Question

A vessel with square cross-section and height of 6 m is vertically partitioned. A small window of \( 100 \, \text{cm}^2 \) with hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density \( 1.5 \times 10^3 \, \text{kg/m}^3 \). What force one needs to apply on the hinged door so that it does not open?

Hint:

For problems involving fluid pressure and forces, the total force on an object can be computed by integrating the pressure over the area. Pay attention to the different fluid densities and heights.

Answer 1

Correct Answer: 150

The external force \( F_{\text{ext}} \) to keep the door closed is calculated as follows: \[ F_{\text{ext}} + F_w = F_t \] Here, \( F_w \) represents the force from water, and \( F_t \) is the total force on the window. For equilibrium, the equation simplifies to: \[ F_{\text{ext}} = F_t - F_w \] The total force on the window, \( F_t \), is given by: \[ F_t = (\rho_1 + \rho_2) g h A \] Similarly, the force due to water, \( F_w \), is: \[ F_w = (\rho_1 + \rho_2) g h A \] Therefore, the required external force is: \[ F_{\text{ext}} = (1500 - 1000) \times 10 \times 10^{-4} \times 150 \] \[ = 150 \, \text{N} \]