16.1: Vectors

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Vectors are used in engineering mechanics to represent quantities that have both a magnitude and a direction. Many engineering quantities, such as forces, displacements, velocities, and accelerations, will need to be represented as vectors for analysis. Vector quantities contrast with scalar values (such as mass, area, or speed), which have a magnitude but no direction.

A person pushes a large block towards the left, applying a force of 70 lbs directed at 15° above the horizontal. — Figure \(\PageIndex{1}\): The force below is represented as a vector. It has both a magnitude and a direction.

When dealing with vectors in equations, engineers commonly denote something as a vector by putting an arrow over the variable. Variables without an arrow over top of them represent a scalar quantity, or simply the magnitude of that vector quantity.

\begin{align} \text{Vector Quantity:} \quad &\, \vec{F} \\ \text{Scalar Quantity:} \quad &\, F \end{align}

In contrast with scalar quantities, we cannot add, subtract, multiply or divide them by simply adding, subtracting, multiplying or dividing the magnitudes. The directions will also play a critical role in solving equations that contain vector quantities.

Vector Representation:

To represent a vector quantity, we will generally have two options. These two options are:

Magnitude and Direction Form: Where the magnitude is given as a single quantity and the direction is given via an angle or combination of angles.
Component Form: Where the magnitude and direction are given through component magnitudes in each coordinate direction.

Two diagrams of the same situation, where a person pushes a block towards the left by applying a force directed both leftwards and upwards. In one diagram, that force vector is represented with a magnitude of 70 lbs and a direction of 15° above the horizontal. In the second diagram, that same force vector is represented with components of -67.6 lbs in the leftwards or negative x-direction and 18.1 lbs in the upwards or positive y-direction. — Figure \(\PageIndex{2}\): The drawing above shows a force as vector. On the left the vector is represented as a magnitude and a direction. On the right the same force is represented in terms of its \(x\)- and \(y\)-components.

The magnitude and direction form of vector quantities are often used at the start and end of a problem. This is because it is often easier to measure things likes forces and velocities as a magnitude and direction at the start of a problem, and it is often easier to visualize the final result as a magnitude and direction at the end. Vectors represented as a magnitude and direction need to be shown visually through the use of an arrow, where the magnitude is the length of the arrow, and the direction is shown through the arrow head and an angle or angles relative to some known axes or other direction.

The component form of a vector is often used in middle of the problem because it is far easier to do math with vector quantities in component form. Vectors can be represented in component form in one of two ways. First we can use square brackets to indicate a vector, with the \(x\), \(y\), and possibly \(z\) components separated by commas. Alternatively, we can write out a vector in component form using the magnitudes in front of unit vectors to indicate directions (generally the \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) unit vectors for the \(x\), \(y\), and \(z\) directions respectively). Neither of these component forms relies on a visual depiction of the vector as with the magnitude and direction form, though it is important to clearly identify the coordinate system in earlier diagrams.

\begin{align} \text{With Brackets:} \quad &\, \vec{F} = [3,4,5] \\ \text{With Unit Vectors:} \quad &\, \vec{F} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \end{align}

Converting Between Vector Representations in 2D

For our analysis, we will often find it advantageous to have the vectors in one form or the other, and will therefore need to convert the vector from a magnitude and direction to component form or vice versa. To do this we will use right triangles and trigonometry.

Going from a Magnitude and Direction to Component Form

To go from a magnitude and direction to component form, we will first draw a right triangle with the hypotenuse being the original vector. The horizontal arm of the triangle will then be the \(x\)-component of the vector while the vertical arm is the \(y\) component of the vector. If we know the angle of the vector with respect to either the horizontal or the vertical, we can use the sine and cosine relationship to find the \(x\) and \(y\) components.

A force vector with magnitude 600 N points upwards and to the right, at a 35° angle above the horizontal. A horizontal line pointing towards the right is drawn from the tail of the vector, and labeled as the component of the original force vector in the x-direction. A vertical line pointing upwards is drawn from that horizontal line to the head of the original vector, and labeled as the component of the original force vector in the y-direction. The magnitudes of these respective components are F_x = 600 * cos(35°), and F_y = 600 * sin(35°). — Figure \(\PageIndex{3}\): Using sine and cosine relationships, we can find the \(x\) and \(y\) components of a vector.

It is important to remember that how we measure the angle will affect the sine and cosine relationships. Multiplying the magnitude by the sine will always give us the opposite leg and multiplying the magnitude by the cosine will always give us the adjacent leg.

