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1.1 Introduction
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1.3 Components of vectors
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In grade 10 you learnt about the resultant vector in one dimension, we are going to extend this to two dimensions. As a reminder, if you have a number of vectors (think forces for now) acting at the same time you can represent the result of all of them together with a single vector known as the resultant. The resultant vector will have the same effect as all the vectors adding together.
We will focus on examples involving forces but it is very important to remember that this applies to all physical quantities that can be described by vectors, forces, displacements, accelerations, velocities and more.
The first thing to make a note of is that in Grade 10 we worked with vectors all acting in a line, on a single axis. We are now going to go further and start to deal with two dimensions. We can represent this by using the Cartesian plane which consists of two perpendicular (at a right angle) axes. The axes are a \(x\)-axis and a \(y\)-axis. We normally draw the \(x\)-axis from left to right (horizontally) and the \(y\)-axis up and down (vertically).
We can draw vectors on the Cartesian plane. For example, if we have a force, \(\vec{F}\), of magnitude \(\text{2}\) \(\text{N}\) acting in the positive \(x\)-direction we can draw it as a vector on the Cartesian plane.
Notice that the length of the vector as measured using the axes is \(\text{2}\), the magnitude specified. A vector doesn't have to start at the origin but can be placed anywhere on the Cartesian plane. Where a vector starts on the plane doesn't affect the physical quantity as long as the magnitude and direction remain the same. That means that all of the vectors in the diagram below can represent the same force. This property is know as equality of vectors.
In the diagram the vectors have the same magnitude because the arrows are the same length and they have the same direction. They are all parallel to the \(x\)-direction and parallel to each other.
This applies equally in the \(y\)-direction. For example, if we have a force, \(\vec{F}\), of magnitude \(\text{2,5}\) \(\text{N}\) acting in the positive \(y\)-direction we can draw it as a vector on the Cartesian plane.
Just as in the case of the \(x\)-direction, a vector doesn't have to start at the origin but can be placed anywhere on the Cartesian plane. All of the vectors in the diagram below can represent the same force.
The following diagram shows an example of four force vectors, two vectors that are parallel to each other and the \(y\)-axis as well as two that are parallel to each other and the \(x\)-axis.
To emphasise that the vectors are perpendicular you can see in the figure below that when originating from the same point the vector are at right angles.
Draw the following forces as vectors on the Cartesian plane originating at the origin:
We choose a scale of \(\text{1}\) \(\text{N}\):\(\text{1}\) \(\text{cm}\). Now we sketch the vectors on the Cartesian plane (we can place the vectors anywhere on the Cartesian plane, we will place them starting at the origin):
Draw the following forces as vectors on the Cartesian plane:
We choose a scale of \(\text{1}\) \(\text{N}\):\(\text{1}\) \(\text{cm}\). Now we sketch the vectors on the Cartesian plane (we can place the vectors anywhere on the Cartesian plane):
Draw the following forces as vectors on the Cartesian plane:
We choose a scale of \(\text{1}\) \(\text{N}\):\(\text{1}\) \(\text{cm}\). Now we sketch the vectors on the Cartesian plane (we can place the vectors anywhere on the Cartesian plane):
Draw the following forces as vectors on the Cartesian plane:
We choose a scale of \(\text{1}\) \(\text{N}\):\(\text{1}\) \(\text{cm}\). Now we sketch the vectors on the Cartesian plane (we can place the vectors anywhere on the Cartesian plane):
Vectors in two dimensions are not always parallel to an axis. We might know that a force acts at an angle to an axis so we still know the direction of the force and if we know the magnitude we can draw the force vector. For example, we can draw \(\vec{F}_{1} = \text{2}\text{ N}\) acting at \(\text{45}\)\(\text{°}\) to the positive \(x\)-direction:
We always specify the angle as being anti-clockwise from the positive \(x\)-axis. So if we specified an negative angle we would measure it clockwise from the \(x\)-axis. For example, \(\vec{F}_{1} = \text{2}\text{ N}\) acting at \(-\text{45}\)\(\text{°}\) to the positive \(x\)-direction:
We can use many other ways of specifying the direction of a vector. The direction just needs to be unambiguous. We have used the Cartesian coordinate system and an angle with the \(x\)-axis so far but there are other common ways of specifying direction that you need to be aware of and comfortable to handle.
We can use compass directions when appropriate to specify the direction of a vector. For example, if we were describing the forces of tectonic plates (the sections of the earth's crust that move) to talk about the forces involved in earthquakes we could talk the force that the moving plates exert on each other.
The four cardinal directions are North, South, East and West when using a compass. They are shown in this figure:
When specifying a direction of a vector using a compass directions are given by name, North or South. If the direction is directly between two directions we can combine the names, for example North-East is half-way between North and East. This can only happen for directions at right angles to each other, you cannot say North-South as it is ambiguous.
Another way of using the compass to specify direction in a numerical way is to use bearings. A bearing is an angle, usually measured clockwise from North. Note that this is different to the Cartesian plane where angles are anti- or counter-clockwise from the positive \(x\)-direction.
In grade 10 you learnt about adding vectors together in one dimension. The same principle can be applied for vectors in two dimensions. The following examples show addition of vectors. Vectors that are parallel can be shifted to fall on a line. Vectors falling on the same line are called co-linear vectors. To add co-linear vectors we use the tail-to-head method you learnt in Grade 10. In the figure below we remind you of the approach of adding co-linear vectors to get a resultant vector.
Figure 1.3: Adding co-linear vectors to get a resultant vector.
In the above figure the blue vectors are in the \(y\)-direction and the red vectors are in the \(x\)-direction. The two black vectors represent the resultants of the co-linear vectors graphically.
What we have done is implement the tail-to-head method of vector addition for the vertical set of vectors and the horizontal set of vectors.
Use the graphical head-to-tail method to determine the resultant force on a rugby player if two players on his team are pushing him forwards with forces of \(\stackrel{\to }{{F}_{1}}\) = \(\text{600}\) \(\text{N}\) and \(\stackrel{\to }{{F}_{2}}\) = \(\text{900}\) \(\text{N}\) respectively and two players from the opposing team are pushing him backwards with forces of \(\stackrel{\to }{{F}_{3}}\) = \(\text{1 000}\) \(\text{N}\) and \(\stackrel{\to }{{F}_{4}}\) = \(\text{650}\) \(\text{N}\) respectively.
Let's choose a scale of \(\text{100}\) \(\text{N}\): \(\text{0,5}\) \(\text{cm}\) and for our diagram we will define the positive direction as to the right.
We will start with drawing the vector \(\stackrel{\to }{{F}_{1}}\) = \(\text{600}\) \(\text{N}\) pointing in the positive direction. Using our scale of \(\text{0,5}\) \(\text{cm}\) : \(\text{100}\) \(\text{N}\), the length of the arrow must be \(\text{3}\) \(\text{cm}\) pointing to the right.
