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Chapter 12: Problem 44

Find the components of the vertical force \(\mathbf{F}=\langle 0,-10\rangle\) in the directions parallel to and normal to the following planes. Show that the total force is the sum of the two component forces. A plane that makes an angle of \(\pi / 6\) with the positive \(x\) -axis

Short Answer

Expert verified
Based on the given force vector, \(\mathbf{F} = \langle 0, -10 \rangle\), and the inclined plane at an angle of \(\pi/6\) with the positive x-axis, the components of the force parallel and normal to the inclined plane are: \(\mathbf{F_{\parallel}} = \langle -\frac{5\sqrt{3}}{2}, -\frac{5}{2} \rangle\) \(\mathbf{F_{\perp}} = \langle \frac{5\sqrt{3}}{2}, -\frac{15}{2} \rangle\) These components confirm that the total force is indeed the sum of the two component forces.

Step by step solution

01

Find the unit vector of the inclined plane

The inclined plane makes an angle of \(\pi / 6\) with the positive x-axis. Let's first find the unit vector of the inclined plane, \(\mathbf{u}\): Since there's no mention to which axis the angle is given, we will interpret it to be in the x-y plane. \(\mathbf{u} = \langle \cos(\pi/6), \sin(\pi/6) \rangle = \langle \frac{\sqrt{3}}{2}, \frac{1}{2} \rangle\)
02

Determine the component of the force parallel to the plane

In order to find the component of the force \(\mathbf{F}\) that is parallel to the plane, we will project \(\mathbf{F}\) onto \(\mathbf{u}\) using the following formula: \(\mathbf{F_{\parallel}} = (\mathbf{F} \cdot \mathbf{u}) \, \mathbf{u}\) Compute the dot product of \(\mathbf{F}\) and \(\mathbf{u}\) and then scale \(\mathbf{u}\) by the result: \(\mathbf{F_{\parallel}} = (0 \times \frac{\sqrt{3}}{2} + (-10) \times \frac{1}{2}) \, \langle \frac{\sqrt{3}}{2}, \frac{1}{2} \rangle = (-5) \langle \frac{\sqrt{3}}{2}, \frac{1}{2} \rangle = \langle -\frac{5\sqrt{3}}{2}, -\frac{5}{2} \rangle\)
03

Determine the component of the force normal to the plane

Now, let's find the component of the force \(\mathbf{F}\) that is normal to the plane: \(\mathbf{F_{\perp}} = \mathbf{F} - \mathbf{F_{\parallel}}\) Subtract \(\mathbf{F_{\parallel}}\) from \(\mathbf{F}\): \(\mathbf{F_{\perp}} = \langle 0, -10 \rangle - \langle -\frac{5\sqrt{3}}{2}, -\frac{5}{2} \rangle = \langle \frac{5\sqrt{3}}{2}, -\frac{15}{2} \rangle\)
04

Verify that the total force is the sum of the two component forces

Finally, we will verify that \(\mathbf{F}\) is the sum of \(\mathbf{F_\parallel}\) and \(\mathbf{F_\perp}\). If our calculations are correct, the following equation should hold true: \(\mathbf{F} = \mathbf{F_{\parallel}} + \mathbf{F_{\perp}}\) By adding the calculated components, we get: \(\mathbf{F_{\parallel}} + \mathbf{F_{\perp}} = \langle -\frac{5\sqrt{3}}{2}, -\frac{5}{2} \rangle + \langle \frac{5\sqrt{3}}{2}, -\frac{15}{2} \rangle = \langle 0, -10 \rangle\) Since the sum is equal to the given force vector \(\mathbf{F}\), we have successfully verified that the total force is the sum of the two component forces.

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Key Concepts

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

Unit Vectors
Unit vectors are an essential part of understanding vector directions. These vectors have a magnitude of one and help in indicating direction without affecting the strength of the vector. They serve as a basis for defining any vector direction.
  • A unit vector in a plane can be described using trigonometric functions. For an angle \( \theta \) in the \( x-y \) plane, the unit vector \( \mathbf{u} \) is \( \langle \cos(\theta), \sin(\theta) \rangle \).
  • In the exercise, since the angle \( \pi/6 \) is given with the x-axis, we find the unit vector by calculating \( \cos(\pi/6) = \frac{\sqrt{3}}{2} \) and \( \sin(\pi/6) = \frac{1}{2} \).
  • The resulting unit vector \( \mathbf{u} = \langle \frac{\sqrt{3}}{2}, \frac{1}{2} \rangle \) represents a directional arrow in the plane.

Using unit vectors, we can decompose any vector into components parallel and perpendicular to this directional vector.
Dot Product
The dot product is a way of multiplying two vectors, yielding a scalar. It helps find how much one vector goes in the direction of another.
  • The formula for the dot product of \( \mathbf{A} = \langle a_1, a_2 \rangle \) and \( \mathbf{B} = \langle b_1, b_2 \rangle \) is \( \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 \).
  • In our exercise, we calculated the dot product of \( \mathbf{F} \) and \( \mathbf{u} \). This gave us the amount of the force \( \mathbf{F} \) that goes in the direction of \( \mathbf{u} \).
  • In particular, \( \mathbf{F} \cdot \mathbf{u} = 0 \times \frac{\sqrt{3}}{2} + (-10) \times \frac{1}{2} = -5 \). The negative sign indicates the direction is opposite to \( \mathbf{u} \).

