Lesson Video: Resultant Motion and Force Physics

Video Transcript

In this lesson, we’re going to learn how to find the net result when combining multiple values of a kinematic quantity, such as two displacements or two accelerations. When making such combinations, vectors will be the best mathematical objects to represent our kinematic quantities. So let’s recall how to work with vectors and their components.

Vectors are mathematical objects that have both a magnitude and a direction. We can conveniently represent vectors as arrows, where the length of the arrow is the magnitude of the vector. And the direction the arrow is pointing is the direction of the vector, in this case directly off to the right. Representing vectors as arrows allows us to do certain types of vector calculations geometrically.

To do these calculations, it’s helpful to be able to refer to the two ends of the vector unambiguously. So we’ll call this end the tail and this end the head. We can easily see why vectors are useful for representing kinematic quantities by considering a person pushing a box across the floor. Let’s say our person exerts a force of 10 newtons to push the box. Then 10 newtons is the magnitude of the force. But just knowing this magnitude doesn’t tell us if the box is being pushed to the left or the right.

Now, as we can clearly see from the picture, the direction of the force is towards the right. So the full information about this force includes both its magnitude and direction, which is why vectors are the appropriate way to represent force. As we’ll see later, to describe any of the kinetic quantities that we’re interested in, we’ll need both a magnitude and a direction. So since vectors are the appropriate choice for all of these, let’s continue to work out how to combine vectors.

When we add two vectors together, we’ll need to take into account that each one has a direction. So we’ll need a different procedure than simply adding two numbers. By representing vectors as arrows, we can actually perform this sum quite easily. For example, consider the sum of these two vectors over here.

To actually perform this sum, we need to align the tail of the second vector to the head of the first vector. We do this graphically by simply drawing the second vector with its tail at the head of the first vector. The sum of these two vectors is now the vector whose tail is at the tail of the first vector and whose head is at the head of the second vector. And again, we find this geometrically by simply drawing that arrow.

As we can see, the magnitude and direction of this vector sum is different from the simple arithmetic sum of the magnitudes and directions of our two original vectors. However, like addition of numbers, vector addition is also commutative. This means that the order of summation doesn’t matter. So drawing the blue vector at the head of the orange vector gives the same result in sum vector as drawing the orange vector at the head of the blue vector. Adding a pair of vectors this way to get a single sum suggests another way that we can represent an individual vector.

We’ll start off by drawing our vector with its tail at the origin of some Cartesian axes. Now let’s imagine that we draw a straight vertical line from the head of our vector to the horizontal axis. Then we can draw this vector along the horizontal axis that goes exactly as far to the right as the head of our original vector. We can also do the same for the vertical axis.

Now, this vector along the vertical axis goes exactly as far up as the head of our original vector. But now let’s look at this picture carefully. The vertical vector would fit exactly between the head of the horizontal vector and the head of the vertical vector. The same is true of the horizontal vector. It fits exactly between the head of the vertical vector and the head of our original vector. But let’s look carefully at what we’ve drawn. Whether we start with the vertical vector or the horizontal vector, we have the tail of a second vector drawn at the head of a first vector.

We then have a third vector whose tail is the same as the tail of the first vector and whose head is the same as the head of the second vector. But these are just the three vectors involved in a vector sum. So our original magenta vector can be represented as the sum of a horizontal orange vector and a vertical blue vector. If we call our original vector 𝐕, where we used a half arrow over a letter to represent that we’re talking about a vector, then we call the horizontal and vertical vectors the components of 𝐕. And we usually use the symbol 𝐕 sub 𝑥 to represent the horizontal component and 𝐕 sub 𝑦 to represent the vertical component. Symbolically, we’d write that 𝐕 is the sum of 𝐕 𝑥 and 𝐕 𝑦.

This is a very powerful representation because as we can see from our graph 𝐕 𝑥 and 𝐕 𝑦 are aligned with the axes of our Cartesian system, which means they’re perpendicular. Qualitatively, this allows us to independently talk about the horizontal and vertical components of 𝐕, even though 𝐕 itself is neither horizontal nor vertical. In technical language, we call 𝐕 𝑥 and 𝐕 𝑦 the projection of 𝐕 onto these axes. This just means that 𝐕 𝑥 and 𝐕 𝑦 are the components of 𝐕 that are parallel to the axes of our system.