The next vector is \(\stackrel{\to }{{F}_{2}}\) = \(\text{900}\) \(\text{N}\) in the same direction as \(\stackrel{\to }{{F}_{1}}\). Using the scale, the arrow should be \(\text{4,5}\) \(\text{cm}\) long and pointing to the right.
The next vector is \(\stackrel{\to }{{F}_{3}}\) = \(\text{1 000}\) \(\text{N}\) in the opposite direction. Using the scale, this arrow should be \(\text{5}\) \(\text{cm}\) long and point to the left.
Note: We are working in one dimension so this arrow would be drawn on top of the first vectors to the left. This will get confusing so we'll draw it next to the actual line as well to show you what it looks like.
The fourth vector is \(\stackrel{\to }{{F}_{4}}\) = \(\text{650}\) \(\text{N}\) in the opposite direction. Using the scale, this arrow must be \(\text{3,25}\) \(\text{cm}\) long and point to the left.
We have now drawn all the force vectors that are being applied to the player. The resultant vector is the arrow which starts at the tail of the first vector and ends at the head of the last drawn vector.
The resultant vector measures \(\text{0,75}\) \(\text{cm}\) which, using our scale is equivalent to \(\text{150}\) \(\text{N}\) and points to the left (or the negative direction or the direction the opposing team members are pushing in).
Find the resultant in the \(x\)-direction, \(R_x\), and \(y\)-direction, \(R_y\) for the following forces:
We choose a scale of \(\text{1}\) \(\text{N}\): \(\text{1}\) \(\text{cm}\) and for our diagram we will define the positive direction as to the right.
We will start with drawing the vector \(\stackrel{\to }{{F}_{1}}\) = \(\text{1,5}\) \(\text{N}\) pointing in the positive direction. Using our scale of \(\text{1}\) \(\text{N}\) : \(\text{1}\) \(\text{cm}\), the length of the arrow must be \(\text{1,5}\) \(\text{cm}\) pointing to the right.
The next vector is \(\stackrel{\to }{{F}_{2}}\) = \(\text{1,5}\) \(\text{N}\) in the same direction as \(\stackrel{\to }{{F}_{1}}\). Using the scale, the arrow should be \(\text{1,5}\) \(\text{cm}\) long and pointing to the right.
The next vector is \(\stackrel{\to }{{F}_{3}}\) = \(\text{2}\) \(\text{N}\) in the opposite direction. Using the scale, this arrow should be \(\text{2}\) \(\text{cm}\) long and point to the left.
Note: We are working in one dimension so this arrow would be drawn on top of the first vectors to the left. This will get confusing so we'll draw it next to the actual line as well to show you what it looks like.
We have now drawn all the force vectors that are given. The resultant vector is the arrow which starts at the tail of the first vector and ends at the head of the last drawn vector.
The resultant vector measures \(\text{1}\) \(\text{cm}\) which, using our scale is equivalent to \(\text{1}\) \(\text{N}\) and points to the right (or the positive direction). This is \(\vec{R}_{x}\). For this set of vectors we have no vectors pointing in the \(y\)-direction and so we do not need to find \(\vec{R}_{y}\).
Find the resultant in the \(x\)-direction, \(\vec{R}_x\), and \(y\)-direction, \(\vec{R}_y\) for the following forces:
We choose a scale of \(\text{1}\) \(\text{N}\): \(\text{1}\) \(\text{cm}\) and for our diagram we will define the positive direction as to the right.
Before we draw the vectors we note the lengths of the vectors using our scale:
\begin{align*} \vec{F}_{1} & = \text{2,3}\text{ cm} \\ \vec{F}_{2} & = \text{1}\text{ cm} \\ \vec{F}_{3} & = \text{2}\text{ cm} \\ \vec{F}_{4} & = \text{3}\text{ cm} \end{align*}We also note the direction the vectors are in:
\begin{align*} \vec{F}_{1} & = \text{positive } x \text{-direction} \\ \vec{F}_{2} & = \text{negative } x \text{-direction} \\ \vec{F}_{3} & = \text{positive } y \text{-direction} \\ \vec{F}_{4} & = \text{negative } y \text{-direction} \end{align*}We now look at the two vectors in the \(x\)-direction to find \(\vec{R}_{x}\):
We have now drawn all the force vectors that act in the \(x\)-direction. To find \(\vec{R}_{x}\) we note that the resultant vector is the arrow which starts at the tail of the first vector and ends at the head of the last drawn vector in that direction.
We note that \(\vec{R}_{x}\) is \(\text{1,3}\) \(\text{cm}\) or \(\text{1,3}\) \(\text{N}\) in the positive \(x\)-direction.
We now look at the two vectors in the \(y\)-direction to find \(\vec{R}_{y}\):
We have now drawn all the force vectors that act in the \(y\)-direction. To find \(\vec{R}_{y}\) we note that the resultant vector is the arrow which starts at the tail of the first vector and ends at the head of the last drawn vector in that direction.
We note that \(\vec{R}_{y}\) is \(\text{1}\) \(\text{cm}\) or \(\text{1}\) \(\text{N}\) in the negative \(y\)-direction.
\(\vec{R}_{x}\) = \(\text{1,3}\) \(\text{N}\) and points in the positive \(x\)-direction. \(\vec{R}_{y}\) = \(\text{1}\) \(\text{N}\) and points in the negative \(y\)-direction.
Find the resultant in the \(x\)-direction, \(\vec{R}_x\), and \(y\)-direction, \(\vec{R}_y\) for the following forces:
We choose a scale of \(\text{1}\) \(\text{N}\): \(\text{1}\) \(\text{cm}\) and for our diagram we will define the positive direction as to the right.
Before we draw the vectors we note the lengths of the vectors using our scale:
\begin{align*} \vec{F}_{1} & = \text{3}\text{ cm} \\ \vec{F}_{2} & = \text{1}\text{ cm} \\ \vec{F}_{3} & = \text{2}\text{ cm} \\ \vec{F}_{4} & = \text{3}\text{ cm} \end{align*}We also note the direction the vectors are in:
\begin{align*} \vec{F}_{1} & = \text{positive } x \text{-direction} \\ \vec{F}_{2} & = \text{positive } x \text{-direction} \\ \vec{F}_{3} & = \text{negative } x \text{-direction} \\ \vec{F}_{4} & = \text{positive } y \text{-direction} \end{align*}We now look at the three vectors in the \(x\)-direction to find \(\vec{R}_{x}\):
We have now drawn all the force vectors that act in the \(x\)-direction. To find \(\vec{R}_{x}\) we note that the resultant vector is the arrow which starts at the tail of the first vector and ends at the head of the last drawn vector in that direction.
We note that \(\vec{R}_{x}\) is \(\text{2}\) \(\text{cm}\) or \(\text{2}\) \(\text{N}\) in the positive \(x\)-direction.
We now look at the vectors in the \(y\)-direction to find \(\vec{R}_{y}\). We notice that we only have one vector in this direction and so that vector is the resultant.
