Lesson Explainer: Resultant Motion and Force Physics • 9th Grade

In this explainer, we will learn how to show that motion in directions that are at right angles to each other can be represented by motion in one direction.

A full description of certain physical quantities such as displacement, velocity, acceleration, and force must include information about direction as well as magnitude. As a result, we generally represent these quantities using vectors, mathematical objects that have both magnitude and direction. Using vectors to represent vector quantities enables us to add these quantities together in a meaningful way as well as perform many other useful mathematical functions with them.

Recall that vectors can be added together using a graphical method called the tip-to-tail method. It works like this: to add two vectors (represented by arrows) together, move one of the vector arrows so that its tail touches the tip of the other vector. Then, draw a new vector from the tail of the unmoved vector to the tip of the moved vector. This new vector is the sum of the two original vectors.

For example, we can add together the blue and yellow vectors shown on this grid.

First we move the orange vector so that its tail touches the tip of the blue vector.

Now, we will draw a new vector in pink, going from the tail of the unmoved (blue) vector to the tip of the moved (orange) vector.

This pink vector is equal to the sum of the blue vector and the orange vector. We can equivalently say that it is the resultant of these two vectors.

We can express the relationship between the blue, orange, and pink vectors algebraically using vector notation. Let’s call the blue vector ⃑𝐴, the orange vector ⃑𝐡, and the pink vector ⃑𝐢.

Since we have shown that the pink vector is the sum of the other two vectors, we can write ⃑𝐴+⃑𝐡=⃑𝐢.

We would also get the same result if we used the tip-to-tail method to add the vectors in the opposite order: ⃑𝐡+⃑𝐴=⃑𝐢.

So, we have shown that if we add together a vector with a magnitude of 3 pointing to the right ⃑𝐴 and a vector with a magnitude of 5 pointing upward ⃑𝐡, we get a new vector. This new vector, which we have called ⃑𝐢, points at an angle (it goes up and right), and its magnitude (i.e., its length) appears to be longer than either ⃑𝐴 or ⃑𝐡, but not quite as long as both of them put together. The fact that the magnitude of this vector is not simply the sum of the magnitudes of 𝐴 and 𝐡 shows us that vectors cannot just be added together like normal numbers.

The central idea of this explainer is that vector quantities in physics obey the rules of vector addition. In other words, vector quantities of the same typeβ€”forces, for exampleβ€”can be added together in exactly the same way as the arrows on our graph.

This means that if we have a 3 N force pushing an object to the right (which we could represent with vector ⃑𝐴) and a 5 N force pushing the same object upward (which we could represent with the vector ⃑𝐡), the combined force that acts on the object (in other words, the resultant force) is represented by vector ⃑𝐢.

Next, let’s look at how we can add vector quantities in an accurate way without drawing them.

Adding vectors accurately depends on the idea of components. Recall that we use components to talk about vectors in terms of how far they go in the horizontal direction and how far they go in the vertical direction. Let’s consider two new vectors, ⃑𝐷 and ⃑𝐸.

We can express these vectors as the sums of their components: ⃑𝐷=3+1,⃑𝐸=βˆ’2+2.horizontalverticalhorizontalvertical

Describing vector quantities in terms of their components in the horizontal and vertical directions makes it easy to add them together. If we want to add together vectors ⃑𝐷 and ⃑𝐸, we just need to add together their horizontal components and their vertical components separately: ⃑𝐷+⃑𝐸=(3βˆ’2)+(1+2)=1+3.horizontalverticalhorizontalvertical

We have found that the resultant vector created by adding ⃑𝐷 and ⃑𝐸 has a horizontal component of 1 and a vertical component of 3.

Let’s call this resultant vector ⃑𝐹 and add it to our diagram.

The same relationship holds for vector quantities in physics. For example, a body subjected to a force represented by ⃑𝐷 and another force represented by ⃑𝐸 would experience a resultant force corresponding to ⃑𝐹.

How To: Adding Vectors Together Using Their Components

To add together two vectors ⃑𝐴 and ⃑𝐡,

  • find the horizontal and vertical components of ⃑𝐴 and ⃑𝐡,
  • note that the horizontal component of ⃑𝐴+⃑𝐡 is the sum of the horizontal components of ⃑𝐴 and ⃑𝐡,
  • note that the vertical component of ⃑𝐴+⃑𝐡 is the sum of the vertical components of ⃑𝐴 and ⃑𝐡.

We have now seen two different ways of expressing vectors:

  • Component form: we can express a vector as the sum of its horizontal and vertical components.
  • Magnitude and direction form: we can express a vector in terms of its overall magnitude and direction.