Quantitatively, the fact that our triangle formed by 𝐕 𝑥, 𝐕 𝑦, and 𝐕 is a right triangle means that we can easily apply the Pythagorean theorem and do trigonometric calculations. Let’s let the symbol for a vector without the half arrow on top represent the magnitude of that vector. Then we have, by the Pythagorean theorem applied to this right triangle, that the magnitude of 𝑉 is equal to the square root of the sum of the squares of the magnitude of 𝑉 𝑥 and the magnitude of 𝑉 𝑦. So this relates the magnitude of 𝑉 to 𝑉 𝑥 and 𝑉 𝑦.

Let’s see if we can also relate the direction. Since our vector is drawn on some Cartesian axes, we can represent the direction of the vector by the angle between the vector and the horizontal axis. Let’s call this angle 𝜃. To relate 𝜃 to 𝑉 𝑥 and 𝑉 𝑦, recall that the tangent of an acute angle in a right triangle is equal to the length of the opposite side divided by the length of the adjacent side. In this triangle, 𝐕 𝑦 is opposite 𝜃 and 𝐕 𝑥 is adjacent to 𝜃. So the tan of 𝜃 is the magnitude of 𝑉 𝑦 divided by the magnitude of 𝑉 𝑥.

Alternatively, we can now take the inverse tangent of both sides of this equation. This gives us 𝜃, the direction of 𝑉, is equal to the inverse tangent of the magnitude of 𝑉 𝑦 divided by the magnitude of 𝑉 𝑥. This gives us two equations, one for the magnitude and one for the direction of the vector 𝐕 in terms of only the magnitudes of its components. We don’t need to make any reference to the direction of these components because we know that they’re perpendicular to each other and parallel to the axes of our system.

Conversely, we can also find the magnitudes of 𝑉 𝑥 and 𝑉 𝑦 from the magnitude of 𝑉 and its direction. Recall that the cosine of an acute angle of a right triangle is the length of the adjacent side divided by the length of the hypotenuse. This means that the length of 𝑉 𝑥 is equal to the length of 𝑉 times the cos of 𝜃. Similarly, to find the length of the opposite side of an acute angle, we use the sine of the angle instead of the cosine. So the magnitude of 𝑉 𝑦 is the magnitude of 𝑉 times the sin of 𝜃.

These two equations now give us the magnitude of the components of a vector in terms of the magnitude of the vector itself and the vector’s direction. Note that, again, we’ve made no reference to the direction of these vectors. This is because their directions are predetermined. The vector 𝐕 𝑥 is always along the horizontal direction, and the vector 𝐕 𝑦 is always along the vertical direction.

We started by using our notion of adding vectors to come to the idea of vector components. Now, we can do the reverse. We can use the equations we’ve derived to improve our ability to add vectors.

Let’s now consider a sum of two vectors, vector 𝐕 and vector 𝐔. We already know how to perform this sum by drawing a vector from the tail of vector 𝐕 to the head of vector 𝐔. But before we do that step, let’s draw the horizontal and vertical components of these two vectors. For vector 𝐕, the process is the same as before. We draw a vertical line to the horizontal axis, which tells us how long the horizontal component is. The vertical component is then the vector between the head of the horizontal component and the vector 𝐕 itself.

And thus, just like before, we have that 𝐕 is the sum of 𝐕 𝑥 and 𝐕 𝑦. We can do the same for 𝐔. But we have to account for the fact that the tail of 𝐔 is not located at the origin of our system. This is not a problem. If we recall though, it really defines the components of the vector is that the horizontal component is parallel to the horizontal axis of the system. The vertical component is parallel to the vertical axis of the system. And the sum of the components equals the vector itself. In other words, the horizontal and vertical components of a vector form the horizontal and vertical legs of the right triangle that has the vector as the hypotenuse.

For the vector 𝐔, those two legs are these dotted lines. So the horizontal leg is 𝐔 𝑥, and the vertical leg is 𝐔 𝑦. So as we can see, we can still draw the components of 𝐔 even though 𝐔 is not at the origin.

All right, let’s now draw the vector that’s the sum of the vectors 𝐕 and 𝐔. Let’s call this vector 𝐖. We’re now looking for the horizontal and vertical components of 𝐖. The components of 𝐖 will form the legs of the right triangle with 𝐖 as the hypotenuse. On the horizontal axis, this will be the vector 𝐕 𝑥 plus the vector represented by this orange dotted arrow. On the vertical axis, this will be the vector represented by this blue dotted arrow plus the vector 𝐔 𝑦.

