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Chapter 3: Problem 77

Vector A points in the negative \(x\) direction. Vector \(\overrightarrow{\mathrm{B}}\) points at an angle of \(30.0^{\circ}\) above the positive \(x\) axis. Vector \(C\) has a magnitude of \(15 \mathrm{m}\) and points in a direction \(40.0^{\circ}\) below the positive \(x\) axis. Given that \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}=0,\) find the magnitudes of \(\overline{\mathbf{A}}\) and \(\overline{\mathbf{B}}\).

Short Answer

Expert verified
Magnitude of \(A\) is 28.91 m; magnitude of \(B\) is 19.284 m.

Step by step solution

01

Understand the Vector Directions and Components

Vector \(\overrightarrow{\mathrm{A}}\) points in the negative \(x\) direction, which means its components can be represented as \(\overrightarrow{\mathrm{A}} = -A\hat{i}\). Vector \(\overrightarrow{\mathrm{B}}\) has an angle \(30.0^{\circ}\) above the positive \(x\) axis, so its components are \(B_x = B \cos(30^{\circ})\) and \(B_y = B \sin(30^{\circ})\). Vector \(\overrightarrow{\mathrm{C}}\) points \(40.0^{\circ}\) below the positive \(x\) axis, so its components are \(C_x = 15 \cos(-40^{\circ})\) and \(C_y = 15 \sin(-40^{\circ})\).
02

Use the Condition \( \overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = 0 \)

Since the sum of the vectors is equal to zero, the sum of their components along the \(x\) and \(y\) axes must also be zero. This gives us two equations: \(-A + B \cos(30^{\circ}) + 15 \cos(-40^{\circ}) = 0\) and \(B \sin(30^{\circ}) + 15 \sin(-40^{\circ}) = 0\).
03

Solve the \(y\)-Component Equation for \(B\)

From the \(y\)-component equation, \(B \sin(30^{\circ}) - 15 \sin(40^{\circ}) = 0\). Solving for \(B\) gives \(B = \frac{15 \sin(40^{\circ})}{\sin(30^{\circ})}\).
04

Calculate \(B\)

Using \(\sin(30^{\circ}) = 0.5\) and \(\sin(40^{\circ}) \approx 0.6428\), substitute to find \(B = \frac{15 \times 0.6428}{0.5} = 19.284\, \mathrm{m}\).
05

Solve the \(x\)-Component Equation for \(A\)

Using the \(x\)-component equation, \(-A + 19.284 \cos(30^{\circ}) + 15 \cos(-40^{\circ}) = 0\). Substitute \(B = 19.284\) into the equation to solve for \(A\).
06

Calculate \(A\)

Using \(\cos(30^{\circ}) \approx 0.866\) and \(\cos(-40^{\circ}) \approx 0.7660\), the equation becomes \(-A + 19.284 \times 0.866 + 15 \times 0.766 = 0\). Solve this to find \(A = 28.91\, \mathrm{m}\).

