# 5.8: Applications of Vector Analysis

**At Grade**Created by: CK-12

## Learning objectives

- Use the dot product to determine the work done on a moving object by a given force.
- Use the cross product to determine the force exerted on a moving charge by a magnetic field.
- Use the cross product to determine the torque exerted by a force on a object.

## Introduction

Vectors can be used by air-traffic controllers when tracking planes, by meteorologists when describing wind conditions, and by computer programmers when they are designing virtual worlds. In this section, we will present three applications of vectors which are commonly used in the study of physics: work, torque, and magnetic force.

## Work

In physics, the term **work** is used to describe energy that is added to or removed from an object or system when a force is applied to it. From experiment, it has been determined that work is maximized when the applied force is parallel to the motion of the object and that no work is done when the force is applied perpendicular to the motion. Therefore, the work done by a force can be described by the dot product of the force vector and the displacement vector. For example, several forces act on the skier in the diagram below.

Sven’s weight pulls downward toward the center of the earth, the snow support’s Sven by pushing upward on his skis perpendicular to the slope, and the friction between Sven’s skis and the snow points in the opposite direction from his motion. The work done by each of these forces can be determined using the dot product of the force and the displacement vector \begin{align*}\overrightarrow{\triangle x}, W = \overrightarrow{F} \times \overrightarrow{\triangle x} = F(\triangle x) \ \mbox{cos} \ \theta\end{align*}

\begin{align*}W_{friction} = \overrightarrow{F_{friction}} \times \overrightarrow{\triangle x} = (F_{friction})(\triangle x)\ \mbox{cos} \ 180^\circ = -F_{friction}\triangle x\end{align*}

\begin{align*}W_{support} = \overrightarrow{F_{support}} \times \overrightarrow{\triangle x} = (F_{support})(\triangle x)\ \mbox{cos} \ 90^\circ = 0\end{align*}

\begin{align*}W_{weight} = \overrightarrow{F_{weight}} \times \overrightarrow{\triangle x} = (F_{weight})(\triangle x)\ \mbox{cos} \ \theta = (F_{weight})(\triangle x)\ \mbox{sin} \ \phi\end{align*}

**Example**: Xiao turns a crank to lower a bucket of water into a well. Determine the total work done on the bucket if the weight of the bucket is 15 N and the tension force in the rope is 13 N. The bucket rises a distance of 4.5 m while he is cranking.

**Solution:** The work done by a force acting on an object is described by the dot product of the force vector and the displacement vector: \begin{align*}W = \overrightarrow{F} \times \overrightarrow{\triangle x} = F(\triangle x)\ \mbox{cos} \ \theta\end{align*}

\begin{align*}W_{rope} = \overrightarrow{F_{rope}} \times \overrightarrow{\triangle x} = F_{rope}(\triangle x)\ \mbox{cos} \ \theta = (13N)(4.5m) \ \mbox{cos} \ 180^\circ = -58.5J\end{align*}

The weight force does positive work on the bucket because the motion and the force are in the same direction.

\begin{align*}W_{weight} = \overrightarrow{F_{weight}} \times \overrightarrow{\triangle x} = F_{weight}(\triangle x)\ \mbox{cos} \ \theta = (15N)(4.5m) \ \mbox{cos} \ 0^\circ =\end{align*}

The total work is the sum of the two individual amounts of work.

\begin{align*}W_{total} = W_{rope} + W_{weight} = -58.5 + 67.5J=9.0J\end{align*}

A total of 9.0 J of work is done on the bucket as the bucket moves downward into the well.

## Magnetic Force

The force that a magnetic field exerts on a charged particle is strongest when the particle moves perpendicular to the field and the magnetic force on the particle is equal to zero when it moves parallel to the field. Therefore the magnetic force can be described using the cross-product of the field strength vector and the particle’s velocity vector: \begin{align*}\overrightarrow{F} = q\overrightarrow{v} \times \overrightarrow{B}\end{align*}*q* is the charge of the particle, \begin{align*}\overrightarrow{v}\end{align*}

