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In unit-vector notation, what is the torque about the origin on a particle located at coordinates \((0,-4.0 \mathrm{~m}, 3.0 \mathrm{~m})\) if that torque is due to (a) force \(\vec{F}_{1}\) with components \(F_{1 x}=2.0 \mathrm{~N}, F_{1 y}=F_{1 z}=0\), and (b) force \(\vec{F}_{2}\) with components \(F_{2 x}=0, F_{2 y}=2.0 \mathrm{~N}, F_{2 z}=4.0 \mathrm{~N} ?\)

Short Answer

Expert verified
(a) \( \vec{\tau}_1 = 6.0\hat{j} + 8.0\hat{k} \); (b) \( \vec{\tau}_2 = -22.0\hat{i} \).

Step by step solution

01

Identify the position and force vectors

The particle is located at the coordinates \((0, -4.0, 3.0)\). This gives us the position vector \( \vec{r} = 0\hat{i} - 4.0\hat{j} + 3.0\hat{k} \). For force \( \vec{F}_1 \), the components are \( F_{1x} = 2.0 \mathrm{~N}, F_{1y} = 0, F_{1z} = 0 \), resulting in the force vector \( \vec{F}_1 = 2.0\hat{i} + 0\hat{j} + 0\hat{k} \).
02

Compute the torque for \( \vec{F}_1 \)

The torque \( \vec{\tau}_1 \) is given by the cross product \( \vec{\tau}_1 = \vec{r} \times \vec{F}_1 \). Compute the cross product: \[\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \0 & -4.0 & 3.0 \2.0 & 0 & 0\end{vmatrix} = \hat{i}(0 \cdot 0 - 3.0 \cdot 0) - \hat{j}(0 \cdot 0 - 3.0 \cdot 2.0) + \hat{k}(0 \cdot 0 - (-4.0 \cdot 2.0))\]\[= 0\hat{i} + 6.0\hat{j} + 8.0\hat{k}.\]Thus, \( \vec{\tau}_1 = 0\hat{i} + 6.0\hat{j} + 8.0\hat{k} \).
03

Identify the force vector for \( \vec{F}_2 \)

For force \( \vec{F}_2 \), the components are \( F_{2x} = 0, F_{2y} = 2.0 \mathrm{~N}, F_{2z} = 4.0\mathrm{~N} \), which results in the force vector \( \vec{F}_2 = 0\hat{i} + 2.0\hat{j} + 4.0\hat{k} \).
04

Compute the torque for \( \vec{F}_2 \)

The torque \( \vec{\tau}_2 \) is given by the cross product \( \vec{\tau}_2 = \vec{r} \times \vec{F}_2 \). Compute the cross product: \[\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \0 & -4.0 & 3.0 \0 & 2.0 & 4.0\end{vmatrix} = \hat{i}((-4.0) \cdot 4.0 - 3.0 \cdot 2.0) - \hat{j}(0 \cdot 4.0 - 3.0 \cdot 0) + \hat{k}(0 \cdot 2.0 - (-4.0) \cdot 0)\]\[= -22.0\hat{i} + 0\hat{j} + 0\hat{k}.\]Thus, \( \vec{\tau}_2 = -22.0\hat{i}. \)