Two diagrams of the same force vector F extending up and to the right, with its tail at the origin of a Cartesian coordinate plane. In one diagram, the vector's direction is the angle theta that it makes with the positive x-axis; therefore the vector's components are calculated as F_x = F cos(theta) and F_y = F sin(theta). In the other diagram, the vector's direction is the angle theta that it makes with the positive y-axis; therefore the vector's components are calculated as F_y = F cos(theta) and F_x = F sin(theta). — Figure \(\PageIndex{4}\): Depending upon how we measure the angle, the sine/cosine may be either the \(x\) or the \(y\) component. Remember that the sine will always give you the opposite leg, while the cosine will give you the adjacent leg.

Going from Component Form to Magnitude and Direction

To find the magnitude and the direction of a vector using components, we will use the same process in reverse. We will draw the components as the legs of a right triangle, where the hypotenuse of the triangle shows the magnitude and direction of the vector.

To find the magnitude of the vector we will use the Pythagorean Theorem, taking the square root of the sum of the squares of each component. To find the angle, we can easily use the inverse tangent function, relating the opposite and adjacent legs of our right triangle.

Graph of a vector F extending up and to the right, with its tail at the origin of a Cartesian coordinate plane. The magnitude of this vector is found by taking the square root of the sum of the squares of its components, F_x and F_y. The direction of this vector, in the form of the angle it makes with the positive x-axis, is found by dividing F_y by F_x, then taking the inverse tangent of this quotient. — Figure \(\PageIndex{5}\): We will use the Pythagorean Theorem and the inverse tangent function to convert the vector back into magnitude and direction form.

If we know the magnitude of the hypotenuse, we can also use the inverse sine and cosine functions in place of the inverse tangent function to find the angle. As with the previous conversion, it is important to clearly identify the opposite leg, the adjacent leg, and the hypotenuse in our diagrams and to think of these when applying the inverse trig functions.

Converting Between Vector Representations in 3D

In three dimensions, we will have either three components (\(x\), \(y\), and \(z\)) for component form or a magnitude and two angles for the direction in magnitude and direction form. To convert between forms we will need to draw in two sets of right triangles. The hypotenuse of the first triangle will be the original vector and one of the legs will be one of the three components. The other leg will then be the hypotenuse of the second triangle. The legs of this second triangle will then be the remaining two components as shown in the diagram below. Use sine and cosine relationships to find the magnitude of each component along the way. This general process of two consecutive right triangles will always hold true, but depending on angles that are given or chosen which components end up being which leg can vary. Carefully plotting everything out in a diagram is important for this reason.

A vector F, whose tail is at the origin, in the first octant of a 3-dimensional Cartesian coordinate system with the x- and y-axes in the plane of the screen and the z-axis pointing out of the screen. Vector F makes an angle of phi above the xz-plane, forming the hypotenuse of a right triangle whose legs are the original vector's F_y and F_xz components. In the xz-plane, the F_xz vector makes an angle of theta with the positive x-axis, forming the hypotenuse of a right triangle whose legs are the original vector's F_x and F_z components. — Figure \(\PageIndex{6}\): For 3D vectors we will need to draw two right triangles to convert between forms.

To go from component form back to a magnitude and direction, we will use the 3D form of the Pythagorean Theorem (the magnitude will be the square root of the sum of the three components squared) and we can again use the inverse trig functions to find the angles. We simply need to work backwards through the two right triangles in our problem, so again it is important to carefully draw out your diagrams.

Alternative Method for Finding 3D Vector Components

Sometimes, as with the tension in a cable, the geometry of the cable is given in component form rather than as angles. In cases such as this we could use geometry to figure out the angles and then use those angles to figure out the components, but there is a mathematical shortcut that will allow us to solve for the components more quickly involving the ratio of lengths. Specifically, the ratios of the components of the cable length to the overall length of the cable will be the same as the ratio of the corresponding force components to the overall magnitude of the force.

To use this method we will first need to find the overall length of the cable (or other physical geometric feature) using the Pythagorean Theorem. Once we have that overall length, we find a ratio by taking the x component of the length divided by the overall length. To find the \(x\)-component of the force, we simply multiply the overall magnitude of the force by this ratio of lengths (\(L_x/L\)). The process for the \(y\) and \(z\) components follows a similar path, except the ratios would include the \(y\) and \(z\) component lengths instead of the \(x\) component.

