Solution to Question 13
1. Kinematic Analysis
Let $a$ be the magnitude of the horizontal acceleration of wedge A. Since there is no external horizontal force on the system, the center of mass remains stationary horizontally.
- Wedge A: Moves to the left with acceleration $a_A = a$.
- Wedge B: Slides down the face of A. Let its horizontal acceleration be $a_{Bx}$ and vertical be $a_{By}$.
- Block C: Rests on the frictionless horizontal top of B. Since there is no horizontal force acting on C (friction is zero), its horizontal acceleration is zero. It only moves vertically with B.
Momentum Conservation (Horizontal): The total horizontal momentum of the system is conserved and is zero. $$M(-a) + M(a_{Bx}) + m(0) = 0 \implies a_{Bx} = a$$ So, wedge B moves to the right with a horizontal acceleration $a$.
Constraint Relation (A and B): The relative acceleration between A and B must be along their contact plane, which makes an angle of $45^\circ$ with the horizontal. The acceleration of B relative to A is $\vec{a}_{B/A} = \vec{a}_B – \vec{a}_A$. $$\vec{a}_{B/A} = (a \hat{i} + a_{By} \hat{j}) – (-a \hat{i}) = (2a) \hat{i} + a_{By} \hat{j}$$ For this vector to be along the $45^\circ$ slope (which has a slope of $\tan(135^\circ) = -1$), the ratio of its vertical to horizontal components must be $-1$: $$\frac{a_{By}}{2a} = -1 \implies a_{By} = -2a$$ Thus, wedge B has a vertical acceleration of $2a$ downwards. Since block C stays in contact with the horizontal top of B, its acceleration is $\vec{a}_C = 2a$ downwards.
2. Dynamic Analysis (Force Equations)
Let $N_1$ be the normal force between A and B, and $N_2$ be the normal force between B and C.
For Wedge A (Horizontal Motion): The normal force $N_1$ acts on A perpendicular to the incline ($45^\circ$). Its horizontal component pushes A to the left. $$N_1 \sin 45^\circ = Ma$$ $$\frac{N_1}{\sqrt{2}} = Ma \implies N_1 = \sqrt{2}Ma$$
For Block C (Vertical Motion): The forces on C are gravity $mg$ (down) and the normal force $N_2$ from B (up). Its acceleration is $2a$ (down). $$mg – N_2 = m(2a)$$ $$N_2 = m(g – 2a)$$
For Wedge B (Vertical Motion): The forces on B are gravity $Mg$ (down), the normal force $N_2$ from C (down), and the vertical component of $N_1$ from A (up). Its acceleration is $2a$ (down). $$Mg + N_2 – N_1 \cos 45^\circ = M(2a)$$ $$Mg + N_2 – \frac{N_1}{\sqrt{2}} = 2Ma$$
3. Solving for Acceleration
Now, substitute the expressions for $N_1$ and $N_2$ into the vertical force equation for B: $$Mg + m(g – 2a) – \frac{\sqrt{2}Ma}{\sqrt{2}} = 2Ma$$ $$Mg + mg – 2ma – Ma = 2Ma$$ Group the terms with $g$ and $a$: $$(M+m)g = 3Ma + 2ma$$ $$(M+m)g = a(3M + 2m)$$ Solve for $a$: $$a = \frac{(M+m)g}{3M + 2m}$$
The question asks for the acceleration of block C. As determined in the kinematic analysis, $a_C = 2a$: $$a_C = 2 \left( \frac{(M+m)g}{3M + 2m} \right)$$ $$a_C = \frac{2(M+m)g}{3M + 2m}$$