Problem 73 - Physics for Scientists and Engineers with Modern Physics

Question

A plumber steps out of his truck, walks \(66 \mathrm{~m}\) east and \(35 \mathrm{~m}\) south, and then takes an elevator \(12 \mathrm{~m}\) into the subbasement of a building where a bad leak is occurring. What is the displacement of the plumber relative to his truck? Give your answer in components; also give the magnitude and angles, with respect to the \(x\) axis, in the vertical and horizontal plane. Assume \(x\) is east, \(y\) is north, and \(z\) is up.

Step-by-Step Solution

Answer

The plumber's displacement is (66 m, -35 m, -12 m) with magnitude 75.68 m; angles are approximately 29.52° with x, 117.92° with y, and 98.95° with z axis.

Step 1: Identify Coordinate Directions

The problem specifies directions: east as positive along the x-axis, north as positive along the y-axis, and upwards as positive along the z-axis. Therefore, moving south implies moving in the negative y-direction, and moving downwards into the subbasement implies moving in the negative z-direction.

Step 2: Determine Displacement Vectors

The plumber moves 66 m east, which corresponds to a vector component of \( +66 \) m along the x-axis. The next movement is 35 m south, which corresponds to \( -35 \) m along the y-axis. Finally, the plumber moves 12 m into the subbasement, giving a vector component of \( -12 \) m along the z-axis.

Step 3: Write Displacement Vector in Component Form

Combine the component movements into a displacement vector \( \vec{d} \) in three-dimensional space: \( \vec{d} = (66 \text{ m}, -35 \text{ m}, -12 \text{ m}) \). This vector describes the plumber's movement in three-dimensional space in terms of its x, y, and z components.

Step 4: Calculate the Magnitude of Displacement

Use the three-dimensional distance formula to calculate the magnitude of the displacement vector: \[|\vec{d}| = \sqrt{(66^2) + (-35^2) + (-12^2)}\]Calculate: \[|\vec{d}| = \sqrt{4356 + 1225 + 144} = \sqrt{5725} \approx 75.68 \text{ m}\]

Step 5: Calculate Angles with Respect to Axes

Calculate the angles the displacement vector makes with each axis. Use the formulas to find angles:For the x-axis:\[\cos(\theta_x) = \frac{66}{75.68}\]For the y-axis:\[\cos(\theta_y) = \frac{-35}{75.68}\]For the z-axis:\[\cos(\theta_z) = \frac{-12}{75.68}\]Calculate these values to get angles:\[\theta_x = \cos^{-1}\left(\frac{66}{75.68}\right) \approx 29.52^\circ\]\[\theta_y = \cos^{-1}\left(\frac{-35}{75.68}\right) \approx 117.92^\circ\]\[\theta_z = \cos^{-1}\left(\frac{-12}{75.68}\right) \approx 98.95^\circ\]

Key Concepts

Vector Components

Understanding vector components is crucial in solving problems related to displacement in three-dimensional space. When you move in a given direction, you can describe your motions using three components: one for each axis in the coordinate system—x, y, and z. For the plumber's problem, eastward movement is along the x-axis, southward movement is along the y-axis (but in the negative direction), and moving downwards into the subbasement is along the negative z-axis.

To derive the displacement vector, you insert the movements into a vector. Here, the vector is given by

• x-component: 66 m (eastward)
• y-component: -35 m (southward)
• z-component: -12 m (downward)

Thus, the displacement vector is written as \( \vec{d} = (66, -35, -12) \). This tells you exactly how much the plumber moved in each direction from his initial position.

Magnitude of Displacement

The magnitude of displacement is a measure of how far the plumber moves from his starting point. Even though the plumber took various paths, the magnitude is like taking a shortcut from start to finish—a straight line distance.

To calculate this, you use the distance formula in three dimensions, which is an extension of the Pythagorean theorem. Normally used in two dimensions, the theorem is expanded by adding squares of the differences along the three axes:\[|\vec{d}| = \sqrt{(x^2) + (y^2) + (z^2)}\]Inserting the values for the plumber's movements:\[|\vec{d}| = \sqrt{(66^2) + (-35^2) + (-12^2)}\]This calculates to:\[|\vec{d}| = \sqrt{4356 + 1225 + 144} = \sqrt{5725} \approx 75.68 \text{ m}\]This number covers the entire path in space that the plumber took, summarized in one straight line movement from start to finish.

Displacement Angle Calculation

Determining the angle of displacement with respect to the axes shows how the direction of movement relates to standard directional axes. These angles are calculated using trigonometric functions, specifically the cosine: cosine of the angle equals the adjacent over hypotenuse.

For each axis, consider the component of the vector along that axis and the magnitude you just calculated. The formulas are:

• For the x-axis: \( \cos(\theta_x) = \frac{66}{75.68} \)
• For the y-axis: \( \cos(\theta_y) = \frac{-35}{75.68} \)
• For the z-axis: \( \cos(\theta_z) = \frac{-12}{75.68} \)

By calculating the inverse cosine (cos^-1), you find the angles:

• \( \theta_x \approx 29.52^\circ \)
• \( \theta_y \approx 117.92^\circ \)
• \( \theta_z \approx 98.95^\circ \)

These angles help in visualizing the plumber's path in relation to each axis, providing a comprehensive understanding of how he moved in a 3D space.