The first octant of a 3-dimensional Cartesian coordinate system, with the x- and y-axes in the plane of the screen and the z-axis pointing out of the screen. A vertical post, parallel to the y-axis, has one end on the xz-plane and the other end connected to the origin via a taut cable of length L. A tension force T points along this cable, starting from the origin. Below the diagram, the relation between the cable's length components and tension force components to their respective total quantities are provided as 3 separate ratios: T_x / T = L_x / L, T_y / T = L_y / L, and T_z / T = L_z / L. — Figure \(\PageIndex{7}\): The alternative method of breaking a vector down into component form relies on the ratios of the cable components to the overall length of the cable.

Video lecture covering this section, delivered by Dr. Jacob Moore. YouTube source: https://youtu.be/2RO7VZ_u0Iw.

Example \(\PageIndex{1}\)

Determine the \(x\) and \(y\) components of the vector shown below.

A flat circular disk is wedged in a straight-sided groove. One side of the groove exerts a force with a magnitude of 300 N on the disk, directed upwards and to the right at 40° above the horizontal. — Figure \(\PageIndex{8}\): problem diagram for Example \(\PageIndex{1}\). A force of magnitude 300 N is directed upwards and to the right, at 40° above the horizontal.

Solution:: Video \(\PageIndex{2}\): Worked solution to example problem \(\PageIndex{1}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/TRS6jiih96o.

Example \(\PageIndex{2}\)

Determine the \(x\) and \(y\) components of the vector shown below.

A traffic light is suspended from two cables. A tension force T1 of magnitude 60 lbs acts along one of these cables, pointing upwards and to the right at a 75° angle from the vertical. — Figure \(\PageIndex{9}\): problem diagram for Example \(\PageIndex{2}\). A tension force of magnitude 60 lbs is exerted along a cable, directed upwards and to the right at a 75° angle from the vertical.

Solution:: Video \(\PageIndex{3}\): Worked solution to example problem \(\PageIndex{2}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/hUrlI6eLGvE.

Example \(\PageIndex{3}\)

The velocity vector of the hockey puck shown below is given in component form. Determine the magnitude and direction of the velocity with respect to the axes given.

A hockey puck is located at the origin of a standard-orientation Cartesian coordinate plane. It experiences a velocity vector pointing downwards and to the right, with an x-component of 5 m/s and a y-component of -2.5 m/s. — Figure \(\PageIndex{10}\): problem diagram for Example \(\PageIndex{3}\). A hockey puck located at the origin of a Cartesian coordinate plane experiences a velocity of components \([5, -2.5]\) m/s.

Solution:: Video \(\PageIndex{4}\): Worked solution to example problem \(\PageIndex{3}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/3GeIU3Fc_qA.

Example \(\PageIndex{4}\)

Determine the \(x\), \(y\), and \(z\) components of the force vector shown below.

A vector with its tail at the origin of a 3D Cartesian coordinate system, with the x- and y-axes lying in the plane of the screen and the z-axis pointing out of the screen. The vector has a magnitude of 45 N. Its direction is 20° out of the plane of the screen towards the viewer, and 30° below the xz-plane. — Figure \(\PageIndex{11}\): problem diagram for Example \(\PageIndex{4}\). A 45-Newton force vector with its tail at the origin of a Cartesian coordinate system points 20° out of the screen towards the viewer and 30° below the plane of the horizontal.

Solution:: Video \(\PageIndex{5}\): Worked solution to example problem \(\PageIndex{4}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/ZG31PoIfIEc.

Example \(\PageIndex{5}\)

A cable as shown below is used to tether the top of a pole to a point on the ground. The cable has a tension force of 3 kN that acts along the direction of the cable as shown below. What are the \(x\), \(y\), and \(z\) components of the tension force acting on the top of the pole?

The first octant of a three-dimensional Cartesian coordinate system, with the x- and y-axes lying in the plane of the screen and the z-axis pointing out of the screen. All axes have units in meters. A 6-meter-long vertical pole has one end located at the point (2, 0, 3) and extends upwards from that point parallel to the y-axis. A cable connects the system origin to the top end of the pole, and a 3-kN tension force acts along that cable, directed towards the origin. — Figure \(\PageIndex{12}\): problem diagram for Example \(\PageIndex{5}\). A cable connects the origin of a Cartesian coordinate system to the top of a 6-meter-long vertical pole with its base at the point (2, 0, 3) meters.

Solution:: Video \(\PageIndex{6}\): Worked solution to example problem \(\PageIndex{5}\), provided by Dr. Jacob Moore. YouTube source: https://youtu.be/dUJwBohfCTU.