Problem 17 - Fundamentals of Physics

Question

Three vectors \(\vec{a}, \vec{b}\), and \(\vec{c}\) each have a magnitude of \(50 \mathrm{~m}\) and lie in an \(x y\) plane. Their directions relative to the positive direction of the \(x\) axis are \(30^{\circ}, 195^{\circ}\), and \(315^{\circ}\), respectively. What are (a) the magnitude and (b) the angle of the vector \(\vec{a}+\vec{b}+\vec{c}\), and (c) the magnitude and (d) the angle of \(\vec{a}-\vec{b}+\vec{c}\) ? What are the (e) magnitude and (f) angle of a fourth vector \(\vec{d}\) such that \((\vec{a}+\vec{b})-(\vec{c}+\vec{d})=0 ?\)

Step-by-Step Solution

Answer

(a) Magnitude: 30.47 m, (b) Angle: 4.87°, (c) Magnitude: 135.22 m, (d) Angle: 20.41°, (e) Magnitude: 83.51 m, (f) Angle: 118.15°.

Step 1: Decompose each vector into components

To solve for the resultant vector, first break down each given vector into its x and y components. Use the formulas: \[ \vec{v}_x = v \cos(\theta) \] \[ \vec{v}_y = v \sin(\theta) \] where \(v\) is the magnitude of the vector and \(\theta\) is the angle.For \(\vec{a}\) with \(\theta = 30^{\circ}\):\[ \vec{a}_x = 50 \cos(30^{\circ}) = 43.3 \text{ m} \]\[ \vec{a}_y = 50 \sin(30^{\circ}) = 25 \text{ m} \]For \(\vec{b}\) with \(\theta = 195^{\circ}\):\[ \vec{b}_x = 50 \cos(195^{\circ}) = -48.3 \text{ m} \]\[ \vec{b}_y = 50 \sin(195^{\circ}) = 12.94 \text{ m} \]For \(\vec{c}\) with \(\theta = 315^{\circ}\):\[ \vec{c}_x = 50 \cos(315^{\circ}) = 35.36 \text{ m} \]\[ \vec{c}_y = 50 \sin(315^{\circ}) = -35.36 \text{ m} \]

Step 2: Calculate components of the vector sum/difference

Add or subtract components accordingly. For \(\vec{a} + \vec{b} + \vec{c}\):\[ (\vec{a} + \vec{b} + \vec{c})_x = 43.3 + (-48.3) + 35.36 = 30.36 \text{ m} \]\[ (\vec{a} + \vec{b} + \vec{c})_y = 25 + 12.94 + (-35.36) = 2.58 \text{ m} \]For \(\vec{a} - \vec{b} + \vec{c}\):\[ (\vec{a} - \vec{b} + \vec{c})_x = 43.3 - (-48.3) + 35.36 = 126.96 \text{ m} \]\[ (\vec{a} - \vec{b} + \vec{c})_y = 25 - 12.94 - (-35.36) = 47.42 \text{ m} \]

Step 3: Find the magnitude and direction of vectors

Use the Pythagorean theorem to find magnitudes of resultant vectors and trigonometry to find directions. For \(\vec{a} + \vec{b} + \vec{c}\):\[ |\vec{r}| = \sqrt{(30.36)^2 + (2.58)^2} = 30.47 \text{ m} \]\[ \theta = \tan^{-1}\left(\frac{2.58}{30.36}\right) \approx 4.87^{\circ} \]For \(\vec{a} - \vec{b} + \vec{c}\):\[ |\vec{r'}| = \sqrt{(126.96)^2 + (47.42)^2} = 135.22 \text{ m} \]\[ \theta' = \tan^{-1}\left(\frac{47.42}{126.96}\right) \approx 20.41^{\circ} \]

Step 4: Solve for the fourth vector \(\vec{d}\)

Since \((\vec{a} + \vec{b}) - (\vec{c} + \vec{d}) = \vec{0}\), it follows that\[ \vec{c} + \vec{d} = \vec{a} + \vec{b} \]Substitute the already calculated components for \(\vec{a} + \vec{b}\): \[ (\vec{a} + \vec{b})_x = 43.3 + (-48.3) = -5 \text{ m} \]\[ (\vec{a} + \vec{b})_y = 25 + 12.94 = 37.94 \text{ m} \]Equating \(\vec{c} + \vec{d}\) gives:\[ \vec{d}_x = -5 - 35.36 = -40.36 \text{ m} \]\[ \vec{d}_y = 37.94 - (-35.36) = 73.3 \text{ m} \]

Step 5: Magnitude and direction of \(\vec{d}\)

Find the magnitude and direction of \(\vec{d}\) using its components:\[ |\vec{d}| = \sqrt{(-40.36)^2 + (73.3)^2} = 83.51 \text{ m} \]\[ \theta_{d} = \tan^{-1}\left(\frac{73.3}{-40.36}\right) \approx 118.15^{\circ} \]

Key Concepts

Vector Components

Understanding vector components is crucial for solving vector-related problems. A vector has both magnitude and direction, which makes it different from a scalar. To handle vectors algebraically, it is often useful to break them down into components that align with the coordinate axes—usually the x-axis and y-axis.

By using trigonometric functions like sine and cosine, each vector can be split into x and y components. For example:

• The x-component of a vector is calculated using the cosine function: \(v_x = v \cos(\theta)\).
• The y-component uses the sine function: \(v_y = v \sin(\theta)\).

Here, \(v\) is the vector's magnitude and \(\theta\) is the angle it makes with the positive x-axis.
These components make it easier to perform vector addition or subtraction.

Magnitude of Vectors

The magnitude of a vector represents its size or length. When a vector is described by its components, you can visualize it lying between these components like a diagonal of a rectangle.

To find the magnitude of a vector from its components, apply the Pythagorean theorem, which works well because vector components form a right triangle. The formula for the magnitude is:

• \(|\vec{v}| = \sqrt{v_x^2 + v_y^2}\)

Here, \(v_x\) and \(v_y\) are the x and y components of the vector, respectively. This gives you a scalar value representing the vector's length, irrespective of direction. Magnitude is crucial because it quantifies how long the vector is, making it easier to compare with other vectors.

Trigonometry

Trigonometry plays a significant role in understanding vectors and performing calculations with them. Often, vectors are angled, meaning they're not aligned purely along one axis.

In vector addition or subtraction, knowing the angle a vector makes with the axis is critical. You can compute this angle using inverse trigonometric functions. The tangent function is particularly useful in these situations:

• Angle calculation: \(\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)\)

This formula lets you find the directional angle of a resultant vector using its components. Understanding these trigonometric relationships helps determine not just the angle involved but enables you to resolve vectors into components, making calculations manageable.

Pythagorean Theorem

The Pythagorean theorem is a cornerstone in vector analysis, especially when dealing with right triangles formed by vector components. This theorem states that for a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.

In vector terms, if a vector's components \(v_x\) and \(v_y\) form a right triangle with the vector itself as the hypotenuse, the Pythagorean theorem becomes:

• \(|\vec{v}| = \sqrt{v_x^2 + v_y^2}\)

This calculation provides the vector's magnitude and is essential when you combine vectors, as it lets you understand how much length is added or subtracted. The theorem ensures that all angles and components are accounted for accurately in your work. Whether dealing with forces, displacements, or velocities, the Pythagorean theorem is always at play.