Solution
Let us analyze the forces on the glass sphere in the frame of the accelerating vessel. The vessel accelerates to the left ($a$), so there is a pseudo force $ma$ acting to the right on the sphere. The wall is inclined inwards (overhanging) as shown to constrain the sphere against this force.
Forces acting on the sphere:
- Weight: $W = mg = \rho V g$ (Downwards)
- Pseudo force: $F_p = ma = \rho V a$ (Rightwards)
- Buoyant force:
- Vertical: $F_{B,y} = \rho_w V g$ (Upwards)
- Horizontal: $F_{B,x} = \rho_w V a$ (Leftwards, due to pressure gradient)
- Normal reaction from the wall ($N_W$): Since the wall leans to the left (angle $\theta$ with horizontal), the normal force acts perpendicular to it, pointing downwards and to the left.
- Normal reaction from the bottom ($N_B$): Acts vertically upwards.
Equilibrium Equations:
1. Horizontal Equilibrium:
The forces pushing right (pseudo force) must be balanced by forces pushing left (horizontal buoyancy + horizontal component of wall normal).
$$F_p = F_{B,x} + N_W \sin\theta$$
$$\rho V a = \rho_w V a + N_W \sin\theta$$
$$N_W \sin\theta = (\rho – \rho_w) V a$$
$$N_W = \frac{(\rho – \rho_w) V a}{\sin\theta}$$
2. Vertical Equilibrium:
Total upward forces = Total downward forces.
$$N_B + F_{B,y} = mg + N_W \cos\theta$$
Note: The vertical component of $N_W$ is downwards because the wall overhangs the sphere.
$$N_B + \rho_w V g = \rho V g + N_W \cos\theta$$
$$N_B = (\rho – \rho_w) V g + N_W \cos\theta$$
Substituting the expression for $N_W$ into the vertical equation:
$$N_B = (\rho – \rho_w) V g + \left[ \frac{(\rho – \rho_w) V a}{\sin\theta} \right] \cos\theta$$ $$N_B = (\rho – \rho_w) V g + (\rho – \rho_w) V a \cot\theta$$ $$N_B = (\rho – \rho_w) V (g + a \cot\theta)$$