We note that \(\vec{R}_{y}\) is \(\text{3}\) \(\text{cm}\) or \(\text{3}\) \(\text{N}\) in the positive \(y\)-direction.
\(\vec{R}_{x}\) = \(\text{2}\) \(\text{N}\) and points in the positive \(x\)-direction. \(\vec{R}_{y}\) = \(\text{3}\) \(\text{N}\) and points in the positive \(y\)-direction.
Find the resultant in the \(x\)-direction, \(\vec{R}_x\), and \(y\)-direction, \(\vec{R}_y\) for the following forces:
We choose a scale of \(\text{1}\) \(\text{cm}\): \(\text{1}\) \(\text{N}\) and for our diagram we will define the positive direction as to the right.
Before we draw the vectors we note the lengths of the vectors using our scale:
\begin{align*} \vec{F}_{1} & = \text{2}\text{ cm} \\ \vec{F}_{2} & = \text{1,5}\text{ cm} \\ \vec{F}_{3} & = \text{2,5}\text{ cm} \\ \vec{F}_{4} & = \text{3}\text{ cm} \end{align*}We also note the direction the vectors are in:
\begin{align*} \vec{F}_{1} & = \text{positive } y \text{-direction} \\ \vec{F}_{2} & = \text{negative } y \text{-direction} \\ \vec{F}_{3} & = \text{negative } x \text{-direction} \\ \vec{F}_{4} & = \text{positive } y \text{-direction} \end{align*}We look at the vectors in the \(x\)-direction to find \(\vec{R}_{x}\). We notice that we only have one vector in this direction and so that vector is the resultant.
We note that \(\vec{R}_{x}\) is \(\text{2,5}\) \(\text{cm}\) or \(\text{2,5}\) \(\text{N}\) in the negative \(x\)-direction.
We now look at the three vectors in the \(y\)-direction to find \(\vec{R}_{y}\):
We have now drawn all the force vectors that act in the \(y\)-direction. To find \(\vec{R}_{y}\) we note that the resultant vector is the arrow which starts at the tail of the first vector and ends at the head of the last drawn vector in that direction.
We note that \(\vec{R}_{y}\) is \(\text{3,5}\) \(\text{cm}\) or \(\text{3,5}\) \(\text{N}\) in the positive \(y\)-direction.
\(\vec{R}_{x}\) = \(\text{2,5}\) \(\text{N}\) and points in the negative \(x\)-direction. \(\vec{R}_{y}\) = \(\text{3,5}\) \(\text{N}\) and points in the positive \(y\)-direction.
Find a force in the \(x\)-direction, \(F_x\), and \(y\)-direction, \(F_y\), that you can add to the following forces to make the resultant in the \(x\)-direction, \(R_x\), and \(y\)-direction, \(R_y\) zero:
To solve this problem we will draw the vectors on the Cartesian plane and then look at what the resultant vector is. Then we determine what force vector to add so that the resultant vector is \(\text{0}\).
We choose a scale of \(\text{1}\) \(\text{cm}\): \(\text{1}\) \(\text{N}\) and for our diagram we will define the positive direction as to the right.
Before we draw the vectors we note the lengths of the vectors using our scale:
\begin{align*} \vec{F}_{1} & = \text{2,4}\text{ cm} \\ \vec{F}_{2} & = \text{0,7}\text{ cm} \\ \vec{F}_{3} & = \text{2,8}\text{ cm} \\ \vec{F}_{4} & = \text{3,3}\text{ cm} \end{align*}We also note the direction the vectors are in:
\begin{align*} \vec{F}_{1} & = \text{positive } y \text{-direction} \\ \vec{F}_{2} & = \text{negative } y \text{-direction} \\ \vec{F}_{3} & = \text{negative } x \text{-direction} \\ \vec{F}_{4} & = \text{positive } y \text{-direction} \end{align*}We look at the vectors in the \(x\)-direction to find \(\vec{R}_{x}\). We notice that we only have one vector in this direction and so that vector is the resultant.
We note that \(\vec{R}_{x}\) is \(\text{2,8}\) \(\text{cm}\) or \(\text{2,8}\) \(\text{N}\) in the negative \(x\)-direction.
So if we add a force of \(\text{2,8}\) \(\text{N}\) in the positive \(x\)-direction the resultant will be \(\text{0}\):
We now look at the three vectors in the \(y\)-direction to find \(\vec{R}_{y}\):
We have now drawn all the force vectors that act in the \(y\)-direction. To find \(\vec{R}_{y}\) we note that the resultant vector is the arrow which starts at the tail of the first vector and ends at the head of the last drawn vector in that direction.
We note that \(\vec{R}_{y}\) is \(\text{5}\) \(\text{cm}\) or \(\text{5}\) \(\text{N}\) in the positive \(y\)-direction.
So if we add a force of \(\text{5}\) \(\text{N}\) in the negative \(y\)-direction the resultant will be \(\text{0}\):
We must add a force of \(\text{2,8}\) \(\text{N}\) in the positive \(x\)-direction and a force of \(\text{5}\) \(\text{N}\) in the negative \(y\)-direction.
We apply the same principle to vectors that are at right angles or perpendicular to each other.
The tail of the one vector is placed at the head of the other but in two dimensions the vectors may not be co-linear. The approach is to draw all the vectors, one at a time. For the first vector begin at the origin of the Cartesian plane, for the second vector draw it from the head of the first vector. The third vector should be drawn from the head of the second and so on. Each vector is drawn from the head of the vector that preceded it. The order doesn't matter as the resultant will be the same if the order is different.
Let us apply this procedure to two vectors:
We first draw a Cartesian plane with the first vector originating at the origin:
The next step is to take the second vector and draw it from the head of the first vector:
The resultant, \(\vec{R}\), is the vector connecting the tail of the first vector drawn to the head of the last vector drawn:
It is important to remember that the order in which we draw the vectors doesn't matter. If we had drawn them in the opposite order we would have the same resultant, \(\vec{R}\). We can repeat the process to demonstrate this:
We first draw a Cartesian plane with the second vector originating at the origin:
The next step is to take the other vector and draw it from the head of the vector we have already drawn:
The resultant, \(\vec{R}\), is the vector connecting the tail of the first vector drawn to the head of the last vector drawn (the vector from the start point to the end point):
Sketch the resultant of the following force vectors using the tail-to-head method:
First draw the Cartesian plane and force, \(\vec{F}_{1}\) starting at the origin:
Starting at the head of the first vector we draw the tail of the second vector:
Starting at the head of the second vector we draw the tail of the third vector:
Starting at the head of the third vector we draw the tail of the fourth vector:
Starting at the origin draw the resultant vector to the head of the fourth vector:
Sketch the resultant of the following force vectors using the tail-to-head method by first determining the resultant in the \(x\)- and \(y\)-directions:
First draw the Cartesian plane with the vectors in the \(x\)-direction:
Next we draw the Cartesian plane with the vectors in the \(y\)-direction:
To double check, we can replot all the vectors again as we did in the previous worked example to see that the outcome is the same:
Sketch the resultant of the following force vectors using the tail-to-head method:
We first draw a Cartesian plane with the first vector originating at the origin:
The next step is to take the second vector and draw it from the head of the first vector:
The resultant, \(\vec{R}\), is the vector connecting the tail of the first vector drawn to the head of the last vector drawn:
Sketch the resultant of the following force vectors using the tail-to-head method:
We first sketch the vectors on the Cartesian plane. We choose a scale of \(\text{1}\) \(\text{cm}\) = \(\text{2}\) \(\text{N}\). Remember to sketch \(F_{2}\) starting at the head of \(F_{1}\), \(F_{3}\) starting at the head of \(F_{2}\) and \(F_{4}\) starting at the head of \(F_{3}\).
