Question

A rectangular glass block 4.00cm×1.50cm×0.05×0.08 is situated in a liquid with the longest edges of the glass vertical and with the top side 2.00cm below the liquide.The density of the liquid is 0.85g/cm³.Calculate the magnitude of the pressure due to the liquid acting on the top and bottom faces of the block

Answer 1

The third dimension of he block is not clearly stated, so we'll solve the task for the rectangular block with dimensions 4.00 cm×1.50 cm×k cm. We assume that k < 4.0 cm.

The areas of the top and bottom faces are equal to 1.50 cm×k cm = 1.5k cm². The pressure exerted on the top face is $p_t = \rho g h = (850 \text{\,kg/m}^3) \cdot9.81\mathrm{\,N/kg}\cdot 0.02\,\text{m} = 166.8\,\text{Pa}.$

The corresponding force is equal to $p_t\cdot S_t = p_t\cdot 0.015\,\text{m}\cdot k/100\,\text{m}= 0.025k \,\text{N}.$

The pressure exerted on the bottom face is bigger than on top because of the larger depth:$p_b= \rho g (h+a) = (850 \text{\,kg/m}^3) \cdot9.81\mathrm{\,N/kg}\cdot (0.02+0.04)\,\text{m} = 500.3\,\text{Pa}.$

The corresponding force is equal to $p_b\cdot S_b= p_b\cdot 0.015\,\text{m}\cdot k/100\,\text{m}= 0.075k \,\text{N}.$

We can see that the pressure exerted on the bottom face is three times the pressure on the top face.

But if k is the longest side, the pressure on the bottom face will be $p_b= \rho g (h+a) = (850 \text{\,kg/m}^3) \cdot9.81\mathrm{\,N/kg}\cdot (0.02+k\cdot0.01)\,\text{m}$ and it will be bigger with bigger k.