Problem Analysis
A wedge of mass $m$ closes a slit at the bottom right edge. We need to find the maximum height $h$ of water before the wedge slips. We must analyze the pressure forces and friction.
Step 1: Analyze Fluid Forces
The water pressure acts perpendicular to the slanted surface of the wedge. We can decompose the total force exerted by the water into horizontal ($F_H$) and vertical ($F_V$) components.
- Horizontal Force ($F_H$): Acts to the right, pushing the wedge against the wall. $$F_H = \text{Average Pressure} \times \text{Vertical Projected Area} = \left(\frac{1}{2}\rho g h\right) (h \cdot b) = \frac{1}{2} \rho g h^2 b$$
- Vertical Force ($F_V$): Acts UPWARDS. Because the water is under the slant (in the corner configuration), the pressure pushes the wedge up. $$F_V = F_H \tan \theta = \frac{1}{2} \rho g h^2 b \tan \theta$$
Step 2: Force Equilibrium
The wedge tends to move UP due to the vertical water force ($F_V$). The forces opposing this upward motion are:
- Gravity ($mg$) acting down.
- Friction ($f$) acting down (opposing impending upward motion).
For the maximum height $h$, the wedge is on the verge of slipping up:
$$F_{up} = F_{down}$$ $$F_V = mg + f_{max}$$Where limiting friction $f_{max} = \mu N$. Since the normal force $N$ is provided by the horizontal water force $F_H$ pushing against the wall, $f_{max} = \mu F_H$.
$$F_V = mg + \mu F_H$$Step 3: Solve for h
Substitute the expressions for $F_V$ and $F_H$:
$$\frac{1}{2} \rho g h^2 b \tan \theta = mg + \mu \left( \frac{1}{2} \rho g h^2 b \right)$$Rearranging to isolate $h$ terms:
$$\frac{1}{2} \rho g h^2 b \tan \theta – \frac{1}{2} \rho g h^2 b \mu = mg$$ $$\frac{1}{2} \rho g h^2 b (\tan \theta – \mu) = mg$$ $$h^2 = \frac{2mg}{\rho g b (\tan \theta – \mu)}$$ $$h = \sqrt{\frac{2m}{\rho b (\tan \theta – \mu)}}$$