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Chapter 1: Problem 33

Vector \(\vec{A}\) is \(2.80 \mathrm{~cm}\) long and is \(60.0^{\circ}\) above the \(x\) -axis in the first quadrant. Vector \(\vec{B}\) is \(1.90 \mathrm{~cm}\) long and is \(60.0^{\circ}\) below the \(x\) -axis in the fourth quadrant (Fig. E1.33). Use components to find the magnitude and direction of (a) \(\vec{A}+\vec{B} ;\) (b) \(\vec{A}-\vec{B} ;\) (c) \(\vec{B}-\vec{A}\). In each case, sketch the vector addition or subtraction and show that your numerical answers are in qualitative agreement with your sketch.

Short Answer

Expert verified
The resultant vectors for each operation are: (a) \(\vec{A} + \vec{B}\) results in a vector with magnitude roughly 2.41 cm and direction approximately 18.43掳; (b) \(\vec{A} - \vec{B}\) produces a vector of magnitude about 4.09 cm and direction around 83.89掳; (c) \(\vec{B} - \vec{A}\) gives a vector with magnitude around 4.09 cm but in the opposite direction of approximately -83.89掳 (or 296.11掳 when measured counterclockwise from the positive x-axis).

Step by step solution

01

Break Down Vectors into Components

First, you need to determine the x and y components for vectors \(\vec{A}\) and \(\vec{B}\). Since both vectors form a \(60^{\circ}\) angle with the x-axis, they will have the same cosine and sine values. For \(\vec{A}\), the x component is \(A_x = 2.80cm \cos60^{\circ} = 1.40cm\) and the y component is \(A_y = 2.80cm \sin60^{\circ} = 2.42cm\). For \(\vec{B}\), the x component is \(B_x = 1.90cm \cos60^{\circ} = 0.95cm\) and the y component is \(B_y = -1.90cm \sin60^{\circ} = -1.65cm\). Remember the negative sign for \(B_y\) since it's below the x-axis.
02

Addition and Subtraction of Component Vectors

(a) For \(\vec{A}+\vec{B}\), sum up the corresponding components of both vectors: \(R_x = A_x + B_x = 1.40cm + 0.95cm = 2.35cm\); \(R_y = A_y + B_y = 2.42cm - 1.65cm = 0.77cm\). (b) For \(\vec{A}-\vec{B}\), subtract the corresponding components: \(R_x = A_x - B_x = 1.40cm - 0.95cm = 0.45cm\); \(R_y = A_y - B_y = 2.42cm + 1.65cm = 4.07cm\). (c) For \(\vec{B}-\vec{A}\), subtract components accordingly: \(R_x = B_x - A_x = 0.95cm - 1.40cm = -0.45cm\); \(R_y = B_y - A_y = -1.65cm - 2.42cm = -4.07cm\).
03

Compute Magnitude and Direction of Resultant Vectors

Now use Pythagoras' Theorem to calculate each resultant vector's magnitude: \[R = \sqrt{{R_x}^2 + {R_y}^2}\]. Afterwards, find the direction using the inverse tangent function: \[\theta = \arctan(\frac{{R_y}}{{R_x}})\]. If \(R_x\) is negative, add \(180^{\circ}\) to the calculated angle for correct placement. Do this for every resultant vector from step 2.
04