The resultant, \(\vec{R}\), is the vector connecting the tail of the first vector drawn to the head of the last vector drawn:
Sketch the resultant of the following force vectors using the tail-to-head method by first determining the resultant in the \(x\)- and \(y\)-directions:
We first determine \(\vec{R}_{x}\)
Draw the Cartesian plane with the vectors in the \(x\)-direction:
This is \(\vec{R}_{x}\) since it is the only vector in the \(x\)-direction.
Secondly determine \(\vec{R}_{y}\)
Next we draw the Cartesian plane with the vectors in the \(y\)-direction:
Now we draw the resultant vectors, \(\vec{R}_{y}\) and \(\vec{R}_{x}\) head-to-tail:
You can check this answer by using the tail-to-head method without first determining the resultant in the \(x\)-direction and the \(y\)-direction.
Sketch the resultant of the following force vectors using the tail-to-head method by first determining the resultant in the \(x\)- and \(y\)-directions:
We choose a scale of \(\text{1}\) \(\text{cm}\) : \(\text{2}\) \(\text{N}\).
We first determine \(\vec{R}_{x}\)
Draw the Cartesian plane with the vectors in the \(x\)-direction:
This is \(\vec{R}_{x}\) since it is the only vector in the \(x\)-direction.
Secondly determine \(\vec{R}_{y}\)
Next we draw the Cartesian plane with the vectors in the \(y\)-direction:
Now we draw the resultant vectors, \(\vec{R}_{y}\) and \(\vec{R}_{x}\) head-to-tail:
You can check this answer by using the tail-to-head method without first determining the resultant in the \(x\)-direction and the \(y\)-direction.
In this method we draw the two vectors with their tails on the origin. Then we draw a line parallel to the first vector from the head of the second vector and vice versa. Where the parallel lines intersect is the head of the resultant vector that will also start at the origin. We will only deal with perpendicular vectors but this procedure works for any vectors.
When dealing with more than two vectors the procedure is repetitive. First find the resultant of any two of the vectors to be added. Then use the same method to add the resultant from the first two vectors with a third vector. This new resultant is then added to the fourth vector and so on, until there are no more vectors to be added.
Let us apply this procedure to the same two vectors we used to illustrate the head-to-tail method:
We first draw a Cartesian plane with the first vector originating at the origin:
Then we add the second vector but also originating from the origin so that the vectors are drawn tail-to-tail:
Now we draw a line parallel to \(\vec{F}_{1}\) from the head of \(\vec{F}_{2}\):
Next we draw a line parallel to \(\vec{F}_{2}\) from the head of \(\vec{F}_{1}\):
Where the two lines intersect is the head of the resultant vector which will originate at the origin so:
You might be asking what you would do if you had more than 2 vectors to add together. In this case all you need to do is first determine \(\vec{R}_{x}\) by adding all the vectors that are parallel to the \(x\)-direction and \(\vec{R}_{y}\) by adding all the vectors that are parallel to the \(y\)-direction. Then you use the tail-to-tail method to find the resultant of \(\vec{R}_{x}\) and \(\vec{R}_{y}\).
Sketch the resultant of the following force vectors using the tail-to-tail method:
We first draw a Cartesian plane with the first vector originating at the origin:
Then we add the second vector but also originating from the origin so that the vectors are drawn tail-to-tail:
Now we draw a line parallel to \(\vec{F}_{1}\) from the head of \(\vec{F}_{2}\):
Next we draw a line parallel to \(\vec{F}_{2}\) from the head of \(\vec{F}_{1}\):
Where the two lines intersect is the head of the resultant vector which will originate at the origin so:
Sketch the resultant of the following force vectors using the tail-to-tail method by first determining the resultant in the \(x\)- and \(y\)-directions:
We need to determine \(\vec{R}_{x}\) and \(\vec{R}_{y}\) and then use these to find the resultant.
Determine \(\vec{R}_{x}\).
Draw the Cartesian plane with the vectors in the \(x\)-direction:
This is \(\vec{R}_{x}\) since it is the only vector in the \(x\)-direction.
Secondly determine \(\vec{R}_{y}\)
Next we draw the Cartesian plane with the vectors in the \(y\)-direction:
Now we draw the resultant vectors, \(\vec{R}_{y}\) and \(\vec{R}_{x}\) tail-to-tail:
Now we can draw the lines to show us where the head of the resultant must be:
And finally we find the resultant:
Sketch the resultant of the following force vectors using the tail-to-tail method by first determining the resultant in the \(x\)- and \(y\)-directions:
We choose a scale of \(\text{1}\) \(\text{cm}\) = \(\text{2}\) \(\text{N}\).
We first determine \(\vec{R}_{x}\)
Draw the Cartesian plane with the vectors in the \(x\)-direction:
This is \(\vec{R}_{x}\) since it is the only vector in the \(x\)-direction.
Secondly determine \(\vec{R}_{y}\)
Next we draw the Cartesian plane with the vectors in the \(y\)-direction:
Now we draw the resultant vectors, \(\vec{R}_{y}\) and \(\vec{R}_{x}\) tail-to-tail:
Now we can draw the lines to show us where the head of the resultant must be:
And finally we find the resultant:
A closed vector diagram is a set of vectors drawn on the Cartesian using the tail-to-head method and that has a resultant with a magnitude of zero. This means that if the first vector starts at the origin the last vector drawn must end at the origin. The vectors form a closed polygon, no matter how many of them are drawn. Here are a few examples of closed vector diagrams:
In this case there were 3 force vectors. When drawn tail-to-head with the first force starting at the origin the last force drawn ends at the origin. The resultant would have a magnitude of zero. The resultant is drawn from the tail of the first vector to the head of the final vector. In the diagram below there are 4 vectors that also form a closed vector diagram.
In this case with 4 vectors, the shape is a 4-sided polygon. Any polygon made up of vectors drawn tail-to-head will be a closed vector diagram because a polygon has no gaps.
If we wanted to know the resultant of the three blue vectors and the three red vectors in Figure 1.3 we can use the resultant vectors in the \(x\)- and \(y\)-directions to determine this.
Figure 1.4: Finding the resultant.
