### Video Transcript

In this video, we’ll learn how to
solve problems using Newton’s first law of motion, sometimes called the law of
inertia.

So, Isaac Newton presented three
laws of motion in a book, often referred to as the Principia in 1686. These laws described the motion of
bodies and how they interact and the underpinned classical mechanics as we know it
today. And they were groundbreaking at the
time as they allowed mathematicians and physicists to take rather complicated
scenarios and simplify them. By thinking about these bodies as
single point masses, this meant they could consider them as points with no size or
rotation and must ignore factors such as friction or air resistance.

So, we’re interested in Newton’s
first law of motion. This law is sometimes called the
law of inertia. And it states that an object will
remain at rest or in uniform motion in a straight line unless acted upon by an
external force. In other words, if there’s no net
force, essentially the forces acting upon the body cancel one another out, the
object will maintain a constant velocity. If that velocity is zero, then the
object remains at rest. In this case, we say that the
system of forces are in equilibrium. The net sum of those forces is
equal to zero.

Now, there are many real-life
applications of this law. For example, imagine that you’re in
a moving car holding a very full cup of hot coffee. If that car is moving at a constant
speed, the coffee will sit quite nicely in the cup, leaving your clean trousers
untouched. However, apply an external force
from, say, breaking and that coffee is quite simply going to continue moving forward
at the same speed and in the same direction as it was before. Now, it’s not only your lovely
trousers that are in danger, but also the dashboard and even the windscreen of the
car.

So now we have Newton’s first law
and a little context, let’s look at how to apply it to a system of forces acting in
two dimensions.

In the given figure, the body is at
rest under the action of a system of forces. Given that the forces are measured
in newtons, find the magnitudes of 𝐹 and 𝑘.

The diagram shows us a body at
rest, and there are a number of forces acting upon it. And so, we’re going to recall
Newton’s first law of motion. This says that if there’s no net
force, in other words, the forces acting on the body essentially cancel one another
out, the object will maintain a constant velocity. Now if that velocity is zero, then
the object remains at rest. So, for our body to remain at rest,
under the action of this system of forces, we need the sum of those forces to be
equal to zero. This is known as equilibrium.

Now, since our forces are acting in
two dimensions, we could actually consider the horizontal and vertical directions
separately. Or we could think about this in
terms of vectors. In this case, we’d need the vector
sum of our forces to be equal to zero. In fact, we’re just going to
consider both the horizontal and vertical directions separately. We’ll begin with the vertical
direction, and we’re going to take upwards as being positive. We have a force of 57 newtons
acting upwards. And then, we have a force 𝐹 acting
downwards, and so this is acting in the negative direction. And we say the sum of the forces in
the vertical direction is 57 minus 𝐹.

For the body to remain at rest, we
know that the sum of these forces is zero. So 57 minus 𝐹 is equal to
zero. We’ll solve this equation to find
the value of 𝐹 by adding 𝐹 to both sides. And when we do, we find that 57 is
equal to 𝐹 or 𝐹 is equal to 57. And of course, we’re told that
these measurements were in newtons. So, 𝐹 must be 57 newtons.

We’re now going to consider the
horizontal direction. And we’re going to consider the
direction to the right as being positive. In this direction, we have 27 and
66 newtons. Then, in the other direction, in
the negative direction, we have 𝑘. So, the sum of the forces acting
horizontally are 27 plus 66 minus 𝑘. Once again, we know this is equal
to zero. So, we have an equation in terms of
𝑘. 27 plus 66 is 93. So, our equation simplifies to 93
minus 𝑘 equals zero. This time, we solve by adding 𝑘 to
both sides, and we get 93 is equal to 𝑘. Once again, our measurements are in
newtons. So, 𝑘 must be equal to 93
newtons. And so, the magnitudes of 𝐹 and
𝑘, remember, that’s just the size of the forces, are 57 and 93 newtons,
respectively.

We’re now going to see what this
process looks like when the object is in motion.

In the figure, the body is moving
at a constant velocity 𝑣 under the action of a system of forces. Given that the forces are measured
in newtons, find the magnitudes of 𝐹 and 𝑘.

