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Chapter 17: Problem 35

A brick has faces with the following dimensions: face 1 is \(1 \mathrm{cm}\) by \(2 \mathrm{cm} ;\) face 2 is \(2 \mathrm{cm}\) by \(3 \mathrm{cm} ;\) face 3 is \(1 \mathrm{cm}\) by \(3 \mathrm{cm} .\) On which face should the brick be placed if it is to have the smallest change in dimensions due to its own weight? Explain.

Short Answer

Expert verified
Place the brick on Face 2 with dimensions 2 cm by 3 cm.

Step by step solution

01

Identify Brick Faces

We are given three faces of a brick with specific dimensions. These include Face 1: 1 cm by 2 cm, Face 2: 2 cm by 3 cm, and Face 3: 1 cm by 3 cm.
02

Understand the Problem

The problem is asking us to determine which face should be placed at the bottom to minimize the change in dimensions due to the brick's weight. Placing the brick on the face with the largest area will minimize pressure and potentially reduce deformation.
03

Calculate the Area of Each Face

Determine the area of each face:- Face 1: Area = 1 cm \(\times\) 2 cm = 2 cm虏- Face 2: Area = 2 cm \(\times\) 3 cm = 6 cm虏- Face 3: Area = 1 cm \(\times\) 3 cm = 3 cm虏
04

Select the Face with Maximal Area

Among the calculated areas, Face 2 has the largest area of 6 cm虏. Placing the brick on Face 2 will distribute the weight over a larger surface area, minimizing deformation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Area Calculation
Understanding the area of each face of the brick is essential. It influences how the weight of the brick will be distributed and affects deformation. Area calculation is straightforward and follows this formula: Area = length \(\times\) width.
  • Face 1: With dimensions 1 cm by 2 cm, the area is 2 cm虏.
  • Face 2: Here, dimensions are 2 cm by 3 cm, giving us an area of 6 cm虏.
  • Face 3: This face measures 1 cm by 3 cm, resulting in an area of 3 cm虏.
By calculating the area of each face, we identify which face offers the largest contact surface. Larger areas mean more distributed pressure, leading to minimized deformation.
Deformation
Deformation occurs when the physical dimensions of an object change under an applied load - in this case, the brick's own weight. With a given weight, the deformation depends significantly on which surface the brick rests upon. When a brick lies on a smaller area, the pressure (force per unit area) increases, likely leading to greater deformation. Conversely, a larger face area spreads the pressure. Therefore, selecting a face with a greater area reduces the force per unit area, and minimizes deformation.
To visualize, imagine squeezing clay with your palm (a large area) versus a finger (a smaller area). The clay deforms less with the palm. Similarly, resting the brick on the face with the largest area (Face 2) will keep its structure more stable and less deformed.
Brick Dimensions
The dimensions of the brick's faces give us insight into how the brick can best manage its own weight without considerable deformation. The three sets of dimensions provide various options for placement:
  • Face 1: 1 cm by 2 cm
  • Face 2: 2 cm by 3 cm
  • Face 3: 1 cm by 3 cm
When considering brick dimensions, each pair determines a specific face which can bear the load differently. Face 2, having the largest dimension (2 cm by 3 cm), should ideally be the bottom face. When this face is selected, the other two dimensions determine height, minimizing height-related pressures that could lead to a more significant dimension change (deformation).
Ultimately, analyzing these dimensions helps decide the most stable position for the brick, balancing between minimal deformation and structural integrity.

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Most popular questions from this chapter

To stretch a relaxed biceps muscle \(2.5 \mathrm{cm}\) requires a force of \(25 \mathrm{N}\). Find the Young's modulus for the muscle tissue, assuming it to be a uniform cylinder of length \(0.24 \mathrm{m}\) and crosssectional area \(47 \mathrm{cm}^{2}\)

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