The black arrow represents the resultant of the vectors \(\vec{R}_x\) and \(\vec{R}_y\). We can find the magnitude of this vector using the theorem of Pythagoras because the three vectors form a right angle triangle. If we had drawn the vectors to scale we would be able to measure the magnitude of the resultant as well.
What we've actually sketched out already is our approach to finding the resultant of many vectors using components so remember this example when we get there a little later.
The force vectors in Figure 1.3 have the following magnitudes: \(\text{1}\) \(\text{N}\), \(\text{1}\) \(\text{N}\), \(\text{2}\) \(\text{N}\) for the blue ones and \(\text{2}\) \(\text{N}\), \(\text{2}\) \(\text{N}\) and \(\text{1,5}\) \(\text{N}\) for the red ones. Determine the magnitude of the resultant.
The resultant of the vectors parallel to the \(y\)-axis is found by adding the magnitudes (lengths) of three vectors because they all point in the same direction. The answer is \(\vec{R}_y\)=\(\text{1}\) \(\text{N}\) + \(\text{1}\) \(\text{N}\) + \(\text{2}\) \(\text{N}\) = \(\text{4}\) \(\text{N}\) in the positive \(y\)-direction.
The resultant of the vectors parallel to the \(x\)-axis is found by adding the magnitudes (lengths) of three vectors because they all point in the same direction. The answer is \(\vec{R}_x\)=\(\text{2}\) \(\text{N}\) + \(\text{2}\) \(\text{N}\) + \(\text{1,5}\) \(\text{N}\) = \(\text{5,5}\) \(\text{N}\) in the positive \(x\)-direction.
We have a right angled triangle. We also know the length of two of the sides. Using Pythagoras we can find the length of the third side. From what we know about resultant vectors this length will be the magnitude of the resultant vector.
The resultant is: \begin{align*} R_x^{2} + R_y^{2} &= R^{2}\ \text{(Pythagoras' theorem)}\\ (\text{5,5})^{2} + (4)^{2} &= R^{2}\\ R &= \text{6,8} \end{align*}
Magnitude of the resultant: \(\text{6,8}\) \(\text{N}\)
Note: we did not determine the resultant vector in the worked example above because we only determined the magnitude. A vector needs a magnitude and a direction. We did not determine the direction of the resultant vector.
In grade 10 you learnt how to add vectors in one dimension graphically.
We can expand these ideas to include vectors in two-dimensions. The following worked example shows this.
Given two vectors, \(\vec{R}_y\) = \(\text{4}\) \(\text{N}\) in the positive \(y\)-direction and \(\vec{R}_x\) = \(\text{3}\) \(\text{N}\) in the positive \(x\)-direction, use the tail-to-head method to find the resultant of these vectors graphically.
The vectors we have do not have very big magnitudes so we can choose simple scale, we can use \(\text{1}\) \(\text{N}\) : \(\text{1}\) \(\text{cm}\) as our scale for the drawing.
Then we draw axes that the vector diagram should fit in. The largest vector has length \(\text{4}\) \(\text{N}\) and both vectors are in the positive direction so we can draw axes from the origin to \(\text{5}\) and expect the vectors to fit.
The magnitude of \(\vec{R}_x\) is \(\text{3}\) \(\text{N}\) so the arrow we need to draw must be \(\text{3}\) \(\text{cm}\) long. The arrow must point in the positive \(x\)-direction.
The length of \(\vec{R}_y\) is \(\text{4}\) so the arrow we need to draw must be \(\text{4}\) \(\text{cm}\) long. The arrow must point in the positive \(y\)-direction. The important fact to note is that we are implementing the head-to-tail method so the vector must start at the end (head) of \(\vec{R}_x\).
The resultant vector is the vector from the tail of the first vector we drew directly to the head of the last vector we drew. This means we need to draw a vector from the tail of \(\vec{R}_{x}\) to the head of \(\vec{R}_{y}\).
We are solving the problem graphically so we now need to measure the magnitude of the vector and use the scale we chose to convert our answer from the diagram to the actual result. In the last diagram the resultant, \(\vec{R}\) is \(\text{5}\) \(\text{cm}\) long therefore the magnitude of the vector is \(\text{5}\) \(\text{N}\).
The direction of the resultant, \(\theta\), we need to measure from the diagram using a protractor. The angle that the vector makes with the \(x\)-axis is \(\text{53}\)\(\text{°}\).
\(\vec{R}\) is \(\text{5}\) \(\text{N}\) at \(\text{53}\)\(\text{°}\) from the positive \(x\)-direction.
In the case where you have to find the resultant of more than two vectors first apply the tail-to-head method to all the vectors parallel to the one axis and then all the vectors parallel to the other axis. For example, you would first calculate \(\vec{R}_{y}\) from all the vectors parallel to the \(y\)-axis and then \(\vec{R}_{x}\) from all the vectors parallel to the \(x\)-axis. After that you apply the same procedure as in the previous worked example to the get the final resultant.
Given the following three force vectors, determine the resultant force:
\(\vec{F}_{1}\) = \(\text{3,4}\) \(\text{N}\) in the positive \(x\)-direction
\(\vec{F}_{2}\) = \(\text{4}\) \(\text{N}\) in the positive \(x\)-direction
\(\vec{F}_{3}\) = \(\text{3}\) \(\text{N}\) in the negative \(y\)-direction
First we determine the resultant of all the vectors that are parallel to the \(x\)-axis. There are two vectors \(\vec{F}_{1}\) and \(\vec{F}_{2}\) that we need to add. We do this using the tail-to-head method for co-linear vectors.
The single vector, \(\vec{R}_{x}\), that would give us the same outcome is:
There is only one vector in the \(y\)-direction, \(\vec{F}_{3}\), therefore \(\vec{R}_{y}\) = \(\vec{F}_{3}\).
The vectors we have do not have very big magnitudes so we can choose simple scale, we can use \(\text{1}\) \(\text{N}\) : \(\text{1}\) \(\text{cm}\) as our scale for the drawing.
Then we draw axes that the diagram should fit on. The longest vector has length \(\text{7,4}\) \(\text{N}\). We need our axes to extend just further than the vectors aligned with each axis. Our axes need to start at the origin and go beyond \(\text{7,4}\) \(\text{N}\) in the positive \(x\)-direction and further than \(\text{3}\) \(\text{N}\) in the negative \(y\)-direction. Our scale choice of \(\text{1}\) \(\text{N}\) : \(\text{1}\) \(\text{cm}\) means that our axes actually need to extend \(\text{7,4}\) \(\text{cm}\) in the positive \(x\)-direction and further than \(\text{3}\) \(\text{cm}\) in the negative \(y\)-direction
The magnitude of \(\vec{R}_x\) is \(\text{7,4}\) \(\text{N}\) so the arrow we need to draw must be \(\text{7,4}\) \(\text{cm}\) long. The arrow must point in the positive \(x\)-direction.
