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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 final expressions are: $$ma_x = 0$$ $$ma_y = 0$$ $$ma_z = -mg$$ These equations show that there is no acceleration in the x and y directions, and the acceleration in the z direction is equal to the acceleration due to gravity, but with a negative sign (downward direction).

Step by step solution

01

1. Recall Newton's Second Law of Motion in 3D:

Newton's Second Law of Motion states that the vector force (\(\vec{F}\)) acting on an object is equal to its mass (\(m\)) times the acceleration vector (\(\vec{a}\)). The general expression in 3D can be written as: $$\vec{F} = m\vec{a}$$ Now, let's write this equation in terms of its x, y, and z components.
02

2. Write the Force and Acceleration Vectors in Components:

The force vector \(\vec{F}\) can be expressed as the sum of forces in the x, y, and z directions: $$\vec{F} = F_x\hat{i} + F_y\hat{j} +F_z\hat{k}$$Similarly, express the acceleration vector \(\vec{a}\) in terms of its components:$$\vec{a} = a_x\hat{i} + a_y\hat{j} +a_z\hat{k}$$
03

3. Insert the Gravitational Force:

Since we have gravitational force acting only in the z-direction, we can write the gravitational force as: $$F_z = -mg$$Here, \(g\) is the acceleration due to gravity, and we use a negative sign because the force acts downward in the z-direction. Now, we have: $$\vec{F} = F_x\hat{i} + F_y\hat{j} - mg\hat{k}$$
04

4. Apply Newton's Second Law of Motion with Gravitational Force:

Now we can substitute the force vector mentioned above into Newton's Second Law:$$F_x\hat{i} + F_y\hat{j} - mg\hat{k} = m(a_x\hat{i} + a_y\hat{j} + a_z\hat{k})$$
05

5. Equate Each Component:

Now let's equate the components in each direction separately: For x-direction: $$F_x = ma_x$$ For y-direction: $$F_y = ma_y$$ For z-direction: $$-mg = ma_z$$ Since there is no force acting in the x and y directions, \(F_x = 0\) and \(F_y = 0\). Therefore, the equations become: For x-direction: $$0 = ma_x$$ For y-direction: $$0 = ma_y$$ For z-direction: $$-mg = ma_z$$
06

6. Write the Final Expressions:

Now we can write the 3D Newton's Second Law of Motion with only gravitational force acting in the z-direction as: $$ma_x = 0$$ $$ma_y = 0$$ $$ma_z = -mg$$ This shows that there is no acceleration in the x and y directions, and the acceleration in the z direction is equal to the acceleration due to gravity, but with a negative sign (downward direction).

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Key Concepts

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

Gravitational Force
When we talk about gravitational force in the context of Newton’s Second Law of Motion, the key idea is how gravity impacts an object. Gravitational force is the pull that the Earth exerts on objects, pulling them downwards towards its center. It's represented mathematically as \( F_z = -mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth's surface). The negative sign indicates that the force acts in the downward direction, which is conventionally taken as the negative \( z \)-axis in three-dimensional analysis. Essentially, this represents gravity pulling objects down toward the ground.
Three-Dimensional Motion
In three-dimensional motion, any movement can be described using three axes: the x-axis, y-axis, and z-axis. This is a more complex form of motion as opposed to one-dimensional or two-dimensional cases because it requires considering motion in all three directions simultaneously. Newton's Second Law, which states \( \vec{F} = m\vec{a} \), still applies; however, both the force and acceleration need to be expressed in vector form to capture their components in each direction. By resolving forces and accelerations into their x, y, and z components, we can analyze the motion more thoroughly and understand how objects move in three-dimensional space.
Acceleration Components
Acceleration components are crucial when looking at three-dimensional motion. These components describe how acceleration occurs in each of the spatial dimensions: x, y, and z. For example, if we decompose the acceleration vector as \( \vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k} \), each \( a_x, a_y, \) and \( a_z \) represents the acceleration along the respective axis.
  • For the x-direction: \( ma_x = 0 \) means there's no acceleration because there isn't any force acting in this direction.
  • For the y-direction: \( ma_y = 0 \) similarly indicates no acceleration.
  • In the z-direction, the equation \( ma_z = -mg \) shows us that the object is accelerating downward due to gravity, matching the gravitational acceleration \( g \).

In practical terms, this tells us that without any additional forces, an object only accelerates in the direction gravity acts upon it – which, for objects on Earth, is vertically downward.

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Most popular questions from this chapter

Three-dimensional motion Consider the motion of the following objects. Assume the \(x\) -axis points east, the \(y\) -axis points north, the positive z-axis is vertical and opposite g. the ground is horizontal, and only the gravitational force acts on the object unless otherwise stated. a. Find the velocity and position vectors, for \(t \geq 0\). b. Make a sketch of the trajectory. c. Determine the time of flight and range of the object. d. Determine the maximum height of the object. A baseball is hit 3 ft above home plate with an initial velocity of \(\langle 30,30,80\rangle \mathrm{ft} / \mathrm{s} .\) The spin on the baseball produces a horizontal acceleration of the ball of \(5 \mathrm{ft} / \mathrm{s}^{2}\) in the northward direction.

Rule By expressing \(\mathbf{u}\) and \(\mathbf{v}\) in terms of their components, prove that $$\frac{d}{d t}(\mathbf{u}(t)+\mathbf{v}(t))=\mathbf{u}^{\prime}(t)+\mathbf{v}^{\prime}(t)$$

A projectile (such as a bascball or a cannonball) launched from the origin with an initial horizontal velocity \(u_{0}\) and an initial vertical velocity \(v_{0}\) moves in a parabolic trajectory given by $$\mathbf{r}(t)=\left\langle u_{0} t_{1}-\frac{1}{2} g t^{2}+v_{0} t\right\rangle, \quad \text { for } t \geq 0$$ where air resistance is neglected and \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity (see Section 14.3 ). a. Let \(u_{0}=20 \mathrm{m} / \mathrm{s}\) and \(v_{0}=25 \mathrm{m} / \mathrm{s} .\) Assuming the projectile is launched over horizontal ground, at what time does it return to Earth? b. Find the integral that gives the length of the trajectory from launch to landing. c. Evaluate the integral in part (b) by first making the change of variables \(u=-g t+v_{0} .\) The resulting integral is evaluated either by making a second change of variables or by using a calculator. What is the length of the trajectory? d. How far does the projectile land from its launch site?

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\left\langle t^{2}+1, t+1,1\right\rangle$$

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\sqrt{t+4} \mathbf{i}+\frac{t}{t+1} \mathbf{j}-e^{-t^{2}} \mathbf{k}$$

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