We’re told that this body is moving
at a constant velocity under a number of forces. And so, we’re able to recall
Newton’s first law of motion. This says that if there’s no net
force, in other words, the forces acting on the body cancel each other out, the
object will maintain a constant velocity. So, for our body to maintain its
constant velocity under the action of the system of forces, we need the sum of these
forces to be equal to zero. Now, in fact, our forces are acting
in two dimensions. So, we could consider the
horizontal and vertical directions separately or think about it as a vector such
that the vector sum of the forces is equal to zero.

We’re going to think about this as
two directions, and we’re going to consider the vertical direction first. We’ll take upwards to be
positive. Now, it doesn’t really matter which
direction we choose to be positive as long as we’re consistent throughout the
question. The only force that we have acting
in that direction is force 𝐹. If we look carefully, we see that
the force that’s parallel to 𝐹 is 20, and that’s acting in the opposite
direction. So, the sum of these two forces
must be 𝐹 minus 20.

We also have a force of 31 newtons
acting in the negative direction. So, the sum of all of our forces
acting vertically is 𝐹 minus 20 minus 31. And of course, we know that this
must be equal to zero for our object to maintain a constant velocity. So let’s simplify the
expression. Negative 20 minus 31 becomes
negative 51. So, the sum of our forces is 𝐹
minus 51. And our equation is 𝐹 minus 51
equals zero. To solve this equation, to find the
value of 𝐹, we’re simply going to add 51 to both sides. And when we do, we find 𝐹 is equal
to 51. We were told that the forces are
measured in newtons. So, we can say 𝐹 must itself be 51
newtons.

We’re now going to repeat this
process in the horizontal direction. This time moving to the right is
our positive direction. The force given by the arrow acting
in this direction is 𝑘 newtons. Then, in the opposite direction, we
have 79 newtons. So, the sum of the forces acting on
this body in our horizontal direction is 𝑘 minus 79. Once again, we can set this equal
to zero because we’re told that the object has a constant velocity. Then, we solve for 𝑘. And we add 79 to both sides. And when we do, we find 𝑘 is equal
to 79 or 79 newtons. 𝐹 is, therefore, equal to 51
newtons and 𝑘 is equal to 79 newtons.

So, what does this mean if one or
more of the forces acting upon the object is acting at an angle? Well, essentially, it’s the same
thing as long as we’re really careful to consider the direction of those forces. Let’s take a look.

A body of mass 20 kilograms is
pulled along a horizontal plane by a rope that makes an angle 𝜃 with the plane,
where tan of 𝜃 is equal to five twelfths. When the tension in the rope is 91
newtons, the body moves with uniform acceleration. Find the total resistance to motion
𝐹 and the normal reaction 𝑅. Use 𝑔 is equal to 9.8 meters per
square second.

Let’s begin by sketching this
out. The body has a mass of 20
kilograms, and so this means it exerts a downward force of 20𝑔 on the plane. It’s pulled by a rope that makes an
angle 𝜃 with the plane. And then, we’re told that when the
tension is 91 newtons, the body moves with uniform velocity. So, the force that’s actually
pulling the body is 91 newtons.

Now, actually, there is another
force that we’re interested in, and it’s a little bit outside the scope of this
video to investigate this too much. But Newton’s third law of motion
tells us that for every action, there’s an equal and opposite reaction. So, there’s a normal reaction force
of the plane on the body. That’s a result of the force of the
weight of the body on the plane. And that acts upwards and away from
the plane, as shown. Finally, let’s add the resistance
to motion 𝐹. We can assume that this acts
parallel to the plane, as shown. This might be, say, a frictional or
air resistance force.

Now, we have all the forces in our
diagram. And we’re told that the body is
moving with uniform velocity. Now, Newton’s first law of motion
tells us that for this to be the case, the net sum of the forces in both the
horizontal and vertical direction must be equal to zero. So, we’re going to need to compare
forces in the horizontal and vertical direction. This does mean, though, that we
need to be really careful with the tension force that’s acting at an angle. And so, if we add a right-angled
triangle as shown, we see that there are components of this force that act in both
the horizontal and the vertical direction.

We, therefore, need to use
right-angled trigonometry to find those components. The hypotenuse of this triangle is
91 newtons. And then, the component that acts
in a vertical direction is the opposite side. And the horizontal direction is the
adjacent side. And so, we’ll begin by considering
the forces that act in a horizontal direction. Let’s define the adjacent side in
our right-angled triangle to be 𝑥 newtons. If we then take the direction to
the right to be positive, we can say that the sum of the forces acting in this
direction are 𝑥 minus 𝐹.