**Example:** In a cathode-ray-tube EKG monitor, the heart-beat trace on the screen is drawn by a beam of electrons that move from the back of the tube toward the screen. When the electrons hit the screen, the collision transfers energy from the electrons to the phosphor on the inner surface of the screen, which then glows. The point where the electrons hit the screen can be changed by changing the magnetic field inside the tube. A sensor connected to the patient translates the electrical pulses across the heart into the strength of the magnetic field inside the EKG’s cathode ray tube, which then changes the path of the electrons. The force which the magnetic field exerts on the electrons is determined both by the strength of the magnetic field and on the velocity of the electrons. If the electrons move at 25,000 m/s northward through a 5.2 millitesla magnetic field which points west to east across the tube, what is the magnitude and direction of the force exerted on the electron. The electron has a charge of \begin{align*}q_{electron} = -1.6 \times 10^{-19} C\end{align*}

## Torque

When you lift a baseball off a table-top, you are exerting a force that moves the object as a whole. When you apply a force to a doorknob, you cause the door to rotate on its hinges. Scientists use the term **torque** to describe the force-like property that affects the rotation of an object. The torque can be described using the cross-product of the force vector and the lever arm, a vector pointing radially outward from the axis of rotation to the point where the force is applied to the object: \begin{align*}\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}\end{align*}

## Practice Problems

- The work done by a force on an object is given by the dot product of the force vector and the displacement vector representing the motion of the object. The diagram below shows Sanjay pulling a large crate across the floor. The four forces which act on the crate during this process are shown in the diagram below. Which of the four forces exert non-zero force on the crate?
- Le’andra is pulling her toy duck (mass 0.75 kg) at a constant speed of 3.0 m/s. The string she uses to pull the duck makes an angle of 42
^{o}above the horizontal and Le’andra keeps a constant tension in the string of 2.0 N. What is the amount of work done by the tension force when the duck is pulled forward a distance of 2.8 m? - Beuford has once again taken Brynna to the park to play on the slide. If Brynna has a weight of 25 kg and if the slide has a 30
^{o}incline above the horizontal, what work is done by her weight as she slides down the 3.5 m incline? Remember that the weight force in newtons is equal to the product of the mass and the acceleration of gravity, 9.8 m/s^{2}. - The scientists at Fermi Lab in Chicago, IL use magnetic fields to direct beams of protons during their explorations of the submicroscopic structure of atoms and quarks. Both magnetic field strength and the velocity of the protons are vector quantities. Determine the force on a proton moving northward at 4.2 × 10
^{6}m/s through a magnetic field of 2.5 T oriented from east to west. (Note: protons are very tiny and are therefore able to move VERY fast.) - Vector \begin{align*}\overrightarrow{B}\end{align*}
B⃗ represents the magnitude and direction of the magnetic field in a certain region. Vector \begin{align*}\overrightarrow{v}\end{align*}v⃗ represents the velocity of a charged particle,*q*= 3.2 × 10^{-15}C, which moves through the magnetic field. The lengths of the vectors are such that the velocity vector is measured in m/s and the magnetic field vector is measured in tesla. What is the force experienced by the charged particle as it moves through the magnetic field? - If two children, Rudolfo (weight = 210 N) and Jennifer (175 N), sit on either end of a see-saw as shown below. Rudolfo is 1.0 m from the pivot and Jennifer is 1.4 m from the pivot. What torques are exerted by the children on the see-saw?
- Exercise scientists and physical therapists use torque to analyze various exercises such as the triceps exercise shown in the diagram below. Here, the triceps muscle exerts a force on the elbow-end of the forearm, affecting the rotation of the forearm. If the triceps exerts a force of 17N, what torque is applied to the forearm?