The dot product is foundational not only for projection but also valuable in understanding angles between vectors, with an angle \( \theta \) between two vectors being \( \cos^{-1}(\text{dot product of unit vectors}) \).
Force Components
Decomposing a force into components is vital in physics and engineering. It allows us to understand how force affects systems in specific directions.
  • The parallel component can be found by projecting the force onto the unit vector, as done in the exercise. It's calculated using \( \mathbf{F_{\parallel}} = (\mathbf{F} \cdot \mathbf{u}) \mathbf{u} \).
  • For \( \mathbf{F} = \langle 0, -10 \rangle \), projecting onto \( \mathbf{u} = \langle \frac{\sqrt{3}}{2}, \frac{1}{2} \rangle \) gives the parallel component as \( \langle -\frac{5\sqrt{3}}{2}, -\frac{5}{2} \rangle \).
  • The normal component, or that perpendicular to the plane, is found by subtracting the parallel component from the original force: \( \mathbf{F_{\perp}} = \mathbf{F} - \mathbf{F_{\parallel}} \).
  • This results in \( \langle \frac{5\sqrt{3}}{2}, -\frac{15}{2} \rangle \), indicating the rest of the force acting perpendicularly.

Understanding these components allows for analyzing how forces impact objects on inclined planes.
Projection of Vectors
Projection is a technique used to express one vector in terms of another vector. It鈥檚 like casting one vector鈥檚 influence along the direction of another.
  • The projection of vector \( \mathbf{F} \) onto \( \mathbf{u} \) is given by \( \text{Proj}_{\mathbf{u}}\mathbf{F} = \frac{\mathbf{F} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \cdot \mathbf{u} \).
  • This simplifies to \( (\mathbf{F} \cdot \mathbf{u}) \mathbf{u} \) when \( \mathbf{u} \) is a unit vector since \( \mathbf{u} \cdot \mathbf{u} = 1 \).
  • The exercise focuses on how much of the force \( \mathbf{F} \) lies in the plane, resulting in the projection \( \mathbf{F_{\parallel}} \).
  • This concept helps visualize and calculate how effectively a vector acts along a specific direction, which has numerous applications in physics and engineering.

Mastery of vector projection aids in resolving complex vector interactions.

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Most popular questions from this chapter

Prove that the midpoint of the line segment joining \(P\left(x_{1}, y_{1}, z_{1}\right)\) and \(Q\left(x_{2}, y_{2}, z_{2}\right)\) is $$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)$$

Practical formula for \(\mathbf{N}\) Show that the definition of the principal unit normal vector $\mathbf{N}=\frac{d \mathbf{T} / d s}{|d \mathbf{T} / d s|}\( implies the practical formula \)\mathbf{N}=\frac{d \mathbf{T} / d t}{|d \mathbf{T} / d t|} .\( Use the Chain Rule and Note that \)|\mathbf{v}|=d s / d t>0.$

Find the points (if they exist) at which the following planes and curves intersect. $$y=1 ; \mathbf{r}(t)=\langle 10 \cos t, 2 \sin t, 1\rangle, \text { for } 0 \leq t \leq 2 \pi$$

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Show that for real numbers \(u_{1}, u_{2},\) and \(u_{3},\) it is true that \(\left(u_{1}+u_{2}+u_{3}\right)^{2} \leq 3\left(u_{1}^{2}+u_{2}^{2}+u_{3}^{2}\right)\). (Hint: Use the Cauchy-Schwarz Inequality in three dimensions with \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) and choose \(\mathbf{v}\) in the right way.)

A baseball leaves the hand of a pitcher 6 vertical feet above home plate and \(60 \mathrm{ft}\) from home plate. Assume that the coordinate axes are oriented as shown in the figure. a. In the absence of all forces except gravity, assume that a pitch is thrown with an initial velocity of \(\langle 130,0,-3\rangle \mathrm{ft} / \mathrm{s}\) (about \(90 \mathrm{mi} / \mathrm{hr}\) ). How far above the ground is the ball when it crosses home plate and how long does it take for the pitch to arrive? b. What vertical velocity component should the pitcher use so that the pitch crosses home plate exactly \(3 \mathrm{ft}\) above the ground? c. A simple model to describe the curve of a baseball assumes that the spin of the ball produces a constant sideways acceleration (in the \(y\) -direction) of \(c \mathrm{ft} / \mathrm{s}^{2}\). Assume a pitcher throws a curve ball with \(c=8 \mathrm{ft} / \mathrm{s}^{2}\) (one-fourth the acceleration of gravity). How far does the ball move in the \(y\) -direction by the time it reaches home plate, assuming an initial velocity of (130,0,-3) ft/s? d. In part (c), does the ball curve more in the first half of its trip to the plate or in the second half? How does this fact affect the batter? e. Suppose the pitcher releases the ball from an initial position of (0,-3,6) with initial velocity \((130,0,-3) .\) What value of the spin parameter \(c\) is needed to put the ball over home plate passing through the point (60,0,3)\(?\)
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