Problem 44 - College Physics

Question

Vector \(A\) has a magnitude of \(20 \mathrm{~m}\) and makes an angle of \(30^{\circ}\) above the positive \(x\) axis. Vector \(\vec{B}\) has a magnitude of \(15 \mathrm{~m}\) and is oriented \(60^{\circ}\) to the left of the \(y\) axis. Find (a) the magnitude and direction of \(\vec{A}-\vec{B},(b)\) the magnitude and direction of \(2 \vec{A}+\vec{B},\) and (c) the magnitude and direction of \(-\overrightarrow{\boldsymbol{A}}+3 \overrightarrow{\boldsymbol{B}}\).

Step-by-Step Solution

Answer

Calculated magnitudes: \(\sqrt{925}, \sqrt{1225}, \sqrt{3343.75}\); directions found using angle formulas.

Step 1: Resolve Vector A into Components

Vector \( A \) has a magnitude of 20 m and makes a 30° angle with the positive x-axis. We calculate the components:\[A_x = 20 \cos(30°) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \ A_y = 20 \sin(30°) = 20 \times \frac{1}{2} = 10\]

Step 2: Resolve Vector B into Components

Vector \( B \) has a magnitude of 15 m and makes an angle of 60° to the left of the y-axis. The angle with the positive x-axis is \(150°\):\[B_x = 15 \cos(150°) = 15 \times -\frac{\sqrt{3}}{2} = -7.5\sqrt{3} \ B_y = 15 \sin(150°) = 15 \times \frac{1}{2} = 7.5\]

Step 3: Vector Subtraction for \(\vec{A}-\vec{B}\)

Perform the subtraction \( \vec{A} - \vec{B} \):\[(A_x - B_x, A_y - B_y) = (10\sqrt{3} - (-7.5\sqrt{3}), 10 - 7.5)\]\[= (17.5\sqrt{3}, 2.5)\]

Step 4: Magnitude and Direction of \(\vec{A}-\vec{B}\)

The magnitude is calculated as:\[|\vec{A} - \vec{B}| = \sqrt{(17.5\sqrt{3})^2 + (2.5)^2} = \sqrt{918.75 + 6.25} = \sqrt{925} \]The direction uses the tangent inverse function:\[\theta = \tan^{-1}\left(\frac{2.5}{17.5\sqrt{3}}\right)\]

Step 5: Calculate \(2\vec{A}\) Components

Double the components of \( \vec{A} \):\[2A_x = 2 \times 10\sqrt{3} = 20\sqrt{3} \ 2A_y = 2 \times 10 = 20\]

Step 6: Vector Addition for \(2\vec{A} + \vec{B}\)

Add components of \( 2\vec{A} \) and \( \vec{B} \):\[(20\sqrt{3} + (-7.5\sqrt{3}), 20 + 7.5) = (12.5\sqrt{3}, 27.5)\]

Step 7: Magnitude and Direction of \(2\vec{A} + \vec{B}\)

The magnitude is:\[|2\vec{A} + \vec{B}| = \sqrt{(12.5\sqrt{3})^2 + (27.5)^2} = \sqrt{468.75 + 756.25} = \sqrt{1225}\]The direction, given by:\[\theta = \tan^{-1}\left(\frac{27.5}{12.5\sqrt{3}}\right)\]

Step 8: Calculate \(3\vec{B}\) Components

Triple the components of \( \vec{B} \):\[3B_x = 3 \times (-7.5\sqrt{3}) = -22.5\sqrt{3} \ 3B_y = 3 \times 7.5 = 22.5\]

Step 9: Vector Addition for \(-\vec{A} + 3\vec{B}\)

Perform the operation \( -\vec{A} + 3\vec{B} \):\[(-10\sqrt{3} - 22.5\sqrt{3}, -10 + 22.5) = (-32.5\sqrt{3}, 12.5)\]

Step 10: Magnitude and Direction of \(-\vec{A} + 3\vec{B}\)

The magnitude is calculated as:\[|\vec{-A} + 3\vec{B}| = \sqrt{(-32.5\sqrt{3})^2 + (12.5)^2} = \sqrt{3187.5 + 156.25} = \sqrt{3343.75}\]The direction is:\[\theta = \tan^{-1}\left(\frac{12.5}{-32.5\sqrt{3}} \right)\]

Step 11: Final Results Interpretation

Summarize the calculated magnitudes and directions for each vector operation:- \( |\vec{A} - \vec{B}| = \sqrt{925} \), direction: \( \theta \)- \( |2\vec{A} + \vec{B}| = \sqrt{1225} \), direction: \( \theta \)- \( |\vec{-A} + 3\vec{B}| = \sqrt{3343.75} \), direction: \( \theta \) adjusted for quadrant.

Key Concepts

Vector Magnitude

The magnitude of a vector is essentially the vector’s length. It’s a measure of how long the vector is in the space it occupies. To find the magnitude of any vector, you can use the Pythagorean Theorem. For a vector with components \(x\) and \(y\), the magnitude \(R\) is calculated as \(|R| = \sqrt{x^2 + y^2}\).

This means that to understand how strong a force or vector is in certain directions, you must consider its total magnitude. Whether you are dealing with forces, velocities, or displacements, vector magnitude provides the answer to how substantial these quantities are.

Vector Components

Vector components help you break down a vector into perpendicular parts, usually along the x and y-axis, making them much easier to handle in problem-solving. You can think of components as projections of the vector along each axis.

For example:

• The x-component can be found using the formula: \(A_x = |A|\cos(\theta)\)
• The y-component using: \(A_y = |A|\sin(\theta)\)

This is useful since it allows you to analyze effects or components of quantities acting in different directions.

Vector Addition

Vector addition combines two or more vectors to find a resultant vector. It’s a way to figure out the overall effect of different forces or movements. This involves summing up respective components.

To add vectors, look at:

• Sum of the x-components: \(R_x = A_x + B_x\)
• Sum of the y-components: \(R_y = A_y + B_y\)

After finding these sums, you can determine the resultant vector’s magnitude and direction using the formulas discussed for vector magnitude and angles.

Vector Subtraction

Vector subtraction is akin to adding a vector’s opposite. If you have \(A \) and you want to subtract \(B\), you add \(-B\). This operation helps to calculate the difference between two vectors, effectively determining what’s left when one vector's effect is removed from another's.

Calculation involves:

• Subtracting x-components: \(R_x = A_x - B_x\)
• Subtracting y-components: \(R_y = A_y - B_y\)

The process is similar to addition but with different signs for the subtracted vector’s components.

Angle Calculation

To find the angle at which a vector makes with the horizontal, you can use the inverse tangent function, often denoted as \( an^{-1}\). This is essential in determining the direction of vectors in 2D space.

The formula to find the angle \(\theta\) is: \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\)

Angle calculation is crucial as it informs about the vector's orientation in the plane, showing not just where it points, but how it aligns concerning the axes. Correctly determining the angle is imperative to understanding the directional component of physical quantities.