Sketch and Validate Results

Sketch each of the vector additions and subtractions. The numerical solutions should align with your graphs qualitatively. That means the directions should match and the lengths of the arrows should make sense in comparison with each other.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Components
Understanding vector components is fundamental to working with vectors, especially when dealing with vector addition and subtraction. Each vector can be broken down into components, typically along the x and y axes. This helps in simplifying complex vector operations by dealing with one dimension at a time.
  • To find the x-component, we use the cosine of the angle the vector makes with the x-axis, multiplied by the vector's magnitude.
  • For the y-component, we use the sine of the angle, again multiplied by the vector's magnitude.
  • For example, if a vector \(\vec{A}\) is 2.80 cm long and makes a 60掳 angle with the x-axis, its x-component \(A_x\) is \(2.80 \text{ cm} \cos 60掳 = 1.40 \text{ cm}\), and its y-component \(A_y\) is \(2.80 \text{ cm} \sin 60掳 = 2.42 \text{ cm}\).
This breakdown into components allows for straightforward addition and subtraction, as each dimension can be handled independently.
Magnitude of Vectors
The magnitude of a vector represents its length or size and is a scalar quantity. When you have the components of a vector, you can determine its magnitude using the Pythagorean theorem.
  • The formula used is: \[ R = \sqrt{{R_x}^2 + {R_y}^2} \]
  • This calculates the hypotenuse of the right triangle formed by the components.
  • For a resultant vector with components \(R_x = 2.35 \text{ cm}\) and \(R_y = 0.77 \text{ cm}\), its magnitude \(R\) would be calculated as \[ \sqrt{(2.35)^2 + (0.77)^2} \]
  • Magnitude helps in visualizing the strength or intensity of the vector. In physics, magnitude can relate to forces, velocities, or displacements.
Calculating the magnitude is essential for understanding the overall effect of vector components together.
Direction of Vectors
The direction of a vector indicates where the vector is pointing and is usually described using angles. Understanding vector direction is crucial for analyzing vector relationships in physics.
  • Once you have the x and y components of a vector, the direction \(\theta\) can be determined using the inverse tangent function: \[ \theta = \arctan(\frac{{R_y}}{{R_x}}) \]
  • If the x-component is negative, add 180掳 to align the direction accurately because it indicates the vector is in the opposite quadrant.
  • Ultimately, direction gives a complete understanding of a vector's behavior, alongside its magnitude.
Get comfortable with using trigonometry to find and interpret vector directions, as this will aid you immensely in analyzing various physical systems.
Trigonometry in Physics
Trigonometry is an essential tool in physics, especially when dealing with vectors. It allows for the resolution of vectors into components and determination of angles, which are crucial for describing vector quantity behaviors.
  • Key functions such as sine, cosine, and tangent are used to relate the angles and sides of right triangles formed by vector components.
  • Understanding these concepts allows one to predict and determine resultant vectors, which is often a central part of solving physics problems.
  • For example, trigonometric identities enable the calculation of component magnitudes and directions by relating them to angles, aligning with the physical reality that direction and magnitude define vectors.
  • Whether it's for calculating forces, motion, or any physical quantity, mastery of trigonometric principles is invaluable.
With these tools, you'll be equipped to analyze and solve complex vector problems effectively.

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Most popular questions from this chapter

You live in a town where the streets are straight but are in a variety of directions. On Saturday you go from your apartment to the grocery store by driving \(0.60 \mathrm{~km}\) due north and then \(1.40 \mathrm{~km}\) in the direction \(60.0^{\circ}\) west of north. On Sunday you again travel from your apartment to the same store but this time by driving \(0.80 \mathrm{~km}\) in the direction \(50.0^{\circ}\) north of west and then in a straight line to the store. (a) How far is the store from your apartment? (b) On which day do you travel the greater distance, and how much farther do you travel? Or, do you travel the same distance on each route to the store?

Vector \(\vec{A}\) has magnitude \(5.00 \mathrm{~m}\) and lies in the \(x y\) -plane in a direction \(53.0^{\circ}\) from the \(+x\) -axis axis measured toward the \(+y\) -axis. Vector \(\vec{B}\) has magnitude \(8.00 \mathrm{~m}\) and a direction you can adjust. (a) You want the vector product \(\vec{A} \times \vec{B}\) to have a positive \(z\) -component of the largest possible magnitude. What direction should you select for vector \(\overrightarrow{\boldsymbol{B}} ?(\mathrm{~b})\) What is the direction of \(\overrightarrow{\boldsymbol{B}}\) for which \(\overrightarrow{\boldsymbol{A}} \times \overrightarrow{\boldsymbol{B}}\) has the most negative \(z\) -component? (c) What are the two directions of \(\overrightarrow{\boldsymbol{B}}\) for which \(\vec{A} \times \vec{B}\) is zero?

You are camping with Joe and Karl. since all three of you like your privacy, you don't pitch your tents close together. Joe's tent is \(21.0 \mathrm{~m}\) from yours, in the direction \(23.0^{\circ}\) south of east. Karl's tent is \(32.0 \mathrm{~m}\) from yours, in the direction \(37.0^{\circ}\) north of east. What is the distance between Karl's tent and Joe's tent?

A rectangular piece of aluminum is \(7.60 \pm 0.01 \mathrm{~cm}\) long and \(1.90 \pm 0.01 \mathrm{~cm}\) wide. (a) Find the area of the rectangle and the uncertainty in the area. (b) Verify that the fractional uncertainty in the area is equal to the sum of the fractional uncertainties in the length and in the width. (This is a general result.)

The volume of a solid cylinder is given by \(V=\pi r^{2} h,\) where \(r\) is the radius and \(h\) is the height. You measure the radius and height of a thin cylindrical wire and obtain the results \(r=0.036 \mathrm{~cm}\) and \(h=12.1 \mathrm{~cm}\). What do your measurements give for the volume of the wire in \(\mathrm{mm}^{3} ?\) Use the correct number of significant figures in your answer.
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