Problem 64 - Fundamentals of Physics

Question

A room has dimensions \(3.00 \mathrm{~m}\) (height) \(x\) \(3.70 \mathrm{~m} \times 4.30 \mathrm{~m}\). A fly starting at one corner flies around, ending up at the diagonally opposite corner. (a) What is the magnitude of its displacement? (b) Could the length of its path be less than this magnitude? (c) Greater? (d) Equal? (e) Choose a suitable coordinate system and express the components of the displacement vector in that system in unit-vector notation. (f) If the fly walks, what is the length of the shortest path? (Hint: This can be answered without calculus. The room is like a box. Unfold its walls to flatten them into a plane.)

Step-by-Step Solution

Answer

The fly's displacement is approximately 6.42 m. Its walking path can be greater, with the shortest path being 7.67 m.

Step 1: Calculate the Diagonal of the Room

To find the displacement of the fly, we calculate the diagonal of the rectangular box. Use the formula for the diagonal of a rectangular prism: \[\text{Diagonal} = \sqrt{L^2 + W^2 + H^2}\]where \(L = 4.30\, \text{m}\), \(W = 3.70\, \text{m}\), and \(H = 3.00\, \text{m}\). Substitute these values: \[\text{Diagonal} = \sqrt{(4.30)^2 + (3.70)^2 + (3.00)^2}\]\[= \sqrt{18.49 + 13.69 + 9.00}\]\[= \sqrt{41.18}\]\[\approx 6.42\, \text{m}\].This diagonal is the magnitude of the displacement.

Step 2: Analyze Path Length Comparisons

(b) The length of the path could be less than the displacement (the diagonal through the air) only if it is a straight line, which for practical purposes of walking around surfaces, isn't possible. (c) The length of its path could definitely be greater than the diagonal if it doesn't fly directly. (d) The path can be equal only when it is a direct route along the calculated diagonal and no other constraints of surfaces.

Step 3: Express the Displacement in Vector Form

(e) Set a coordinate system with the starting corner at the origin, \((0,0,0)\), and the diagonally opposite corner at \((4.30, 3.70, 3.00)\). The displacement vector is therefore \[\vec{d} = 4.30\hat{i} + 3.70\hat{j} + 3.00\hat{k}\].

Step 4: Calculate Shortest Walk Path

(f) To determine the shortest walking path, imagine unfolding the room into a plane. Consider two possible unfoldings, making adjacent rectangles and checking their diagonals:1. Flatten the room onto a 6.70 m \(\times\) 3.70 m rectangle (summing one dimension with height or width), diagonal calculation: \[\sqrt{6.70^2+3.70^2} \approx 7.67\, \text{m}\].2. Alternatively, flatten the room into a 7.30 m \(\times\) 3.00 m rectangle, diagonal calculation: \[\sqrt{7.30^2+3.00^2} \approx 7.88\, \text{m}\].The shortest path by walking is 7.67 m, as confirmed by the smallest diagonal across the flattened surfaces.

Key Concepts

Vector Displacement

In physics, vector displacement refers to a change in position that has both magnitude and direction. For problems involving three dimensions, such as the movement of a fly inside a room, displacement can be visualized as a vector stretching from an initial to a final position.

In the context of the original problem, this means calculating the straight-line distance from one corner of the room to the diagonally opposite corner. This is essentially finding the diagonal of a rectangular prism.

This displacement doesn't account for any path the fly might take, only the shortest distance "as the crow flies," or through the air, from start to finish. Using the Pythagorean theorem, the displacement vector can be described by: \[ \text{Diagonal} = \sqrt{L^2 + W^2 + H^2} \]where \(L\), \(W\), and \(H\) are the dimensions of the room.

Rectangular Prism

A rectangular prism is a three-dimensional shape with six faces, all of which are rectangles. In simpler terms, it's a box-like structure where every corner angle is a right angle. The room in this exercise is a classic example of a rectangular prism.

Each pair of opposite faces is congruent and parallel, which helps in the calculation of the diagonal for vector displacement. To find the fly's displacement, we treat the room as an enclosed space with all its inner walls forming these simple rectangles.

Understanding these properties can significantly simplify the process of solving problems that require calculating distances across such spaces.

Coordinate System

Choosing a coordinate system is often the first step in solving problems involving vectors. In this exercise, the room is aligned with a Cartesian coordinate system, a grid defined by three perpendicular axes: x, y, and z.

The choice of origin, \((0,0,0)\) corresponds with one corner of the room where the fly starts. Points in space are then expressed as coordinates relating to this origin. The opposite corner, where the fly ends its journey, is located at:\((4.30, 3.70, 3.00)\)This setup allows us to express the displacement as a vector in the form:\(\vec{d} = 4.30\hat{i} + 3.70\hat{j} + 3.00\hat{k}\) This shows both the magnitude in each direction and the overall path.

Path Optimization

Path optimization involves finding the most efficient route or shortest distance within given constraints. For this exercise, the concept is to find the shortest walkable path on the surfaces of the room.

By "unfolding" the walls into a two-dimensional space, we can treat it like a flat plane, revealing the shortest paths as diagonals that represent minimized travel distance.

Two possible unfoldings result in different paths, but by calculating the diagonals of each rectangle: \[ \sqrt{6.70^2 + 3.70^2} \approx 7.67\, \text{m} \]\[ \sqrt{7.30^2 + 3.00^2} \approx 7.88\, \text{m} \]We find that the first flattening method gives us the shortest path, showing just how powerful this conceptual approach can be: it's both visual and exact.