点的速度合成定理是瞬时矢量式，共包括大小‚方向 六个 元素，已知任意四个元素，就能求出其他两个。 二．应用举例 [例1] 桥式吊车 已知：小 车水平运行，速度为v平， 物块A相对小车垂直上升 的速度为v⊥。求物块A的运 行速度
Kinematics Solution: moving point: weight A MCS: small vehicle SCS: earth ground Relative motion: rectilinear Relative velocity v =v, direction convected motion translation convected velocity ve=Vm, direction = 4 Absolute motion: curvilinear Ve=v absolute velocity va magnitude and direction are unknown Employing the theorem of composition of velocities, yields V=1+ ae Draw the parallelogram of velocities shown in the figure the magnitude and direction of the velocity of the weight A can be obtained as 2 V=V A v +V=/Vr+ r ⊥ 6=t 平 29
29 Draw the parallelogram of velocities shown in the figure, the magnitude and direction of the velocity of the weight A can be obtained as 2 2 2 = = + = + ⊥ v v v v v v A a e r 2 平 v 平 v − ⊥ = 1 tg Solution：moving point: weight A MCS: small vehicle SCS: earth ground Relative motion: rectilinear; Relative velocity vr =vv , direction convected motion: translation; convected velocity ve=vh , direction → Absolute motion : curvilinear; Absolute velocity va ,magnitude and direction are unknown Employing the theorem of composition of velocities, yields a e r v =v +v
动学 解:选取动点:物块A 动系:小车 静系:地面 相对运动:直线 相对速度v=v方向个 牵连运动:平动 牵连速度v=v平方向→ = 4 绝对运动:曲线: Ve=v 绝对速度vn的大小,方向待 求 由速度合成定理:=v+vr 作出速度平四边形如图示,则物块A的速度大小和方向为 2 2 νA=vn=Vv+v v十v ⊥ 6=tg 平 30
30 作出速度平四边形如图示，则物块Ａ的速度大小和方向为 2 2 2 = = + = + ⊥ v v v v v v A a e r 2 平 v 平 v − ⊥ = 1 tg 解：选取动点: 物块A 动系: 小车 静系: 地面 相对运动: 直线; 相对速度vr =v⊥ 方向 牵连运动: 平动; 牵连速度ve =v平 方向→ 绝对运动: 曲线; 绝对速度va 的大小,方向待 求 由速度合成定理： a e r v =v +v