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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 \(\theta=\tan ^{-1}\left(\frac{4}{5}\right)\) with the positive \(x\) -axis

Short Answer

Expert verified
Question: Determine the parallel and normal components of the force F = <0, -10> acting on an object placed on a plane inclined at an angle θ = tan⁻¹(4/5). Answer: The parallel component of the force, Fₚₐᵣₐₗₗₑₗ, is given by Fₚₐᵣₐₗₗₑₗ = < -200/41, -160/41 >. The normal component of the force, Fₙₒᵣₘₐₗ, is given by Fₙₒᵣₘₐₗ = < 200/41, -250/41 >.

Step by step solution

01

Determine the unit vector in the direction of the plane

We are given the angle formed between the plane and the positive \(x\)-axis, θ = tan⁻¹(4/5). So, we can determine the unit vector along the direction of the plane using trigonometry. Since θ = tan⁻¹(4/5): $$\sin\theta =\frac{4}{\sqrt{4^2+5^2}}=\frac{4}{\sqrt{41}}$$ $$\cos\theta =\frac{5}{\sqrt{4^2+5^2}}=\frac{5}{\sqrt{41}}$$ The unit vector in the direction of the plane, \(\mathbf{u}=\langle \cos\theta, \sin\theta\rangle\), so: $$\mathbf{u}=\left\langle\frac{5}{\sqrt{41}}, \frac{4}{\sqrt{41}}\right\rangle$$
02

Find the parallel component of the force using vector projection

The projection of vector F onto the plane is the parallel component of the force. $$\mathbf{F}_{\parallel}=(\mathbf{F}\cdot\mathbf{u})\mathbf{u}$$ Here \(\mathbf{F}=\langle 0,-10\rangle\) and \(\mathbf{u}=\left\langle\frac{5}{\sqrt{41}},\frac{4}{\sqrt{41}}\right\rangle\). So, \(\mathbf{F}\cdot\mathbf{u}= -\frac{40}{\sqrt{41}}\). Therefore, $$\mathbf{F}_{\parallel}=\left(-\frac{40}{\sqrt{41}} \right)\left\langle\frac{5}{\sqrt{41}},\frac{4}{\sqrt{41}} \right\rangle= \left\langle -\frac{200}{41},- \frac{160}{41} \right\rangle$$
03

Find the normal component of the force using orthogonal decomposition

Using orthogonal decomposition, we can find the normal component of the force as: $$\mathbf{F}_{\perp}= \mathbf{F}-\mathbf{F}_{\parallel}$$ So, $$\mathbf{F}_{\perp}=\left\langle 0,-10\right\rangle-\left\langle- \frac{200}{41}, -\frac{160}{41}\right\rangle=\left\langle\frac{200}{41},-10+\frac{160}{41}\right\rangle=\left\langle\frac{200}{41},\frac{-250}{41}\right\rangle$$
04

Show that the total force is the sum of the component forces

Adding the parallel and normal components of the force, we get: $$\mathbf{F}_{\parallel}+\mathbf{F}_{\perp}=\left\langle-\frac{200}{41},- \frac{160}{41} \right\rangle + \left\langle\frac{200}{41},-\frac{250}{41}\right\rangle$$ $$\mathbf{F}_{\parallel}+\mathbf{F}_{\perp}=\left\langle -\frac{200}{41}+\frac{200}{41},-\frac{160}{41}+\frac{-250}{41}\right\rangle=\left\langle0,-10\right\rangle$$ So, we can conclude that the total force \(\mathbf{F}\) is the sum of the parallel (\(\mathbf{F}_{\parallel}\)) and normal (\(\mathbf{F}_{\perp}\)) components of the force, as \(\mathbf{F}=\mathbf{F}_{\parallel}+\mathbf{F}_{\perp}=\left\langle0,-10\right\rangle\).

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Key Concepts

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

Trigonometry
Trigonometry is a branch of mathematics that helps us explore the relationships between the angles and sides of triangles. In vector decomposition, trigonometry plays a crucial role because it helps us determine angles and calculate components along specific directions. This is particularly useful when dealing with forces or movements that are not aligned along the standard axes.

In the given problem, trigonometry is used to find the angle of the plane given by \( \theta = \tan^{-1}(4/5) \). This is an inverse tangent operation that helps us determine the angle which is essential to find the direction vectors. Once we know \( \theta \), we can use sine and cosine functions to calculate the components of the unit vector:
  • \( \sin\theta = \frac{4}{\sqrt{41}} \)
  • \( \cos\theta = \frac{5}{\sqrt{41}} \)
This knowledge allows us to decompose forces, like \( \mathbf{F} = \langle 0, -10 \rangle \), accurately along intended directions, using projections and other methods.
Vector Projection
Vector projection is a method used to find a vector component in the direction of another vector. It essentially "projects" one vector onto another, thus helping to identify how much of one vector lies in the direction of another.

In this exercise, vector projection helps us determine the component of the vertical force \( \mathbf{F} = \langle 0, -10 \rangle \) that is parallel to the plane's direction. The formula for vector projection of \( \mathbf{F} \) onto a unit vector \( \mathbf{u} \) is:

\[ \mathbf{F}_{\parallel} = (\mathbf{F} \cdot \mathbf{u}) \mathbf{u} \]
  • The dot product \( \mathbf{F} \cdot \mathbf{u} = -\frac{40}{\sqrt{41}} \)
  • This determines how aligned \( \mathbf{F} \) is with \( \mathbf{u} \)
Thus, the result of the projection, \( \mathbf{F}_{\parallel} = \langle -\frac{200}{41}, -\frac{160}{41} \rangle \), provides the vector component parallel to the plane. This shows how trigonometry and vector operations combine to yield meaningful components of forces.
Unit Vector
A unit vector is a vector of length one used to specify a direction. It helps in converting vectors from one form to another without altering the direction, only scaling magnitude accordingly.