We have seen that component form is useful for adding vectors together. However, it is often more useful and intuitive to express vector quantities in terms of their magnitude and direction. Being able to convert between these two forms is the final step in effectively using vectors in physics. Let’s start with converting from component form to magnitude and direction form.

Let’s consider an object whose velocity in metres per second (m/s) is represented by vector ⃑𝑣. Let’s say that ⃑𝑣 has a horizontal component of 4 and a vertical component of βˆ’3.

What is the magnitude and direction of the object’s velocity?

We can calculate both of these using geometry. First, let’s draw the horizontal and vertical components ⃑𝑣 and βƒ‘π‘£ο˜ onto our diagram.

Here, we can remember that the vector ⃑𝑣 is equivalent to the sum of its components. That is, ⃑𝑣=⃑𝑣+⃑𝑣.ο—ο˜

Remember that in this equation, ⃑𝑣, ⃑𝑣, and βƒ‘π‘£ο˜ are all vectors. This means that we cannot simply add the magnitudes of 𝑣 and π‘£ο˜ to get the magnitude of 𝑣.

Notice that ⃑𝑣 and its components form a right triangle. This enables us to use geometry to calculate the magnitude and direction of the object’s velocity. First, we can calculate the magnitude (i.e., the length) of ⃑𝑣, ‖‖⃑𝑣‖‖, using the Pythagorean theorem. Consider a right triangle with a hypotenuse of length π‘Ž and two other sides of lengths 𝑏 and 𝑐, as shown in the diagram below.

The lengths are related by the equation π‘Ž=βˆšπ‘+𝑐.

Applying this to the triangle formed by the vector ⃑𝑣 and its components, we have ‖‖⃑𝑣‖‖=ο„žβ€–β€–βƒ‘π‘£β€–β€–+‖‖⃑𝑣‖‖,ο—οŠ¨ο˜οŠ¨ where ‖‖⃑𝑣‖‖ and β€–β€–βƒ‘π‘£β€–β€–ο˜ are the magnitudes (lengths) of the horizontal and vertical components respectively. Substituting these lengths into our equation, we have ‖‖⃑𝑣‖‖=√4+9=√16+9=√25=5.

So, the magnitude of the object’s velocityβ€”that is, its speedβ€”is 5 m/s.

Equation: Finding the Magnitude of a 2D Vector Based on its Components

The magnitude of any 2-dimensional vector ⃑𝐴 is given by ‖‖⃑𝐴‖‖=ο„žβ€–β€–βƒ‘π΄β€–β€–+‖‖⃑𝐴‖‖,ο—οŠ¨ο˜οŠ¨ where ⃑𝐴 and βƒ‘π΄ο˜ are the horizontal and vertical components of ⃑𝐴 respectively.

Let’s look at an example where we put this idea into practice.

Example 1: Finding the Magnitude of Two Perpendicular Displacement Vectors

Point 𝐴 is located 8 m horizontally from the base of the wall of a house, and point 𝐡 is located 6 m vertically above the base of the wall, as shown in the diagram. What is the magnitude of the displacement from point 𝐴 to point 𝐡?

Answer

The key to answering this question is to think about the displacement vector from 𝐴 to 𝐡. We can call this vector 𝐴𝐡. It describes the magnitude and direction of the displacement of point 𝐡 from (or relative to) point 𝐴.

Notice that in this case, the horizontal and vertical components of 𝐴𝐡 are labeled in the diagram.

Following the usual convention of defining our horizontal direction as pointing to the right and our vertical direction as pointing up, we can express 𝐴𝐡 as the sum of its componenets as 𝐴𝐡=βˆ’8+6.horizontalvertical

We can calculate the magnitude of any vector using its components. The magnitude of any 2-dimensional vector ⃑𝑣 is given by ‖‖⃑𝑣‖‖=ο„žβ€–β€–βƒ‘π‘£β€–β€–+‖‖⃑𝑣‖‖,ο—οŠ¨ο˜οŠ¨ where ⃑𝑣 and βƒ‘π‘£ο˜ are the horizontal and vertical components of ⃑𝑣 respectively.

In this question, we want to find the magnitude of the vector 𝐴𝐡. Substituting its components into the equation above, we have ‖‖𝐴𝐡‖‖=|βˆ’8|+|6|.

We can then evaluate the magnitudes on the right-hand side (i.e., ignore any negative signs) and then work out the magnitude as follows: ‖‖𝐴𝐡‖‖=√8+6=√64+36=√100=10.

