Solution
Analysis of the System:
We consider the stability of the wedge of mass $M$ as the small block of mass $m$ slides down its inclined face. The wedge rests on a horizontal floor with sufficient friction to prevent sliding. The condition for the wedge to topple is determined by the torques acting about the pivot point.
1. Forces and Geometry
Let the angle of the wedge be $\theta$. The vertical face has height $h$. The base width is $b = h \cot\theta$.
The forces exerted by the block on the wedge at the top corner are:
- Normal Force ($N$): Acts perpendicular to the inclined surface, pushing the wedge down and to the right. $$ N = mg \cos\theta $$ Horizontal component: $N_x = N \sin\theta$ (Right, $\hat{i}$)
- Kinetic Friction ($f$): Since the block slides down, the friction on the block is up the incline. By Newton’s 3rd Law, the friction on the wedge acts down the incline. $$ f = \mu N = \mu mg \cos\theta $$ Horizontal component: $f_x = -f \cos\theta$ (Left, $-\hat{i}$)
2. Torque Analysis
The wedge will topple about its bottom-right corner (the “toe”), denoted as $O$ in the diagram. We calculate torques about this point.
Toppling Torque ($\tau_{topple}$):
The forces from the block are applied at the top corner $(0, h)$ relative to the pivot $(0,0)$. The vertical line of action passes through the pivot, so vertical components create zero torque. Only horizontal components contribute.
- $N_x$ pushes to the right (tries to topple clockwise).
- $f_x$ pushes to the left (opposes toppling, counter-clockwise).
Net horizontal force: $F_{net,x} = N \sin\theta – f \cos\theta$.
$$ \tau_{topple} = h \cdot (N \sin\theta – f \cos\theta) $$
Restoring Torque ($\tau_{restore}$):
The weight of the wedge $Mg$ acts at its center of mass. For a right-triangular wedge, the CM is at a horizontal distance of $b/3$ from the vertical face.
$$ b = h \cot\theta $$
$$ \tau_{restore} = Mg \left( \frac{b}{3} \right) = Mg \frac{h \cot\theta}{3} $$
3. Condition for Stability
For the wedge not to topple, the restoring torque must differ from or exceed the toppling torque:
$$ \tau_{topple} \le \tau_{restore} $$ $$ h (N \sin\theta – f \cos\theta) \le Mg \frac{h \cot\theta}{3} $$Canceling $h$ and substituting $f = \mu N$:
$$ N (\sin\theta – \mu \cos\theta) \le \frac{Mg \cot\theta}{3} $$Substitute $N = mg \cos\theta$:
$$ mg \cos\theta (\sin\theta – \mu \cos\theta) \le \frac{Mg \cos\theta}{3 \sin\theta} $$Assuming $\theta \neq 90^\circ$, we cancel $g \cos\theta$:
$$ m (\sin\theta – \mu \cos\theta) \le \frac{M}{3 \sin\theta} $$Solving for $m$:
Note: This solution is valid when $\mu < \tan\theta$. If $\mu \ge \tan\theta$, the term in the denominator becomes non-positive, implying the system is stable for any mass $m$ (physically, the friction dominates or sliding doesn't occur in the same way).