## Solutions

- The diagram below shows Sanjay pulling a large crate across the floor. The four forces which act on the crate during this process are shown in the diagram below. Which of the four forces exert non-zero force on the crate? The dot product is defined by \begin{align*}\overrightarrow{A} \times \overrightarrow{B} = |A||B| \ \mbox{cos} \ \theta\end{align*}
A⃗×B⃗=|A||B| cos θ therefore only forces which have at least some component parallel to the motion will do non-zero work on the object. The angle θ between the displacement and forces perpendicular to the motion is 90^{o}so \begin{align*}\overrightarrow{A} \times \overrightarrow{B} = |A||B| \ \mbox{cos} \ \theta = 0\end{align*}A⃗×B⃗=|A||B| cos θ=0 . The force from the floor and the weight of the crate do no work, since both of these forces are perpendicular to the motion of the crate. The rope does positive work on the crate since the force of the rope on the crate has a non-zero x-component. The friction does negative work on the crate since it is in the opposite direction from the displacement. - De’andra is pulling her toy duck (mass 0.75 kg) at a constant speed of 3.0 m/s. The string she uses to pull the duck makes an angle of 42
^{o}above the horizontal and De’andra keeps a constant tension in the string of 2.0 N. What is the amount of work done by the tension force when the duck is pulled forward a distance of 2.8 m? The work done by De’andra on the duck depends on the force she uses to pull the duck and on the distance the duck moves while she pulls. It also depends on the angle between the pulling force and the displacement vector. \begin{align*}\overrightarrow{F_{pull}} \times \overrightarrow{x} = |\overrightarrow{F_{pull}}| |\overrightarrow{x}| \ \mbox{cos} \ \theta = (2.0N)(2.8m) \ \mbox{cos} \ 42^\circ = 4.16Nm\end{align*}Fpull−→−×x⃗=|Fpull−→−||x⃗| cos θ=(2.0N)(2.8m) cos 42∘=4.16Nm The N represents “newton”, the unit of force. The m represents “meters,” the unit of displacement. (1.0 N)(1.0 m) = 1.0 J where J represents “joules,” the unit of work and energy. - Beuford has once again taken Brynna to the park to play on the slide. If Brynna has a weight of 25 kg and if the slide has a 30
^{o}incline above the horizontal, what work is done by her weight as she slides down the 3.5 m incline? Remember that the weight force in newtons is equal to the product of the mass and the acceleration of gravity, 9.8 m/s^{2}. If the slide is inclined at 30^{o}above the horizontal, then θ = 60^{o}from the vertical. The work done by a force on an object is given by the dot product of the force and the displacement of the object. Here \begin{align*}\overrightarrow{F_{weight}} = mg = (25kg)(9.8m/s^2) = 245N\end{align*}Fweight−→−−−=mg=(25kg)(9.8m/s2)=245N . Therefore, \begin{align*}W = \overrightarrow{F_{weight}} \times \overrightarrow{d} = |\overrightarrow{F_{weight}}| |\overrightarrow{d}| \ \mbox{cos} \ \theta = (245N)(3.5m)\ \mbox{cos} \ 60^\circ = 428.75J\end{align*}W=Fweight−→−−−×d⃗=|Fweight−→−−−||d⃗| cos θ=(245N)(3.5m) cos 60∘=428.75J - The scientists at Fermi Lab in Chicago, IL use magnetic fields to direct beams of protons during their explorations of the submicroscopic structure of atoms and quarks. Both magnetic field strength and the velocity of the protons are vector quantities. Determine the force on a proton moving northward at 4.2 × 10
^{6}m/s through a magnetic field of 2.5 T oriented from east to west. (Note: protons are very tiny and are therefore able to move VERY fast.) Define a coordinate system where eastward is the +z direction, northward is the +y direction, and upward is the +z direction. In this coordinate system, \begin{align*}\overrightarrow{v} = \left \langle 0, 4.2 \times 10^6, 0 \right \rangle\end{align*}v⃗=⟨0,4.2×106,0⟩ and \begin{align*}\overrightarrow{B} = \left \langle -2.5, 0, 0 \right \rangle\end{align*}. Therefore, \begin{align*}|\overrightarrow{F}| = q |\overrightarrow{v} \times \overrightarrow{B}| = q \left ( |\overrightarrow{v}| |\overrightarrow{B}| \ \mbox{sin} \ \theta \right )\end{align*}. Since the velocity is northward and the magnetic field is westward, the angle between the two vectors is 90^{o}. \begin{align*}|\overrightarrow{F}| = q |\overrightarrow{v} \times \overrightarrow{B}| = q \left ( | \overrightarrow{v}| |\overrightarrow{B}|\ \mbox{sin}\ \theta \right ) = (1.6 \times 10^{-19}) \left ((4.2 \times 10^6)(2.5) \ \mbox{sin} \ 90^\circ \right ) =\end{align*} \begin{align*}16.8 \times 10^{-13}N\end{align*} Using the right hand