But now let’s look closely at these two dotted vectors. The orange dotted vector is the same as the vector 𝐔 𝑥. Similarly, the blue dotted vector is the same as the vector 𝐕 𝑦. Both of these are true because the two solid vectors and the two dotted vectors form a rectangle and opposite sides of a rectangle are congruent.

Anyway, we can now easily see that since the tail of 𝐔 𝑥 lined up to the head of 𝐕 𝑥 makes the vector 𝐖 𝑥, 𝐖 𝑥 equals 𝐕 𝑥 plus 𝐔 𝑥. By the same logic and using the commutativity of vector addition, we have that 𝐖 𝑦 is equal to 𝐕 𝑦 plus 𝐔 𝑦. Remember though that both of our equations for the components of 𝐖 came from our original relationship that 𝐖 is equal to 𝐕 plus 𝐔. This leads us to a very important conclusion about the sum of two vectors. The components of a vector that is the sum of two other vectors are each individually equal to the sum of the corresponding component of those two vectors.

So as we’ve already seen, the horizontal component of 𝐖 is the sum of the horizontal components of 𝐕 and 𝐔. Now, we might reasonably wonder why this helps us. It seems like we’ve just replaced one vector equation with two vector equations. However, we’ve actually gained a lot. We can’t evaluate the sum of 𝐕 plus 𝐔 without considering the direction of these vectors.

However, recall that every horizontal component has the same direction. When we add two vectors with the same direction, we simply add their magnitudes and keep the direction. This means that we can replace both of our equations for the horizontal and vertical components of 𝐖 with their nonvector equivalents. So the magnitude of 𝑊 𝑥 is equal to the magnitude of 𝑉 𝑥 plus the magnitude of 𝑈 𝑥. And the same is true of the magnitude of 𝑊 𝑦. It’s the magnitude of 𝑉 𝑦 plus the magnitude of 𝑈 𝑦.

Again, the only reason we were able to convert these vector equations into scalar equations is because all of the vectors in each equation have the same direction. What this means is that by using our previous equations that related the magnitude and direction of vectors to the sides of their components, we can evaluate this vector sum as these two usually much easier scalar sums.

Okay, this has been a whole lot of abstract math. But it’s been worth it because we can now apply this one set of ideas to any of our kinematic quantities. Each of the quantities we’re interested in can be represented by a vector.

Displacement is the distance from a reference point in a particular direction. So since displacement has a magnitude and a direction, we can represent it as a vector. Similarly, velocity is the speed of motion in a particular direction. So it, too, has a magnitude and a direction and can be represented with a vector. Acceleration is the rate of change of velocity. Since velocity has both magnitude and direction, the change in velocity can either be a change in magnitude or direction or both. So the acceleration has to carry information about magnitude and direction. So it, too, can be represented as a vector. Finally, force, by Newton’s second law, is equal to mass times acceleration. We know acceleration is a vector. So mass, a scalar, times a vector is just another vector.

Furthermore, as we saw in our example of a person pushing a box, in order to fully describe a force, we need to supply both a size and a direction. Because each of these quantities can be represented with a vector, we can also work with combinations of these quantities the same way we’d work with combinations of vectors. Let’s see some examples of this.

A length of road stretches north for 10 kilometers from the edge of a town to where it intersects an eastward road. A car is broken down on the eastward road. And the displacement of the edge of the town from the car has a magnitude of 24 kilometers. How far east of the intersection is the car to the nearest kilometer?

To start answering this question, let’s draw a diagram. We have a northbound road, an eastbound road, a broken-down car, and a town. Here’s our northbound road, our eastbound road, our car, and our town. We’re told that the intersection of these roadways is 10 kilometers north of the town. We’re also told that the magnitude of the displacement between the car and the town is 24 kilometers. Our task is to find this distance from the intersection to the car to the nearest kilometer.

If we view the displacement as a vector, we can see that the distance we’re looking for is just the size of the horizontal projection of the displacement. But we can easily find the length of this vector component by applying the Pythagorean theorem to the right triangle formed by the northbound road, the eastbound road, and the displacement. This is a right triangle because, by definition, the direction of north and the direction of east are perpendicular.