The magnitude of \(\vec{R}_y\) is \(\text{3}\) \(\text{N}\) so the arrow we need to draw must be \(\text{3}\) \(\text{cm}\) long. The arrow must point in the negative \(y\)-direction. The important fact to note is that we are implementing the head-to-tail method so the vector must start at the end (head) of \(\vec{R}_{x}\).
The resultant vector is the vector from the tail of the first vector we drew directly to the head of the last vector we drew. This means we need to draw a vector from the tail of \(\vec{R}_{x}\) to the head of \(\vec{R}_{y}\).
We are solving the problem graphically so we now need to measure the magnitude of the vector and use the scale we chose to convert our answer from the diagram to the actual result. In the last diagram the resultant, \(\vec{R}\) is \(\text{8,0}\) \(\text{cm}\) long therefore the magnitude of the vector is \(\text{8,0}\) \(\text{N}\).
The direction of the resultant we need to measure from the diagram using a protractor. The angle that the vector makes with the \(x\)-axis is \(\text{22}\)\(\text{°}\).
\(\vec{R}\) is \(\text{8,0}\) \(\text{N}\) at \(-\text{22}\)\(\text{°}\) from the positive \(x\)-direction.
Given the following three force vectors, determine the resultant force:
\(\vec{F}_{1}\) = \(\text{2,3}\) \(\text{N}\) in the positive \(x\)-direction
\(\vec{F}_{2}\) = \(\text{4}\) \(\text{N}\) in the positive \(y\)-direction
\(\vec{F}_{3}\) = \(\text{3,3}\) \(\text{N}\) in the negative \(y\)-direction
\(\vec{F}_{4}\) = \(\text{2,1}\) \(\text{N}\) in the negative \(y\)-direction
There is only one vector in the \(x\)-direction, \(\vec{F}_{1}\), therefore \(\vec{R}_{x}\) = \(\vec{F}_{1}\).
Then we determine the resultant of all the vectors that are parallel to the \(y\)-axis. There are three vectors \(\vec{F}_{2}\), \(\vec{F}_{3}\) and \(\vec{F}_{4}\) that we need to add. We do this using the tail-to-head method for co-linear vectors.
The single vector, \(\vec{R}_{y}\), that would give us the same effect is:
We choose a scale \(\text{1}\) \(\text{N}\) : \(\text{1}\) \(\text{cm}\) for the drawing.
Then we draw axes that the diagram should fit in. We need our axes to extend just further than the vectors aligned with each axis. Our axes need to start at the origin and go beyond \(\text{2,3}\) \(\text{N}\) in the positive \(x\)-direction and further than \(\text{1,4}\) \(\text{N}\) in the negative \(y\)-direction. Our scale choice of \(\text{1}\) \(\text{N}\) = \(\text{1}\) \(\text{cm}\) means that our axes actually need to extend \(\text{2,3}\) \(\text{cm}\) in the positive \(x\)-direction and further than \(\text{1,4}\) \(\text{cm}\) in the negative \(y\)-direction
The magnitude of \(\vec{R}_x\) is \(\text{2,3}\) \(\text{N}\) so the arrow we need to draw must be \(\text{2,3}\) \(\text{cm}\) long. The arrow must point in the positive \(x\)-direction.
The magnitude of \(\vec{R}_y\) is \(\text{1,4}\) \(\text{N}\) so the arrow we need to draw must be \(\text{1,4}\) \(\text{cm}\) long. The arrow must point in the negative \(y\)-direction. The important fact to note is that we are implementing the head-to-tail method so the vector must start at the end (head) of \(\vec{R}_{x}\).
The resultant vector is the vector from the tail of the first vector we drew directly to the head of the last vector we drew. This means we need to draw a vector from the tail of \(\vec{R}_{x}\) to the head of \(\vec{R}_{y}\).
We are solving the problem graphically so we now need to measure the magnitude of the vector and use the scale we chose to convert our answer from the diagram to the magnitude of the vector. In the last diagram the resultant, \(\vec{R}\) is \(\text{2,7}\) \(\text{cm}\) long therefore the magnitude of the vector is \(\text{2,7}\) \(\text{N}\).
The direction of the resultant we need to measure from the diagram using a protractor. The angle that the vector makes with the \(x\)-axis is \(\text{31}\) degrees.
\(\vec{R}\) is \(\text{2,7}\) \(\text{N}\) at \(-\text{31}\)\(\text{°}\) from the positive \(x\)-direction.
A number of tugboats are trying to manoeuvre a submarine in the harbour but they are not working as a team. Each tugboat is exerting a different force on the submarine.
Given the following force vectors, determine the resultant force:
\(\vec{F}_{1}\) = \(\text{3,4}\) \(\text{kN}\) in the positive \(x\)-direction
\(\vec{F}_{2}\) = \(\text{4 000}\) \(\text{N}\) in the positive \(y\)-direction
\(\vec{F}_{3}\) = \(\text{300}\) \(\text{N}\) in the negative \(y\)-direction
\(\vec{F}_{4}\) = \(\text{7}\) \(\text{kN}\) in the negative \(y\)-direction
To use the graphical method of finding the resultant we need to work in the same units. Strictly speaking in this problem all the vectors are in newtons but they have different factors which will affect the choice of scale. These need to taken into account and the simplest approach is to convert them all to a consistent unit and factor. We could use kN or N, the choice does not matter. We will choose kN. Remember that k represents a factor of \(\times 10^{3}\).
\(\vec{F}_{1}\) and \(\vec{F}_{4}\) do not require any adjustment because they are both in kN. To convert N to kN we use: \begin{align*} \text{kN} &= \times 10^{3}\\ \frac{\text{N}}{\text{kN}} &= \frac{1}{\times 10^{3}}\\ \text{N} &= \times 10^{-3}\ \text{kN} \end{align*}
To convert the magnitude of \(\vec{F}_{2}\) to kN: \begin{align*} F_2 &= \text{4 000}\text{ N}\\ F_2 &= \text{4 000} \times \text{10}^{-\text{3}}\text{ kN}\\ F_2 &= \text{4}\text{ kN} \end{align*} Therefore \(\vec{F}_{2}\) = \(\text{4}\) \(\text{kN}\) in the positive \(y\)-direction.
To convert the magnitude of \(\vec{F}_{3}\) to kN: \begin{align*} F_3 &= \text{300}\text{ N}\\ F_3 &= \text{300} \times \text{10}^{-\text{3}}\text{ kN}\\ F_3 &= \text{0,3}\text{ kN} \end{align*} Therefore \(\vec{F}_{3}\) = \(\text{0,3}\) \(\text{kN}\) in the negative \(y\)-direction. So:
\(\vec{F}_{1}\) = \(\text{3,4}\) \(\text{kN}\) in the positive \(x\)-direction
\(\vec{F}_{2}\) = \(\text{4}\) \(\text{kN}\) in the positive \(y\)-direction
\(\vec{F}_{3}\) = \(\text{0,3}\) \(\text{kN}\) in the negative \(y\)-direction
\(\vec{F}_{4}\) = \(\text{7}\) \(\text{kN}\) in the negative \(y\)-direction
The vectors we have do have very big magnitudes so we need to choose a scale that will allow us to draw them in a reasonable space, we can use \(\text{1}\) \(\text{kN}\) : \(\text{1}\) \(\text{cm}\) as our scale for the drawings.
