For hydraulic systems, the total force exerted by water pressure on a surface is the product of the water pressure...

1. TRANSLATE the problem information

• Given information:
  • Two square plates with larger side = \(3 \times \mathrm{smaller\;side}\)
  • Water pressure = 18.0 pounds per square inch on both plates
  • Total force on both plates = 1,800 pounds
  • Need: Force on larger plate only

• What this tells us: We need to work with areas and use the relationship Force = Pressure × Area

2. TRANSLATE to set up variables

• Let \(\mathrm{s}\) = side length of smaller square plate (inches)
• Then larger square plate side length = \(3\mathrm{s}\) inches

3. INFER the areas and strategy

• Area of smaller plate = \(\mathrm{s}^2\) square inches
• Area of larger plate = \((3\mathrm{s})^2 = 9\mathrm{s}^2\) square inches
• Total area = \(\mathrm{s}^2 + 9\mathrm{s}^2 = 10\mathrm{s}^2\) square inches
• Strategy: Use total force to find \(\mathrm{s}^2\), then calculate force on larger plate

4. SIMPLIFY to find \(\mathrm{s}^2\)

• Total force = Pressure × Total area
• \(1,800 = 18.0 \times 10\mathrm{s}^2\)
• \(1,800 = 180\mathrm{s}^2\)
• \(\mathrm{s}^2 = 1,800 \div 180 = 10\)

5. SIMPLIFY to find force on larger plate

• Force on larger plate = Pressure × Area of larger plate
• \(\mathrm{Force\;on\;larger\;plate} = 18.0 \times 9\mathrm{s}^2\)
• \(\mathrm{Force\;on\;larger\;plate} = 18.0 \times 9 \times 10 = 1,620\;\mathrm{pounds}\)

Answer: 1620

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students struggle to set up the relationship between the plate sizes correctly. They might think the larger plate has area \(3\mathrm{s}^2\) instead of \(9\mathrm{s}^2\) (forgetting to square the 3 when going from side length to area).

This leads them to calculate total area as \(\mathrm{s}^2 + 3\mathrm{s}^2 = 4\mathrm{s}^2\) instead of \(10\mathrm{s}^2\). Using the equation \(1,800 = 18.0 \times 4\mathrm{s}^2\) gives \(\mathrm{s}^2 = 25\), leading to force on larger plate = \(18.0 \times 3 \times 25 = 1,350\;\mathrm{pounds}\). This causes them to get stuck since 1350 is not the answer, leading to guessing.

Second Most Common Error:

Poor INFER reasoning: Students calculate \(\mathrm{s}^2\) correctly as 10, but then get confused about what they're solving for. Instead of calculating force on the larger plate specifically, they might calculate the force per unit area or try to find some other quantity.

This leads to confusion about the final calculation step and random answer selection.

The Bottom Line:

This problem tests whether students can correctly translate the "3 times larger" relationship to area (which becomes 9 times larger), not just the linear dimension scaling.