GSO002 Problem 2

1. Definition of Pressure and Understanding the Problem

In physics, when a force of magnitude $F$ acts perpendicular to a surface of area $S$ , the pressure $P$ is defined as:

P = \frac{F}{S}

The goal is to indent the sponge as deeply as possible. Since the sponge "indents more deeply the greater the pressure," we need to find the conditions that maximize pressure.

2. Identifying the Condition for Maximum Pressure

In the formula $P = \frac{F}{S}$ , the block weight $F = W$ is constant. Therefore, pressure $P$ is maximized when the contact area $S$ is minimum.

The three edge lengths are $a = 0.25 \, \text{m}$ , $b = 0.20 \, \text{m}$ , $c = 0.13 \, \text{m}$ . The three possible face areas are:

Face formed by edges $a$ and $b$ : $S_1 = 0.25 \times 0.20 = 0.050 \, \text{m}^2$
Face formed by edges $a$ and $c$ : $S_2 = 0.25 \times 0.13 = 0.0325 \, \text{m}^2$
Face formed by edges $b$ and $c$ : $S_3 = 0.20 \times 0.13 = 0.026 \, \text{m}^2$

Comparing these, the smallest area is $S_3 = 0.026 \, \text{m}^2$ .

3. Numerical Calculation

Using the minimum area $S_{min} = 0.026 \, \text{m}^2$ , compute the maximum pressure $P_{max}$ :

P_{max} = \frac{W}{S_3} = \frac{39}{0.026}

Converting to a ratio of integers:

P_{max} = \frac{39000}{26}

Since both $39$ and $26$ are multiples of $13$ ( $39 = 13 \times 3$ , $26 = 13 \times 2$ ):

P_{max} = \frac{3 \times 13000}{2 \times 13} = \frac{3000}{2} = 1500 \, \text{Pa}

Therefore, the value of the required pressure is $1500$ .

Answer: 1500

GSO002 Problem 2

Sponge Quality Inspection: Measuring Maximum Pressure with a Rectangular Block

Problem Statement

Constraints

Input Format

Solution

1. Definition of Pressure and Understanding the Problem

2. Identifying the Condition for Maximum Pressure

3. Numerical Calculation