rule, we can determine the direction of the force on the proton. If you point your thumb northward along the velocity vector and your fore-finger westward along the magnetic field vector, your palm and your extended middle-finger point upward. Therefore the force which the magnetic field exerts on the proton is in the +z direction: \begin{align*}|\overrightarrow{F}| = q\overrightarrow{v} \times \overrightarrow{B} = \left \langle 0, 0, 16.8 \times 10^{-13} N \right \rangle\end{align*}. - Vector \begin{align*}\overrightarrow{B}\end{align*} represents the magnitude and direction of the magnetic field in a certain region. Vector \begin{align*}\overrightarrow{v}\end{align*} represents the velocity of a charged particle,
*q*= 3.2 × 10^{-15}C, which moves through the magnetic field. The lengths of the vectors are such that the velocity vector is measured in m/s and the magnetic field vector is measured in tesla. What is the force, \begin{align*}\overrightarrow{F}\end{align*}, experienced by the charged particle as it moves through the magnetic field? We saw in a previous problem that \begin{align*}|\overrightarrow{F}| = q |\overrightarrow{v} \times \overrightarrow{B}|\end{align*}, therefore can use the component version of the cross product equation to solve this problem. \begin{align*}\overrightarrow{F} = q \left (\overrightarrow{v} \times \overrightarrow{B} \right ) = q \left \langle (v_yB_z - v_zB_y), (v_zB_x - v_xB_z), (v_xB_y - v_yB_x) \right \rangle\end{align*} - If two children, Rudolfo (weight = 210 N) and Jennifer (weight = 175 N), sit on either end of a see-saw as shown below. Rudolfo is 1.0 m from the pivot and Jennifer is 1.4 m from the pivot. What torques are exerted by the children on the see-saw? The coordinate system has been defined such that the weight-force vectors are parallel to the y-axis and the lever-arm vectors are parallel to the x-axis. First determine the component form of each vector equation and then use the component version of the cross-product equation to determine the torque exerted by each child. \begin{align*}\overrightarrow{\tau_R} = \overrightarrow{r_R} \times \overrightarrow{F_{weight,R}} = \left \langle -1.0m, 0, 0 \right \rangle \times \left \langle 0, -210N, 0 \right \rangle\end{align*} \begin{align*}\overrightarrow{r} \times \overrightarrow{F} = \left \langle (r_yF_z - r_zF_y), (r_zF_x - r_xF_z), (r_xF_y - r_yF_x) \right \rangle\end{align*} \begin{align*}\overrightarrow{r_R} \times \overrightarrow{F_R} = \left \langle \left ((0*0) - (0*-210)\right), \left ((0*0) - (-1*0)\right ), \left ((-1*-210) - (0*0)\right ) \right \rangle =\end{align*} \begin{align*}\left \langle 0, 0, 210mN \right \rangle\end{align*} \begin{align*}\overrightarrow{\tau_J} = \overrightarrow{r_J} \times \overrightarrow{F_{weight,J}} = \left \langle 1.4m, 0, 0 \right \rangle \times \left \langle 0, -170N, 0 \right \rangle\end{align*} \begin{align*}\overrightarrow{r} \times \overrightarrow{F} = \left \langle (r_yF_z - r_zF_y), (r_zF_x - r_xF_z), (r_xF_y - r_yF_x) \right \rangle\end{align*} \begin{align*}\overrightarrow{r_J} \times \overrightarrow{F_J} = \left \langle \left ((0*0) - (0*-170)\right ), \left ((0*0) - (1.4*0)\right ), \left ((1.4*-170) - (0*0)\right ) \right \rangle =\end{align*} \begin{align*}\left \langle 0, 0, -238mN \right \rangle\end{align*}
- Exercise scientists and physical therapists use torque to analyze various exercises such as the triceps exercise shown in the diagram below. Here, the triceps muscle exerts a force on the elbow-end of the forearm, affecting the rotation of the forearm. If the triceps exerts a force of 17N, what torque is applied to the forearm? A close-up of the triceps force and lever-arm is shown below. Since the forearm is positioned at an angle of 15
^{o}to the vertical, the angle between the two vectors is 90^{o}– 15^{o}= 75^{o}. The magnitude of the lever-arm vector is the distance from the elbow-pivot to the point where the triceps pulls on the bone, \begin{align*}|\overrightarrow{r}| = 2.5cm\end{align*}. Similarly, the magnitude of the force vector is the strength of the force, \begin{align*}|\overrightarrow{F}| = 17N\end{align*}. Since we know the magnitudes of both vectors and the angle between them, we can use the angle-version of the cross-product equation to determine the magnitude of the torque. \begin{align*}|\overrightarrow{\tau}| = rF \ \mbox{sin} \ \theta = (2.5cm)(17N) \ \mbox{sin} \ 75 = 41.05cm \cdot N\end{align*}

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