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Key Concepts

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

Cross Product
In physics, the cross product is a mathematical operation that is crucial for determining quantities like torque, which involves vector quantities. The cross product of two vectors results in a third vector that is perpendicular to the plane containing the initial two vectors. The magnitude of this resultant vector is given by the product of the magnitudes of the two original vectors and the sine of the angle between them.
The formula for the cross product of vectors \( \vec{a} \) and \( \vec{b} \) is:
  • \( \vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin(\theta) \hat{n} \)
where \( |\vec{a}| \) and \( |\vec{b}| \) are the magnitudes of the vectors, \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \), and \( \hat{n} \) is a unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \).
In the case of torque, the cross product is used between the position vector \( \vec{r} \) and the force vector \( \vec{F} \), yielding the torque vector \( \vec{\tau} \). This helps in determining the rotational effect of the force applied at a distance from the pivot. The direction of torque is given by the right-hand rule, which is a useful mnemonic for determining the direction of the cross product. Overall, understanding the cross product helps in comprehending how torques operate, especially when analyzing forces that cause rotation.
Vectors in Physics
Vectors are essential components in physics, as they represent quantities that have both magnitude and direction. Common examples of vector quantities include displacement, velocity, acceleration, force, and torque.
A vector is typically represented with an arrow, where the length indicates the magnitude and the orientation indicates the direction. Mathematically, a vector is expressed in a coordinate system using basis vectors such as \( \hat{i}, \hat{j}, \) and \( \hat{k} \), which correspond to the \( x \), \( y \), and \( z \) directions respectively.
When dealing with vectors, operations such as addition, subtraction, and multiplication (dot product and cross product) are used to solve various physical problems. For instance, in the given exercise, the position vector was defined as \( \vec{r} = 0\hat{i} - 4.0\hat{j} + 3.0\hat{k} \), which indicates the particle's location relative to an origin.
  • Vectors can be meaningfully added or subtracted by combining their respective components.
  • The magnitude of a vector \( \vec{v} \) with components \( (v_x, v_y, v_z) \) is given by \( |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \).
  • Understanding vector mathematics is crucial for analyzing how different forces or physical quantities interact, particularly in physics where directionality plays an essential role.
Mastering vector operations is fundamental for solving physical problems, especially those involving forces and motion, as vectors accurately represent these phenomena in a mathematical form.
Newton's Laws of Motion
Newton's Laws of Motion form the foundation of classical mechanics, outlining how objects respond under the influence of forces. These three laws provide a framework for understanding how objects move and interact.
The First Law, often called the Law of Inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. This law highlights the importance of forces in changing the state of motion of objects.
The Second Law establishes the relationship between force, mass, and acceleration. It is usually expressed as \( \vec{F} = m \vec{a} \), indicating that the force \( \vec{F} \) applied to an object is equal to the mass \( m \) of the object multiplied by its acceleration \( \vec{a} \). This law demonstrates how force influences the acceleration of an object.
  • It shows that more force leads to more acceleration.
  • Heavier objects require more force to achieve the same acceleration as lighter ones.
The Third Law states that for every action, there is an equal and opposite reaction. This means that if an object A exerts a force on object B, then object B will exert a force of equal magnitude and opposite direction upon object A.
Understanding Newton's Laws of Motion is crucial for comprehending how forces affect objects. In torque problems, especially, these laws aid in analyzing the rotational effects an applied force may have, linking it to the overall motion and stability of objects in physics.

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Most popular questions from this chapter

A hollow sphere of radius \(0.15 \mathrm{~m}\), with rotational inertia \(I=0.040 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about a line through its center of mass, rolls without slipping up a surface inclined at \(30^{\circ}\) to the horizontal. At a certain initial position, the sphere's total kinetic energy is \(20 \mathrm{~J}\). (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved \(1.0 \mathrm{~m}\) up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

A uniform solid ball rolls smoothly along a floor, then up a ramp inclined at \(15.0^{\circ} .\) It momentarily stops when it has rolled \(1.50 \mathrm{~m}\) along the ramp. What was its initial speed?

A uniform wheel of mass \(10.0 \mathrm{~kg}\) and radius \(0.400 \mathrm{~m}\) is mounted rigidly on a massless axle through its center (Fig. \(11-62\) ). The radius of the axle is \(0.200\) \(\mathrm{m}\), and the rotational inertia of the wheel-axle combination about its central axis is \(0.600 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The wheel is initially at rest at the top of a surface that is inclined at angle \(\theta=\) \(30.0^{\circ}\) with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along the surface smoothly and without slipping. When the wheel-axle combination has moved down the surface by \(2.00 \mathrm{~m}\), what are (a) its rotational kinetic energy and (b) its translational kinetic energy?

At one instant, force \(\vec{F}=4.0 \hat{\mathrm{j}} \mathrm{N}\) acts on a \(0.25 \mathrm{~kg}\) object that has position vector \(\vec{r}=(2.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{k}}) \mathrm{m}\) and velocity vector \(\vec{v}=(-5.0 \hat{\mathrm{i}}+5.0 \hat{\mathrm{k}}) \mathrm{m} / \mathrm{s}\). About the origin and in unit-vector notation, what are (a) the object's angular momentum and (b) the torque acting on the object?

A uniform solid sphere rolls down an incline. (a) What must be the incline angle if the linear acceleration of the center of the sphere is to have a magnitude of \(0.10 g ?\) (b) If a frictionless block were to slide down the incline at that angle, would its acceleration magnitude be more than, less than, or equal to \(0.10 \mathrm{~g}\) ? Why?

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