There is only one vector in the \(x\)-direction, \(\vec{F}_{1}\), therefore \(\vec{R}_{x}\) = \(\vec{F}_{1}\).
Then we determine the resultant of all the vectors that are parallel to the \(y\)-axis. There are three vectors \(\vec{F}_{2}\), \(\vec{F}_{3}\) and \(\vec{F}_{4}\) that we need to add. We do this using the tail-to-head method for co-linear vectors.
The single vector, \(\vec{R}_{y}\), that would give us the same outcome is:
Then we draw axes that the diagram should fit on. We need our axes to extend just further than the vectors aligned with each axis. Our axes need to start at the origin and go beyond \(\text{3,4}\) \(\text{kN}\) in the positive \(x\)-direction and further than \(\text{3,3}\) \(\text{kN}\) in the negative \(y\)-direction. Our scale choice of \(\text{1}\) \(\text{kN}\) : \(\text{1}\) \(\text{cm}\) means that our axes actually need to extend \(\text{3,4}\) \(\text{cm}\) in the positive \(x\)-direction and further than \(\text{3,3}\) \(\text{cm}\) in the negative \(y\)-direction
The length of \(\vec{R}_x\) is \(\text{3,4}\) \(\text{kN}\) so the arrow we need to draw must be \(\text{3,4}\) \(\text{cm}\) long. The arrow must point in the positive \(x\)-direction.
The length of \(\vec{R}_y\) is \(\text{3,3}\) \(\text{kN}\) so the arrow we need to draw must be \(\text{3,3}\) \(\text{cm}\) long. The arrow must point in the negative \(y\)-direction. The important fact to note is that we are implementing the head-to-tail method so the vector must start at the end (head) of \(\vec{R}_{x}\).
The resultant vector is the vector from the tail of the first vector we drew directly to the head of the last vector we drew. This means we need to draw a vector from the tail of \(\vec{R}_{x}\) to the head of \(\vec{R}_{y}\).
We are solving the problem graphically so we now need to measure the magnitude of the vector and use the scale we chose to convert our answer from the diagram to the actual result. In the last diagram the resultant, \(\vec{R}\) is \(\text{4,7}\) \(\text{cm}\) long therefore the magnitude of the vector is \(\text{4,7}\) \(\text{kN}\).
The direction of the resultant we need to measure from the diagram using a protractor. The angle that the vector makes with the \(x\)-axis is \(\text{44}\)\(\text{°}\).
\(\vec{R}\) is \(\text{4,7}\) \(\text{kN}\) at \(-\text{44}\)\(\text{°}\) from the positive \(x\)-direction.
In grade 10 you learnt about addition and subtraction of vectors in one dimension. The following worked example provides a refresher of the concepts.
A force of \(\text{5}\) \(\text{N}\) to the right is applied to a crate. A second force of \(\text{2}\) \(\text{N}\) to the left is also applied to the crate. Calculate algebraically the resultant of the forces applied to the crate.
A simple sketch will help us understand the problem.
Remember that force is a vector. Since the forces act along a straight line (i.e. the \(x\)-direction), we can use the algebraic technique of vector addition.
Choose the positive direction to be to the right. This means that the negative direction is to the left.
Rewriting the problem using the choice of a positive direction gives us a force of \(\text{5}\) \(\text{N}\) in the positive \(x\)-direction and force of \(\text{2}\) \(\text{N}\) in the negative \(x\)-direction being applied to the crate.
Thus, the resultant force is:
\begin{align*} \vec{F}_{1}+\vec{F}_{2}& = (5)+(-2) \\ & = \text{3}\text{ N} \end{align*}Remember that in this case a positive force means to the right: \(\text{3}\) \(\text{N}\) to the right.
We can now expand on this work to include vectors in two dimensions.
A force of \(\text{40}\) \(\text{N}\) in the positive \(x\)-direction acts simultaneously (at the same time) to a force of \(\text{30}\) \(\text{N}\) in the positive \(y\)-direction. Calculate the magnitude of the resultant force.
As before, the rough sketch looks as follows:
Note that the triangle formed by the two force vectors and the resultant vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let \(R\) represent the length of the resultant vector. Then: \begin{align*} F_x^{2} + F_y^{2} &= R^{2}\ \text{Pythagoras' theorem}\\ (40)^{2} + (30)^{2} &= R^{2}\\ R &= \text{50}\text{ N} \end{align*}
The magnitude of the resultant force is then \(\text{50}\) \(\text{N}\).
For two dimensional vectors we have only covered finding the magnitude of vectors algebraically. We also need to know the direction. For vectors in one dimension this was simple. We chose a positive direction and then the resultant was either in the positive or in the negative direction. In grade 10 you learnt about the different ways to specify direction. We will now look at using trigonometry to determine the direction of the resultant vector.
We can use simple trigonometric identities to calculate the direction. We can calculate the direction of the resultant in the previous worked example.
A force of \(\text{40}\) \(\text{N}\) in the positive \(x\)-direction acts simultaneously (at the same time) to a force of \(\text{30}\) \(\text{N}\) in the positive \(y\)-direction. Calculate the magnitude of the resultant force.
We determined the magnitude of the resultant vector in the previous worked example to be \(\text{50}\) \(\text{N}\). The sketch of the situation is:
To determine the direction of the resultant force, we calculate the angle α between the resultant force vector and the positive \(x\)-axis, by using simple trigonometry: \begin{align*} \tan\alpha &= \frac{\text{opposite side}} {\text{adjacent side}} \\ \tan\alpha &= \frac{\text{30}}{\text{40}} \\ \alpha &= \tan^{-1}(\text{0,75}) \\ \alpha &= \text{36,87}\text{°} \end{align*}
The resultant force is then \(\text{50}\) \(\text{N}\) at \(\text{36,9}\)\(\text{°}\) to the positive \(x\)-axis.
A force of \(\text{17}\) \(\text{N}\) in the positive \(x\)-direction acts simultaneously (at the same time) to a force of \(\text{23}\) \(\text{N}\) in the positive \(y\)-direction. Calculate the resultant force.
We draw a rough sketch:
Now we determine the length of the resultant.s
We note that the triangle formed by the two force vectors and the resultant vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let \(R\) represent the length of the resultant vector. Then: \begin{align*} F_x^{2} + F_y^{2} &= R^{2}\ \text{Pythagoras' theorem}\\ (17)^{2} + (23)^{2} &= R^{2}\\ R &= \text{28,6}\text{ N} \end{align*}
To determine the direction of the resultant force, we calculate the angle α between the resultant force vector and the positive \(x\)-axis, by using simple trigonometry: \begin{align*} \tan\alpha &= \frac{\text{opposite side}} {\text{adjacent side}} \\ \tan\alpha &= \frac{\text{23}}{\text{17}} \\ \alpha &= \tan^{-1}(\text{1,353}) \\ \alpha &= \text{53,53}\text{°} \end{align*}
The resultant force is then \(\text{28,6}\) \(\text{N}\) at \(\text{53,53}\)\(\text{°}\) to the positive \(x\)-axis.
