How much work is done by a 10kg ball rolling down a 25* ^{o}* slope for 10 yards?

How can the force which propels a proton through a magnetic field be calculated?

How could you calculate the torque exerted by children on a see-saw?

This lesson is all about real-world uses of vector analysis.

### Watch This

James Sousa: Graphing Quadratics

### Guidance

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}\end{align*}.

\begin{align*}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*}

*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*} where \begin{align*}\overrightarrow{F}\end{align*} is the force on the particle, *q* is the charge of the particle, \begin{align*}\overrightarrow{v}\end{align*} is the velocity of the particle, and \begin{align*}\overrightarrow{B}\end{align*} is the vector representing the magnetic field. If the velocity is measured in m/s and if the magnetic field is measured in tesla, the force will be measured in newtons, the metric base-unit of force.

*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*}, where \begin{align*}\overrightarrow{\tau}\end{align*} is the torque, \begin{align*}\overrightarrow{r}\end{align*} is the lever arm, and \begin{align*}\overrightarrow{F}\end{align*} is the applied force.

#### Example A

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*}. In this case, the rope does negative work on the bucket because the motion and the force are in opposite directions. If the force is measured in newtons and the displacement in meters, the work is measured in Joules.

\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*} \begin{align*}67.5J\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.

#### Example B

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?

*Solution:*

The dot product is defined by \begin{align*}\overrightarrow{A} \times \overrightarrow{B} = |A||B| \ \mbox{cos} \ \theta\end{align*} 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*}. 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.

#### Example C

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?

*Solution:*

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*}

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.

-->

### Guided Practice

1) 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}.

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.)

3) 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?

- 4) 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?

- 5) 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?

*Answers*

1) 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*}. 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*}

2) 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*} 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*}.

3) 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*}

4) 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*}

- 5) 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*}

### Explore More

- A fishing boat sails due west at a constant 15 km/h. A current of 3 km/h is running due south. What is the velocity of the boat relative the sea floor?
- Two bulldozers pull on a large tree to move it out of the way of a road. One bulldozer pulls 3000N west and the other bulldozer 2500N in a southerly direction because of the terrain. Determine the resultant force acting on the tree. If the tree has a mass of 1000 kg calculate the acceleration of the tree due to the resultant force acting on it.
- Three coplanar horizontal forces each of magnitude 10 N act on a body of mass 5 kg as shown below. Determine the magnitude of the net force acting on the body and the magnitude of the resultant acceleration.
- If vector A = 5 N north and vector B = 10 N east, find the resultant of vector A – vector B.
- A submarine is travelling at 20 km/h due east. Shortly after, it is travelling due north at 15 km/h. Calculate the change in velocity.
- An fighter jet has a true airspeed of 1000 km/h due east. There is a cross wind blowing 60 degrees east of south at 100 km/h. Calculate the velocity of the jet relative to the ground.
- A pallet of mass 10 kg sits on a horizontal surface, a force is exerted on the pallet by means of a chain inclined at 60° to the horizontal. If the tension in the chain is 150 N and the frictional force between the pallet and the horizontal surface is 55 N, will the pallet move under these conditions? Explain. If the pallet does move determine the size of the acceleration with which it moves.
- A broken down car with a total mass of 300 kg is pulled by a single tow line attached to a tow-truck with a mass of 1000 kg. If the drag on the truck and the car is one-tenth of their respective weights, and the total forward force exerted by the tow-truck is 5130 kg force (ie 5130 x 9.8 N), find the magnitude of: a. the total force resisting the forward motion of the truck and car; b. the acceleration of the truck/car system; c. the unbalanced accelerating force on the broken down car; d. the tension in the towing line.
- A fighter pilot with a mass of MA = 80 kg sits in the cock-pit with his back horizontal to the ground. His jet is moving vertically with acceleration a. If the acceleration due to gravity on the pilot is g = 9.8 ms-2, write a mathematical expression for and calculate the value of the reaction force R between the pilot and the back of his seat in the jet when: a. a = 0; b. a = 8 ms-2 upwards; c. a = 8 ms-2 downwards.
- A plane is flying 691mph headed 13° east of north. A tail wind is blowing 67° west of north at 92 mph. Determine the actual speed of the plane relative the ground.
- A bird is headed 40° east of north at 67 mph. A tail wind is blowing 45° west of south at 68 mph. Determine the direction of the bird.
- A UFO is gliding along at 516 mph, headed 32° east of south. Determine the direction of the UFO if a tail wind is blowing 42° west of north at 29 mph.
- A rocket is flying 63° west of north at 575 mph. Determine the actual direction of the rocket if a tail wind is blowing 22° west of south at 78 mph.
- What is the resultant direction of a speed boat racing along at 66.5mph headed 80° west of south at 611 mph, if a tail wind is pushing it toward a direction 89° east of south at 4.5 mph?
- Several evil villians are holding Spiderman in place with ropes. If Flat-Nose Frankie is pulling at a constant force of 19 pounds, 44° to the horizontal, Green-Toe Gary is pulling at a constant force of 29 pounds along 31° to the horizontal, and Jelly-Knee Jennifyr is pulling at a constant force of 26 pounds at an angle of 33°, How hard and in what direction is Orange-Chin Oswald pulling if Spiderman is stuck in place?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 5.11.