Because the lengths we used were expressed in metres, our answer is expressed in metres too. So we obtain a final value for the magnitude of the displacement from 𝐴 to 𝐡 of 10 m.

Let’s get back to our velocity vector ⃑𝑣. Now that we have found its magnitude, let’s find its direction. We can do this using trigonometry.

Recall that for any right triangle, the angle πœƒ, the length of the side opposite that angle O, and the length of the side adjacent to that angle A are related by the equation tanOAπœƒ=.

We can apply this equation to our vector ⃑𝑣.

Note that the horizontal component ⃑𝑣 is adjacent to the angle πœƒ and the vertical component of βƒ‘π‘£ο˜ is opposite it. This gives us the equation tanπœƒ=‖‖⃑𝑣‖‖‖‖⃑𝑣‖‖.ο˜ο—

To find the angle of our velocity vector below the horizontal, we just need to rearrange the equation to make theta the subject: πœƒ=‖‖⃑𝑣‖‖‖‖⃑𝑣‖‖.arctanο˜ο—

Substituting the magnitudes (lengths) of ⃑𝑣 and βƒ‘π‘£ο˜ into the equation gives us πœƒ=ο€Ό34=37.87….arctan∘

So, we know that ⃑𝑣 points at an angle of 37.87β€¦βˆ˜ below the horizontal axis.

Let’s look at an example question where we need to calculate the direction of a vector based on its components.

Example 2: Finding the Direction of a Displacement Vector

A bird flies along a line that displaces it 450 m east and 350 m north of its starting point, as shown in the diagram. What angle must the bird turn toward the west to change direction and fly directly north? Give your answer to the nearest degree.

Answer

In the diagram, the blue arrow represents the bird’s displacement vector. The black arrows, pointing 450 m to the east and 350 m to the north, effectively act as horizontal and vertical axes respectively. These measurements tell us that the horizontal component of the bird’s displacement is 450 m to the east, and the vertical component of the bird’s displacement is 350 m to the north.

Let’s start by modifying our diagram to show the angle πœƒ through which the bird would have to turn in order to point north.

This is the angle that we are trying to find. Notice that this is the same as the angle of the bird’s displacement vector from the vertical (north) axis.

This means we are effectively looking to find the angle at which the bird was flying, as measured clockwise from the vertical axis. We could equivalently say that we are looking for the direction of the bird’s displacement vector, measured clockwise from the vertical axis.

We can work out the direction in which a vector points using trigonometry. First, we arrange the bird’s displacement vector and its vertical and horizontal components to form a right triangle.

The length of the hypotenuse is the magnitude of the bird’s displacement vector, the length of the side adjacent to πœƒ is the magnitude of the vertical component of the displacement vector, and the length of the side opposite πœƒ is the magnitude of the horizontal component of the displacement vector.

Let’s label these sides H, A, and O respectively.

We know O and A, and we need to work out the angle πœƒ. We can do this using the following trigonometric relationship: tanOA(πœƒ)=.

To make πœƒ the subject, we take the inverse tangent of both sides: πœƒ=.arctanOA

Now, we just need to substitute in the lengths O and A: πœƒ=450350=97=52.13….arctanmmarctan∘

So, the bird needs to turn this far in the counterclockwise direction and it will face north. All we need to do now is round this to nearest degree to get our final answer, which is 52∘.

Now, let’s see how we can do things the other way around. We can calculate the components of a vector if we are given its magnitude and direction. Consider a force vector ⃑𝐹 with magnitude 8 N acting at an angle of 30∘ above the horizontal to the left.

Let’s show this vector on a grid along with its horizontal and vertical components.

Now, we can see that the horizontal component of ⃑𝐹, ⃑𝐹, is negative, and it appears to have a magnitude just less than 7. The vertical component, βƒ‘πΉο˜, is positive, and it looks like it has a magnitude of 4 units.

To find the exact lengths of these components, we can use trigonometry.

Recall that for any right triangle, the angle πœƒ, the length of the side opposite that angle O, the length of the side adjacent to that angle A, and the length of the hypotenuse H are related by two trigonometric equations: sinOHcosAHπœƒ=,πœƒ=.