A force of \(\text{23,7}\) \(\text{N}\) in the negative \(x\)-direction acts simultaneously to a force of \(\text{9}\) \(\text{N}\) in the positive \(y\)-direction. Calculate the resultant force.
We draw a rough sketch:
Now we determine the length of the resultant.
We note that the triangle formed by the two force vectors and the resultant vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let \(R\) represent the length of the resultant vector. Then: \begin{align*} F_x^{2} + F_y^{2} &= R^{2}\ \text{Pythagoras' theorem}\\ (\text{23,7})^{2} + (9)^{2} &= R^{2}\\ R &= \text{25,4}\text{ N} \end{align*}
To determine the direction of the resultant force, we calculate the angle α between the resultant force vector and the positive \(x\)-axis, by using simple trigonometry: \begin{align*} \tan\alpha &= \frac{\text{opposite side}} {\text{adjacent side}} \\ \tan\alpha &= \frac{\text{9}}{\text{23,7}} \\ \alpha &= \tan^{-1}(\text{0,3797}) \\ \alpha &= \text{20,79}\text{°} \end{align*}
The resultant force is then \(\text{25,4}\) \(\text{N}\) at \(\text{20,79}\)\(\text{°}\) to the \(x\)-axis. However we actually give this as \(\text{159,21}\)\(\text{°}\) (i.e. \(\text{180}\)\(\text{°}\) minus \(\text{20,79}\)\(\text{°}\)) to keep with the convention we have defined of giving vector directions.
Four forces act simultaneously at a point, find the resultant if the forces are:
\(\vec{F}_{1}\) = \(\text{2,3}\) \(\text{N}\) in the positive \(x\)-direction
\(\vec{F}_{2}\) = \(\text{4}\) \(\text{N}\) in the positive \(y\)-direction
\(\vec{F}_{3}\) = \(\text{3,3}\) \(\text{N}\) in the negative \(y\)-direction
\(\vec{F}_{4}\) = \(\text{2,1}\) \(\text{N}\) in the negative \(y\)-direction
We note that we have more than two vectors so we must first find the resultant in the \(x\)-direction and the resultant in the \(y\)-direction.
In the \(x\)-direction we only have one vector and so this is the resultant.
In the \(y\)-direction we have three vectors. We can add these algebraically to find \(\vec{R}_{y}\):
\begin{align*} \vec{F}_{1} & = \text{+4}\text{ N} \\ \vec{F}_{2} & = -\text{3,3}\text{ N} \\ \vec{F}_{3} & = -\text{2,1}\text{ N} \end{align*}Thus, the resultant force is:
\begin{align*} \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3} & = (4) + (-\text{3,3}) + (-\text{2,1}) \\ & = -\text{1,4} \end{align*}Now we use \(\vec{R}_{x}\) and \(\vec{R}_{y}\) to find the resultant.
We note that the triangle formed by \(\vec{R}_{x}\), \(\vec{R}_{y}\) and the resultant vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let \(R\) represent the length of the resultant vector. Then: \begin{align*} F_x^{2} + F_y^{2} &= R^{2}\ \text{Pythagoras' theorem}\\ (\text{2,3})^{2} + (-\text{1,4})^{2} &= R^{2}\\ R &= \text{2,69}\text{ N} \end{align*}
To determine the direction of the resultant force, we calculate the angle α between the resultant force vector and the positive \(x\)-axis, by using simple trigonometry: \begin{align*} \tan\alpha &= \frac{\text{opposite side}} {\text{adjacent side}} \\ \tan\alpha &= \frac{\text{1,4}}{\text{2,3}} \\ \alpha &= \tan^{-1}(\text{0,6087}) \\ \alpha &= \text{31,33}\text{°} \end{align*}
The resultant force is then \(\text{2,69}\) \(\text{N}\) at \(\text{31,33}\)\(\text{°}\) to the \(x\)-axis. However we actually give this as \(\text{328,67}\)\(\text{°}\) to keep with the convention we have defined of giving vector directions.
The following forces act simultaneously on a pole, if the pole suddenly snaps in which direction will it be pushed:
\(\vec{F}_{1}\) = \(\text{2,3}\) \(\text{N}\) in the negative \(x\)-direction
\(\vec{F}_{2}\) = \(\text{11,7}\) \(\text{N}\) in the negative \(y\)-direction
\(\vec{F}_{3}\) = \(\text{6,9}\) \(\text{N}\) in the negative \(y\)-direction
\(\vec{F}_{4}\) = \(\text{1,9}\) \(\text{N}\) in the negative \(y\)-direction
To determine the answer we need to find the magnitude and direction of the resultant. This is then the direction in which the pole will be pushed.
We note that we have more than two vectors so we must first find the resultant in the \(x\)-direction and the resultant in the \(y\)-direction.
In the \(x\)-direction we only have one vector and so this is the resultant.
In the \(y\)-direction we have three vectors. We can add these algebraically to find \(\vec{R}_{y}\):
\begin{align*} \vec{F}_{1} & = -\text{11,7}\text{ N} \\ \vec{F}_{2} & = -\text{6,9}\text{ N} \\ \vec{F}_{3} & = -\text{1,9}\text{ N} \end{align*}Thus, the resultant force is:
\begin{align*} \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3} & = (-\text{11,7}) + (-\text{6,9}) + (-\text{1,9}) \\ & = -\text{20,5} \end{align*}Now we use \(\vec{R}_{x}\) and \(\vec{R}_{y}\) to find the resultant.
We note that the triangle formed by \(\vec{R}_{x}\), \(\vec{R}_{y}\) and the resultant vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let \(R\) represent the length of the resultant vector. Then: \begin{align*} F_x^{2} + F_y^{2} &= R^{2}\ \text{Pythagoras' theorem}\\ (-\text{2,3})^{2} + (-\text{20,5})^{2} &= R^{2}\\ R &= \text{20,63}\text{ N} \end{align*}
A rough sketch will help to determine the direction.
To determine the direction of the resultant force, we calculate the angle α between the resultant force vector and the positive \(x\)-axis, by using simple trigonometry: \begin{align*} \tan\alpha &= \frac{\text{opposite side}} {\text{adjacent side}} \\ \tan\alpha &= \frac{\text{2,3}}{\text{20,5}} \\ \alpha &= \tan^{-1}(\text{0,112}) \\ \alpha &= \text{6,40}\text{°} \end{align*}
The resultant force acts in a direction of \(\text{6,40}\)\(\text{°}\) to the \(x\)-axis. However we actually give this as \(\text{186,4}\)\(\text{°}\) to keep with the convention we have defined of giving vector directions.
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1.3 Components of vectors
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