In our problem, the unit vector \( \mathbf{u} \) helps us determine the direction along the plane. Since the unit vector has a magnitude of 1, it is ideal for use in projection formulas without altering the vector it "touches."

The unit vector \( \mathbf{u} \) is derived using trigonometry through its components:\[ \mathbf{u} = \left\langle \frac{5}{\sqrt{41}}, \frac{4}{\sqrt{41}} \right\rangle \]
  • \( \mathbf{u} \) keeps the direction consistent while scaling the force appropriately
  • Essential in decomposing vectors into parallel and perpendicular components
Without such vectors, calculating the direction-specific components would be much more complex when dealing with angles and inclined planes.
Orthogonal Decomposition
Orthogonal decomposition refers to breaking down a vector into two perpendicular components. These components are generally parallel and normal (perpendicular) relative to a specified direction.

In the context of vector forces acting on an inclined plane, orthogonal decomposition helps us find two distinct segments of the original vector \( \mathbf{F} = \langle 0, -10 \rangle \). These segments are the parallel component to the plane (found using vector projection) and the normal component, which is the leftover after removing the parallel projection.

The normal component \( \mathbf{F}_{\perp} \) is determined by subtracting the parallel component \( \mathbf{F}_{\parallel} \) from \( \mathbf{F} \): \[\mathbf{F}_{\perp}= \mathbf{F}-\mathbf{F}_{\parallel}\]
  • This isolates the vector portion orthogonally diametric to the plane's orientation
  • Makes solving physical problems involving inclines more manageable and systematic
By doing so, we can confidently dissect the total force to understand each component's role, indispensable in fields like mechanical physics and engineering. This is how we show that \( \mathbf{F} = \mathbf{F}_{\parallel} + \mathbf{F}_{\perp} \), by confirming their sum returns the original force vector.

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Most popular questions from this chapter

An object moves along an ellipse given by the function \(\mathbf{r}(t)=\langle a \cos t, b \sin t\rangle,\) for \(0 \leq t \leq 2 \pi,\) where \(a > 0\) and \(b > 0\) a. Find the velocity and speed of the object in terms of \(a\) and \(b\) for \(0 \leq t \leq 2 \pi\) b. With \(a=1\) and \(b=6,\) graph the speed function, for \(0 \leq t \leq 2 \pi .\) Mark the points on the trajectory at which the speed is a minimum and a maximum. c. Is it true that the object speeds up along the flattest (straightest) parts of the trajectory and slows down where the curves are sharpest? d. For general \(a\) and \(b\), find the ratio of the maximum speed to the minimum speed on the ellipse (in terms of \(a\) and \(b\) ).

Carry out the following steps to determine the (smallest) distance between the point \(P\) and the line \(\ell\) through the origin. a. Find any vector \(\mathbf{v}\) in the direction of \(\ell\) b. Find the position vector u corresponding to \(P\). c. Find \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). d. Show that \(\mathbf{w}=\mathbf{u}-\) projy \(\mathbf{u}\) is a vector orthogonal to \(\mathbf{v}\) whose length is the distance between \(P\) and the line \(\ell\) e. Find \(\mathbf{w}\) and \(|\mathbf{w}| .\) Explain why \(|\mathbf{w}|\) is the distance between \(P\) and \(\ell\). \(P(1,1,-1) ; \ell\) has the direction of $$\langle-6,8,3\rangle$$.

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. Verify that the Cauchy-Schwarz Inequality holds for \(\mathbf{u}=\langle 3,-5,6\rangle\) and \(\mathbf{v}=\langle-8,3,1\rangle\).

Assume that \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in \(\mathrm{R}^{3}\) that form the sides of a triangle (see figure). Use the following steps to prove that the medians intersect at a point that divides each median in a 2: 1 ratio. The proof does not use a coordinate system. a. Show that \(\mathbf{u}+\mathbf{v}+\mathbf{w}=\mathbf{0}\) b. Let \(\mathbf{M}_{1}\) be the median vector from the midpoint of \(\mathbf{u}\) to the opposite vertex. Define \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) similarly. Using the geometry of vector addition show that \(\mathbf{M}_{1}=\mathbf{u} / 2+\mathbf{v} .\) Find analogous expressions for \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) c. Let \(a, b,\) and \(c\) be the vectors from \(O\) to the points one-third of the way along \(\mathbf{M}_{1}, \mathbf{M}_{2},\) and \(\mathbf{M}_{3},\) respectively. Show that \(\mathbf{a}=\mathbf{b}=\mathbf{c}=(\mathbf{u}-\mathbf{w}) / 3\) d. Conclude that the medians intersect at a point that divides each median in a 2: 1 ratio.

Zero curvature Prove that the curve $$ \mathbf{r}(t)=\left\langle a+b t^{p}, c+d t^{p}, e+f t^{p}\right\rangle $$ where \(a, b, c, d, e,\) and \(f\) are real numbers and \(p\) is a positive integer, has zero curvature. Give an explanation.

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