We can apply these formulas to our force vector, ⃑𝐹, to determine the magnitudes of its components. In our diagram, H=‖‖⃑𝐹‖‖, O=β€–β€–βƒ‘πΉβ€–β€–ο˜, and A=‖‖⃑𝐹‖‖. Substituting these quantities into our trigonometric equations, we have sincosπœƒ=‖‖⃑𝐹‖‖‖‖⃑𝐹‖‖,πœƒ=‖‖⃑𝐹‖‖‖‖⃑𝐹‖‖.ο˜ο—

Multiplying both sides of each equation by the magnitude of ⃑𝐹 gives us ‖‖⃑𝐹‖‖=β€–β€–βƒ‘πΉβ€–β€–πœƒ,‖‖⃑𝐹‖‖=β€–β€–βƒ‘πΉβ€–β€–πœƒ.ο˜ο—sincos

All we need to do now is substitute in the values ‖‖⃑𝐹‖‖=8 and πœƒ=30∘ to obtain the magnitudes of the π‘₯- and 𝑦-components of ⃑𝐹: ‖‖⃑𝐹‖‖=830‖‖⃑𝐹‖‖=830‖‖⃑𝐹‖‖=4,‖‖⃑𝐹‖‖=4√3.ο˜ο—ο˜ο—sincos

We have now found that the magnitude of the vertical component of ⃑𝐹 is 4, and the magnitude of the horizontal component of ⃑𝐹 is 4√3 (which, expressed as a decimal and rounded to 2 decimal places, is 6.93).

Looking back at the previous diagram, we can see that the vertical component of ⃑𝐹 points in the positive vertical direction, while the horizontal component of ⃑𝐹 points in the negative horizontal direction. When we express ⃑𝐹 as the sum of its components, we make this clear by our use of positive and negative signs: ⃑𝐹=βˆ’4√3+4.horizontalvertical

In other words, the 8 N force represented by the vector ⃑𝐹 is equivalent to a force of 4√3 N acting to the left and a 4 N force acting upward. Converting vectors into their component form is essential if we want to be able to add vectors together.

Example 3: Finding the Components of a Displacement Vector

A surveyor walks through a field, as shown in the diagram. How much further east does the surveyor walk than he walks north? Round your answer to the nearest metre.

Answer

In the diagram, the surveyor’s displacement vector is represented by the red arrow. We can see that he walks 450 m at an angle of 30∘, measured counterclockiwse from the east direction.

The question asks how much further the surveyor walks east than he walks north. In order to answer this, we can work out the components of displacement in the east and north directions.

Let’s start by considering the right triangle formed by the displacement vector and its components. We will call the surveyor’s displacement vector ⃑𝑠, and we will call its vertical (north) and horizontal (east) components βƒ‘π‘ ο˜ and ⃑𝑠 respectively.

For a right triangle of sides A and O and a hypotenuse H, if πœƒ is the angle between the side A and the hypotenuse H, then the length of A can be given by AHcosOHsin=πœƒ,=πœƒ.

In both cases, H is the length of the hypotenuse.

In this case, H is the magnitude of the surveyor’s displacement vector ⃑𝑠 (450 m), A is the magnitude of the horizontal component of this vector, and O is the vertical component of this displacement vector. We also know that the angle, πœƒ, is 30∘. So, ⃑𝑠=45030⃑𝑠=45030⃑𝑠=225,⃑𝑠=389.71….ο˜βˆ˜ο—βˆ˜ο˜ο—sincosmm

So, we can see that the horizontal component (i.e., displacement in the east direction) is larger than vertical component (i.e., displacement in the north direction). To find out exactly how much further the surveyor walked to the east than to the north, we just need to find the difference between these two components: 389.71β€¦βˆ’225=164.71….mmm

Rounding this to the nearest metre, we have a final answer of 165 m.

Now, let’s look at an example where we need to add together two vectors acting at different angles. We first need to convert these vectors into component form so we can add them, then we convert the resultant vector back into magnitude and direction form.

Example 4: Finding the Resultant of Two Forces Acting at Different Angles

The force ⃑𝐹 is the resultant of the two force vectors shown in the diagram. What is the magnitude of ⃑𝐹 to the nearest newton?

Answer

In the diagram, we can see that the two red vectors have been arranged so that they are tip to tail. The vector ⃑𝐹 spans from the tail of one vector to the tip of the other. This is a graphical way of showing that the vector ⃑𝐹 is equal to the sum (or the β€œresultant”) of the two red vectors.

If we call the 70 N force vector βƒ‘π‘“οŠ­οŠ¦ and the 60 N force vector βƒ‘π‘“οŠ¬οŠ¦, then we can express this relationship as an equation: ⃑𝐹=⃑𝑓+⃑𝑓.

