Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Write Newton's Second Law of Motion for three-dimensional motion with only the gravitational force (acting in the \(z\) -direction).

Short Answer

Expert verified
Answer: The equations of motion for this case are: \(a_x = 0\), \(a_y = 0\), and \(a_z = -g\), where \(a_x\), \(a_y\), and \(a_z\) are the accelerations in the x, y, and z directions, respectively, and g is the acceleration due to gravity.

Step by step solution

01

Recalling Newton's Second Law of Motion

Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, it is expressed as: $$ \vec{F} = m\vec{a} $$ where \(\vec{F}\) is the net force acting on an object, \(m\) is the mass of the object, and \(\vec{a}\) is the acceleration of the object.
02

Considering the three-dimensional motion

In three-dimensional motion, the net force and acceleration both have x, y, and z components. We can represent Newton's Second Law of Motion in component form as follows: $$ F_x = m a_x $$ $$ F_y = m a_y $$ $$ F_z = m a_z $$
03

Identifying the given conditions

In this problem, it is given that the only force acting on the object is the gravitational force and it is acting in the z-direction. Therefore, the force components in the x and y directions (\(F_x\) and \(F_y\)) are zero. The gravitational force acting in the z-direction is given by: $$ F_z = -mg $$ where \(g\) is the acceleration due to gravity and the negative sign indicates that the force is acting in the negative z-direction (downwards).
04

Applying the given conditions to Newton's Second Law

Now that we know the force components, we can apply the given condition to the component form of Newton's Second Law which we derived in step 2: $$ F_x = 0 = m a_x $$ $$ F_y = 0 = m a_y $$ $$ F_z = -mg = m a_z $$
05

Simplifying the equations

Finally, we can simplify these equations to get the equation of motion for this specific case: $$ a_x = 0 $$ $$ a_y = 0 $$ $$ a_z = -g $$ These equations represent Newton's Second Law of Motion for three-dimensional motion with only the gravitational force acting in the z-direction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the 12 vectors that have their tails at the center of a (circular) clock and their heads at the numbers on the edge of the clock. a. What is the sum of these 12 vectors? b. If the 12: 00 vector is removed, what is the sum of the remaining 11 vectors? c. By removing one or more of these 12 clock vectors, explain how to make the sum of the remaining vectors as large as possible in magnitude. d. Consider the 11 vectors that originate at the number 12 at the top of the clock and point to the other 11 numbers. What is the sum of the vectors?

Use vectors to show that the midpoint of the line segment joining \(P\left(x_{1}, y_{1}\right)\) and \(Q\left(x_{2}, y_{2}\right)\) is the point \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\) (Hint: Let \(O\) be the origin and let \(M\) be the midpoint of \(P Q\). Draw a picture and show that $$\left.\overrightarrow{O M}=\overrightarrow{O P}+\frac{1}{2} \overrightarrow{P Q}=\overrightarrow{O P}+\frac{1}{2}(\overrightarrow{O Q}-\overrightarrow{O P}) \cdot\right)$$

Direction angles and cosines Let \(\mathbf{v}=\langle a, b, c\rangle\) and let \(\alpha, \beta\) and \(\gamma\) be the angles between \(\mathbf{v}\) and the positive \(x\) -axis, the positive \(y\) -axis, and the positive \(z\) -axis, respectively (see figure). a. Prove that \(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\) b. Find a vector that makes a \(45^{\circ}\) angle with \(\mathbf{i}\) and \(\mathbf{j}\). What angle does it make with \(\mathbf{k} ?\) c. Find a vector that makes a \(60^{\circ}\) angle with i and \(\mathbf{j}\). What angle does it make with k? d. Is there a vector that makes a \(30^{\circ}\) angle with \(\mathbf{i}\) and \(\mathbf{j} ?\) Explain. e. Find a vector \(\mathbf{v}\) such that \(\alpha=\beta=\gamma .\) What is the angle?

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and \(f\) are real numbers. It can be shown that this curve lies in a plane. Assuming the curve lies in a plane, show that it is a circle centered at the origin with radius \(R\) provided \(a^{2}+c^{2}+e^{2}=b^{2}+d^{2}+f^{2}=R^{2}\) and \(a b+c d+e f=0\).

Let $$\mathbf{u}(t)=2 t^{3} \mathbf{i}+\left(t^{2}-1\right) \mathbf{j}-8 \mathbf{k} \text { and } \mathbf{v}(t)=e^{t} \mathbf{i}+2 e^{-t} \mathbf{j}-e^{2 t} \mathbf{k}$$ Compute the derivative of the following functions. $$\mathbf{v}(\sqrt{t})$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free