### Statement ### Statement $ 2.1.50. $ On a smooth horizontal plane there is a wedge with an angle $ α $ at the base. A$ 2.1.50. $ On a smooth horizontal plane there is a wedge with an angle $ α $ at the base. Abody of mass $ m $ placed on a wedge descends with acceleration directed at an body of mass $ m $ placed on a wedge descends with acceleration directed at an angle $ β > α $ to the horizontal. Determine the mass of the wedge. angle $ β > α $ to the horizontal. Determine the mass of the wedge.   ### Solution ### Solution   #### 1) Kinematics #### 1) Kinematics Let us consider the instantaneous kinematics of the system.\ Let us consider the instantaneous kinematics of the system.\ Let $ M $ be the mass of the wedge.\ Let $ M $ be the mass of the wedge.\ Let $ V $ be the velocity of the wedge. \ Let $ V $ be the velocity of the wedge. \ Let $ v $ be the velocity of the block relative to the wedge. Let $ v $ be the velocity of the block relative to the wedge. \ \ The velocity components of the block relative to the ground are: The velocity components of the block relative to the ground are: $$ v_x = v \cos { \alpha } -|V|, \qquad v_y= v \sin { \alpha } $$ $$ v_x = v \cos { \alpha } -|V|, \qquad v_y= v \sin { \alpha } $$ From there we obtain \begin {equation} From there we obtain \begin {equation} \tan {\beta }=\frac {v\sin {\alpha }}{v\cos {\alpha }-|V|} \tan {\beta }=\frac {v\sin {\alpha }}{v\cos {\alpha }-|V|} \end {equation}\end {equation}#### 2) Displacement of CM #### 2) Displacement of CM Since no external force acts on the system horizontally, the horizontal displacement of the center of mass $ \Delta x_ { cm_x } $ must equal to 0, therefore we can write: Since no external force acts on the system horizontally, the horizontal displacement of the center of mass $ \Delta x_ { cm_x } $ must equal to 0, therefore we can write: $$ \Delta x_ { cm_x } = \frac { m \vec { \Delta x } _ { mx } +M \vec { \Delta x } _ { Mx }}{ m +M } $$ $$ \Delta x_ { cm_x } = \frac { m \vec { \Delta x } _ { mx } +M \vec { \Delta x } _ { Mx }}{ m +M } $$ $$ 0 = \frac { mv_x \Delta t - M|V| \Delta t }{ m+M } $$ $$ 0 = \frac { mv_x \Delta t - M|V| \Delta t }{ m+M } $$ \begin {equation} \begin {equation} M|V|=m(v \cos {\alpha }-|V|) M|V|=m(v \cos {\alpha }-|V|) \end {equation} \end {equation} $$ |V|= \frac { mv \cos { \alpha }}{ M+m } $$ $$ |V|= \frac { mv \cos { \alpha }}{ M+m } $$
@@ -36,10 +36,10 @@2) Displacement of CM
#### 3) Substitution #### 3) Substitution
Now we substitute V into first equation:
Now we substitute V into first equation:
$$ \tan { \beta } = \frac { v }{ v- \frac { mv }{ M+m }} \tan { \alpha } $$ $$ \tan { \beta } = \frac { v }{ v- \frac { mv }{ M+m }} \tan { \alpha } $$
And obtain the following: \boldmath $$ M= \frac { m \tan { \alpha }}{ \tan { \beta } + \tan { \alpha }} $$
And obtain the following: \boldmath $$ M= \frac { m \tan { \alpha }}{ \tan { \beta } - \tan { \alpha }} $$
#### Answer #### Answer
$$ M= \frac { m \tan { \alpha }}{ \tan { \beta } + \tan { \alpha }} $$ $$ M= \frac { m \tan { \alpha }}{ \tan { \beta } - \tan { \alpha }} $$