Let’s call the length of this eastbound road segment 𝑙. Then the Pythagorean theorem tells us that 10 kilometers squared plus 𝑙 squared, the sum of the squares of the lengths of the legs of the triangle, is equal to 24 kilometers squared, the square of the length of the hypotenuse. If we subtract 10 kilometers squared from both sides, 10 kilometers squared minus 10 kilometers squared on the left-hand side is zero. And we’re left with 𝑙 squared is equal to 24 kilometers squared minus 10 kilometers squared. 24 squared is 576, and 10 squared is 100. So we can rewrite the right-hand side as 576 kilometers squared minus 100 kilometers squared. 576 minus 100 is 476, so 𝑙 squared is equal to 476 kilometers squared.

To find 𝑙, we simply take the square root of both sides. The square root of kilometers squared is just kilometers. So 𝑙 is equal to the square root of 476 kilometers. The square root of 476 is approximately 21.8. Rounding to the nearest kilometer, we look to the first place to the right of the decimal point, which is eight. Eight is greater than five, so one rounds up to two. So, to the nearest kilometer, the distance from the broken-down car to the intersection is 22 kilometers.

Great, let’s see another example.

A bird flies along a line that displaces it 450 meters east and 350 meters 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.

All right, so we have a diagram where the cardinal directions are labeled such that east is toward the right and north is toward the top. The blue line represents the displacement vector of the bird. And the two black lines represent the vertical and horizontal components of this displacement. As stated in the question, these are 350 meters north and 450 meters east, respectively. The question asks us what angle the bird must turn to change its direction and fly directly towards the north.

To help us figure out this angle, let’s draw the bird’s northward trajectory on the diagram. This magenta line represents a northward trajectory since it points directly up. On the other hand, if the bird were to continue along its current trajectory, it would follow this blue dotted line. So in order to change direction and fly directly north, the bird has to turn through this angle between the two trajectories.

This is the angle we need to calculate. But since we don’t know anything else about the dotted or magenta vectors, let’s try to find another part of our diagram with the same angle. The magenta vector and the northward part of the displacement both point directly north, so they’re parallel. But this means that the bird’s trajectory is a straight line intersecting two parallel lines. So corresponding angles at the two intersections have the same measure.

In our diagram, this angle between the northern component of the displacement and the displacement itself corresponds to the angle we drew before, since both of these angles are to the right of one of the parallel lines and above the displacement line. So if we can find the measure of this angle, we’ll know the angle that the bird needs to turn.

Now, recall that by adding the horizontal and vertical components of a vector, we get the vector itself. Therefore, if we draw 450 meters to the east between the head of the northern vector and the head of the displacement vector, we’ll get a right triangle formed by the two components and the displacement. These two vectors form a right angle because north and east are perpendicular. Let’s call our angle 𝜃.

Now, recall that the tangent of an angle is equal to the length of the opposite side divided by the length of the adjacent side. So tan of 𝜃 is equal to 450 meters divided by 350 meters. Meters divided by meters is just one. So the right-hand side of this equation is just a number without units. Now, we can take the inverse tangent of both sides to find this angle. Plugging into a calculator 𝜃 is equal to arctan of 450 divided by 350, we get 𝜃 is very close to 52.125 degrees.

Now, remember, this is the same angle that we were looking for to answer the question. So all we need to do is round 𝜃 to the nearest degree. Looking to the first digit right of the decimal point, one is less than five, so two rounds to two. So, to the nearest degree, the bird needs to turn 52 degrees toward the west to fly directly north.

Okay, now that we’ve seen some examples, let’s review the key points we’ve learned in this lesson. In this video, we learned that a full description of the kinematic quantities displacement, velocity, acceleration, and force must include a magnitude and a direction. This makes vectors the perfect mathematical object to represent these quantities.

We also saw that a vector representing any one of these quantities could be represented as a sum of two vectors, each one parallel to one of the axes in a Cartesian system. Using the Pythagorean theorem and trigonometry rules for right triangles, we were then able to relate the magnitude and direction of our original vector to the magnitudes of its components, where we’ve used the same symbols as the vectors, but without the half arrow on top to represent the magnitude of each vector.

We were also able to determine the magnitude of the components in terms of the magnitude and direction of the original vector. This is sufficient to completely determine the components because each component’s direction is already known. Using components, we also converted a vector sum into two scalar sums, one for each component. Using these sums is often quite easy, which is very useful for us because combining two values for kinematic quantity can be represented by vector addition.