Problem 5 - Physics for Scientists and Engineers with Modern Physics

Question

(II) \(\hat{\mathbf{v}}\) is a vector 24.8 units in magnitude and points at an angle of \(23.4^{\circ}\) above the negative \(x\) axis. \((a)\) Sketch this vector. (b) Calculate \(V_{x}\) and \(V_{y}-(c)\) Use \(V_{x}\) and \(V_{y}\) .to obtain (again) the magnitude and direction of \(\vec{\mathbf{v}}\)[Note. Part \((c)\) is a good way to check if you've resolved your vector correctly.

Step-by-Step Solution

Answer

Sketch vector pointing 23.4° above negative \(x\) axis; calculate \(V_x = -22.89\), \(V_y = 9.87\); verify magnitude and direction.

Step 1: Sketch the Vector

To sketch \(\hat{\mathbf{v}}\), begin by drawing the negative \(x\) axis.Then draw a vector pointing 23.4° above this axis, with a length proportional to 24.8 units.The vector forms an angle of 23.4° with the negative \(x\) axis.

Step 2: Calculate V_x

To find \(V_x\), use the cosine component of the vector:\[V_x = -|\hat{\mathbf{v}}| \cos(23.4°) = -24.8 \cos(23.4°)\]This is negative because the vector points along the negative \(x\) direction.

Step 3: Calculate V_y

To find \(V_y\), use the sine component of the vector:\[V_y = |\hat{\mathbf{v}}| \sin(23.4°) = 24.8 \sin(23.4°)\]This component is positive because the vector is pointing upwards.

Step 4: Verify Magnitude with V_x and V_y

Use the Pythagorean theorem to ensure these components resolve back into the original magnitude:\[|\hat{\mathbf{v}}| = \sqrt{V_x^2 + V_y^2}\]Calculate to check consistency with the given magnitude of 24.8 units.

Step 5: Verify Direction with V_x and V_y

Compute the direction angle using the tangent:\[\theta = \tan^{-1}\left(\frac{V_y}{V_x}\right)\]Check to ensure this angles measures 23.4° above the negative \(x\) axis.

Key Concepts

Magnitude and Direction

Understanding the concept of magnitude and direction is crucial when dealing with vectors. A vector is not just a number; it has both a size and a direction. When we talk about the magnitude of a vector, we are referring to its length. In the context of our exercise, the magnitude is 24.8 units. This tells us how far the vector extends in space without any concern for its direction.

The direction, on the other hand, describes where this vector is pointing. This is often measured in degrees, radians, or as an angle with respect to a reference line or axis. In our exercise, the vector makes an angle of 23.4° above the negative x-axis. This means it doesn’t simply extend horizontally or vertically, but at an angle that needs to be carefully considered, especially when breaking it down into components along the x and y axes.

Trigonometric Functions

Trigonometric functions are essential tools for resolving vectors into their components. These functions relate the angles and sides of triangles, and in vector resolution, they help us break down a vector into orthogonal components: horizontal (x-axis) and vertical (y-axis).

To find the x-component of the vector, we use the cosine function. The x-component, or \(V_x\), can be calculated as \(V_x = |\hat{\mathbf{v}}| \cos(23.4°)\). Of course, since this vector points along the negative x-axis, we should remember to include the negative sign.

Similarly, to find the y-component, we use the sine function. Therefore, the y-component, or \(V_y\), is given by \(V_y = |\hat{\mathbf{v}}| \sin(23.4°)\). Since the vector is pointing upwards, the \(V_y\) component will be positive. Trigonometry thus provides a way to accurately break down vectors into manageable pieces.

Pythagorean Theorem

The Pythagorean theorem is a mathematical principle that helps us verify the accuracy of the vector components we calculated. Originally used to find the length of the sides of right-angled triangles, this theorem is widely applicable. In the context of vectors, it allows us to ensure that our resolved components maintain the original magnitude of the vector when combined.

The theorem states that for a right-angled triangle, the sum of the squares of the two shorter sides equals the square of the longest side, known as the hypotenuse. For our vector, this means that \(|\hat{\mathbf{v}}| = \sqrt{V_x^2 + V_y^2}\). By plugging in the values of \(V_x\) and \(V_y\) obtained from trigonometric functions, we can solve the equation and reconfirm the magnitude of 24.8 units.

This step is not only a check on our work but also a way to appreciate how interrelated trigonometric identities and geometric principles are in simplifying complex vector calculations.