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Key Concepts

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

Vector Components
Vectors can be understood through their components, often along the x and y axes. Each vector can be broken down into horizontal (x) and vertical (y) parts:
  • Vector A: Since it points in the negative x direction, the component form is \(\overrightarrow{A} = -A\hat{i}\).
  • Vector B: Given its angle of \(30^{\circ}\) above the x-axis, use trigonometric functions to find its components: \(B_x = B \cos(30^{\circ})\), and \(B_y = B \sin(30^{\circ})\).
  • Vector C: With a direction \(40^{\circ}\) below the x-axis, the components are: \(C_x = 15 \cos(-40^{\circ})\), and \(C_y = 15 \sin(-40^{\circ})\).
Breaking vectors into components simplifies the process of combining them. Especially useful when dealing with equilibrium.
Trigonometric Functions
Trigonometric functions like sine and cosine are crucial in handling vector problems. They help convert between vector angles and components. Here鈥檚 how we used them:
  • The cosine function \(\cos\) helps find the adjacent side of the triangle formed by the vector, i.e., the x-component. This is why \(B_x = B \cos(30^{\circ})\) and \(C_x = 15 \cos(-40^{\circ})\).
  • The sine function \(\sin\) provides the opposite side of the triangle, or the y-component. Thus, \(B_y = B \sin(30^{\circ})\), and \(C_y = 15 \sin(-40^{\circ})\).
Using sine and cosine allows you to project the vectors into horizontal and vertical lines. This projection helps in assembling and disassembling vectors, aiding in overcoming complex force interactions.
Equilibrium of Forces
In physics, equilibrium refers to the state where the net force acting on an object is zero. This means all individual forces cancel each other out. In this exercise:
  • The given condition \(\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C} = 0\) signifies an equilibrium where vector sums result in zero net movement.
  • This condition allows us to set up two separate equations for the x and y components since equilibrium in each direction is required. Thus: - \(-A + B \cos(30^{\circ}) + 15 \cos(-40^{\circ}) = 0\) for the x-component and \(B \sin(30^{\circ}) + 15 \sin(-40^{\circ}) = 0\) for the y-component.
To solve such equilibrium problems, breakdown all forces into their components and apply equilibrium conditions separately along each axis. By resolving these component equations, we can deduce the required magnitudes for the vectors involved.

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Most popular questions from this chapter

Suppose we orient the \(x\) axis of a two-dimensional coordinate system along the beach at Waikiki. Waves approaching the beach have a velocity relative to the shore given by \(\mathbf{v}_{w s}=(1.3 \mathrm{m} / \mathrm{s}) \hat{\mathbf{y}} .\) Surfers move more rapidly than the waves, but at an angle to the beach. The angle is chosen so that the surfers approach the shore with the same speed as the waves. (a) If a surfer has a speed of \(7.2 \mathrm{m} / \mathrm{s}\) relative to the water, what is her direction of motion relative to the positive \(x\) axis? (b) What is the surfer's velocity relative to the wave? \((c)\) If the surfer's speed is increased, will the angle in part (a) increase or decrease? Explain.

Find the direction and magnitude of the vectors. (a) \(\overrightarrow{\mathbf{A}}=(25 \mathrm{m}) \hat{\mathrm{x}}+(-12 \mathrm{m}) \hat{\mathrm{y}}\) (b) \(\overrightarrow{\mathbf{B}}=(2.0 \mathrm{m}) \hat{\mathrm{x}}+(15 \mathrm{m}) \hat{\mathrm{y}},\) and (c) \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\)

In its daily prowl of the neighborhood, a cat makes a displacement of \(120 \mathrm{m}\) due north, followed by a \(72-\mathrm{m}\) displacement due west. (a) Find the magnitude and direction of the displacement required for the cat to return home. (b) If, instead, the cat had first prowled \(72 \mathrm{m}\) west and then \(120 \mathrm{m}\) north, how would this affect the displacement needed to bring it home? Explain.

You drive a car 680ft to the east, then 340ft to the north. (a) What is the magnitude of your displacement? (b) Using a sketch, estimate the direction of your displacement. (c) Verify your estimate in part (b) with a numerical calculation of the direction.

The Position of the Moon Relative to the center of the Earth, the position of the Moon can be approximated by $$\begin{aligned} \overrightarrow{\mathbf{r}}=&\left(3.84 \times 10^{8} \mathrm{m}\right)\left\\{\cos \left[\left(2.46 \times 10^{-6} \mathrm{radians} / \mathrm{s}\right) \mathrm{f}\right] \hat{\mathrm{x}}\right.\\\ &\left.+\sin \left[\left(2.46 \times 10^{-6} \mathrm{radians} / \mathrm{s}\right) t\right] \hat{\mathrm{y}}\right\\} \end{aligned}$$ where \(t\) is measured in seconds. (a) Find the magnitude and direction of the Moon's average velocity between \(t=0\) and \(t=7.38\) days. (This time is one- quarter of the 29.5 days it takes the Moon to complete one orbit.) (b) Is the instantaneous speed of the Moon greater than, less than, or the same as the average speed found in part (a)? Explain.
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