In order to find the magnitude of ⃑𝐹, we need to add βƒ‘π‘“οŠ¬οŠ¦ and βƒ‘π‘“οŠ­οŠ¦ together. To do this, we will have to first express βƒ‘π‘“οŠ¬οŠ¦ and βƒ‘π‘“οŠ­οŠ¦ in component form.

Let’s start with βƒ‘π‘“οŠ­οŠ¦. We can draw a diagram where βƒ‘π‘“οŠ­οŠ¦ and its horizontal and vertical components (represented by the blue and pink arrows respectively) form a right triangle.

We can use trigonometry to tell us the magnitudes of the side opposite the angle, O, and the side adjacent to the angle, A, in terms of the magnitude of the hypotenuse H and the angle πœƒ: OHsinAHcos=πœƒ,=πœƒ.

We know that the magnitude of the hypotenuse, H, is 70 N and the angle πœƒ is 20∘: OsinAcosOA=7020=7020=23.94…,=65.78….∘∘

These β€œside lengths” are the magnitudes of our horizontal and vertical components expressed in newtons. Rounding these to two decimal places, we can express the 70 N force as the sum of its components: ⃑𝑓=65.78+23.94.horizontalvertical

Let’s keep note of this. Next we will find the horizontal and vertical components of the 60 N force βƒ‘π‘“οŠ¬οŠ¦.

Once again, we can draw the horizontal and vertical components of this force onto our diagram.

And we can use the same trigonometric equations to figure out the magnitudes of the horizontal and vertical components. This time, H is 60 and πœƒ is 70∘: OHsinAHcosOsinAcosOA=πœƒ=πœƒ=6070=6070=56.38…,=20.52….

So, we can write βƒ‘π‘“οŠ¬οŠ¦ as the sum of its components (rounded to 2 decimal places) too as follows: ⃑𝑓=20.52+56.38.horizontalvertical

Now, we have the horizontal and vertical components of both red vectors.

Finding the components of these two vectors means we can easily find the components of the resultant, ⃑𝐹. We can see that the horizontal component of ⃑𝐹 is the sum of the horizontal components of βƒ‘π‘“οŠ¬οŠ¦ and βƒ‘π‘“οŠ­οŠ¦. Similarly, the vertical component of ⃑𝐹 is the sum of the vertical components of βƒ‘π‘“οŠ¬οŠ¦ and βƒ‘π‘“οŠ­οŠ¦: ⃑𝐹=(65.78+20.52)+(23.94+56.38).horizontalvertical

Rounded to 2 decimal places, we have ⃑𝐹=86.30+80.32.horizontalvertical

Now that we have found the components of the resultant force ⃑𝐹, we can calculate its magnitude. Recall that the magnitude of any 2-dimensional vector ⃑𝑣 is given by ‖‖⃑𝑣‖‖=ο„žβ€–β€–βƒ‘π‘£β€–β€–+‖‖⃑𝑣‖‖,ο—οŠ¨ο˜οŠ¨ where ⃑𝑣 and βƒ‘π‘£ο˜ are the horizontal and vertical components of ⃑𝑣 respectively.

In this case, we have ‖‖⃑𝐹‖‖=√86.30+80.32=√13898.99…=117.89….N

Rounding this to the nearest newton gives us a final answer of 118 N.

Key Points

  • A full description of certain physical quantities such as displacement, velocity, acceleration, and force must include information about both magnitude and direction. This makes vectors the perfect mathematical tool to describe them.
  • Any 2D vector ⃑𝑣 can be expressed as the sum of its horizontal ⃑𝑣 and vertical ο€Ήβƒ‘π‘£ο…ο˜ components: ⃑𝑣=⃑𝑣+⃑𝑣.ο—ο˜
  • Given the components of a vector, its magnitude and direction are given by ‖‖⃑𝑣‖‖=ο„žβ€–β€–βƒ‘π‘£β€–β€–+β€–β€–βƒ‘π‘£β€–β€–πœƒ=‖‖⃑𝑣‖‖‖‖⃑𝑣‖‖.ο—οŠ¨ο˜οŠ¨ο˜ο—arctan
  • Given the magnitude ‖‖⃑𝑣‖‖ of a vector and the angle πœƒ at which it acts, we can determine the magnitudes of the horizontal ⃑𝑣 and vertical ο€Ήβƒ‘π‘£ο…ο˜ components of that vector using trigonometry. In this case, ‖‖⃑𝑣‖‖=β€–β€–βƒ‘π‘£β€–β€–πœƒ,‖‖⃑𝑣‖‖=β€–β€–βƒ‘π‘£β€–β€–πœƒ.ο—ο˜cossin

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