Then, since the body moves with
uniform velocity, we can say that the sum of these forces is equal to zero. When we solve for 𝐹 by adding 𝐹
to both sides, we find 𝐹 is equal to 𝑥. We’re actually able to work out the
value of 𝑥 by using the cosine ratio, since we know the hypotenuse and we’re trying
to find the adjacent. We can say that cos of 𝜃 is 𝑥
divided by 91. So, multiplying by 91 gives us 𝑥
equals 91 cos 𝜃.

But we haven’t yet used the fact
that tan of 𝜃 is five twelfths. And so, since tan of 𝜃 is opposite
over adjacent, we can construct a more general triangle. In this triangle, the length of the
side opposite to 𝜃 is five units and its side adjacent is 12 units. We have the Pythagorean triple five
squared plus 12 squared equals 13 squared, and so, the hypotenuse must be 13. And so, cos of 𝜃 for our angle
which is adjacent over hypotenuse must be 12 over 13. And so, 𝑥 is 91 times 12 over 13,
and that’s equal to 84. Since 𝐹 is equal to 𝑥, we can say
that 𝐹 must also be equal to 84. And all our measurements are in
newtons, so 𝐹 is 84 newtons.

We’ll need to perform a similar
process, but this time in a vertical direction. And that will allow us to calculate
the value of 𝑅. We’re going to define upwards to be
positive. And we’re also going to say that
the length of the side in our right-angled triangle that’s opposite the angle 𝜃 is
equal to 𝑦. This is also acting upwards. So, in an upwards direction, we
have 𝑅 plus 𝑦. And then, we have 20𝑔 acting in
the opposite direction. So, the sum of our forces is 𝑅
plus 𝑦 minus 20𝑔. And once again, that must be equal
to zero. We’re going to add 20𝑔 to both
sides of this equation and subtract 𝑦, and we get 𝑅 is equal to 20𝑔 minus 𝑦.

But we now need to work out the
value of 𝑦. And so, once again, we’re going to
use right-angled trigonometry. This time, we use the sin ratio
since sine is opposite over hypotenuse. So, sin 𝜃 is 𝑦 over 91. And so, solving for 𝑦, we get 𝑦
is 91 sin 𝜃. But let’s go back to our more
general triangle. We know the opposite side in this
triangle is five and its hypotenuse is 13. And so, sin of 𝜃 must be five
thirteenths and 𝑦 is 91 times five thirteenths. And that’s equal to 35.

Our earlier equation, therefore,
becomes 𝑅 equals 20𝑔 minus 35. But of course, we were told that 𝑔
is 9.8. So, this becomes 20 times 9.8 minus
35, and that’s 161 or 161 newtons. The resistance to motion 𝐹 is then
84 newtons, and the normal reaction 𝑅 is 161 newtons.

In our final example, we’ll look at
how we can use Newton’s first law of motion to calculate a speed.

A soldier jumped out of a plane
with a parachute. After he had opened his parachute,
the resistance to his movement was directly proportional to the cube of his
speed. When his speed was 19 kilometers
per hour, the resistance to his motion was a twenty-seventh of the combined weight
of him and his parachute. Determine the maximum speed of his
descent.

Let’s sketch a little diagram. Here is our soldier plummeting
through the sky. The downwards force of his weight
is what’s causing him to do so. Then, we have a resistance to the
motion of the parachute, that’s 𝑅. And it acts in the opposite
direction. We’re told that the resistance is
directly proportional to the cube of his speed. So, let’s let his speed be equal to
𝑣, and that’s kilometers per hour. For this to be the case, 𝑅 must be
equal to 𝑘 times 𝑣 cubed, where 𝑘 is known as the constant of
proportionality. Let’s find an expression for 𝑘 by
using the rest of the information in this question.

When his speed was 19, the
resistance was a twenty-seventh of the combined weight of him and his parachute. So, when 𝑣 is 19, 𝑅 is one
twenty-seventh times 𝑤. So, we can say that one
twenty-seventh times 𝑤 is 𝑘 times 19 cubed. This means 𝑘 is a twenty-seventh
𝑤 divided by 19 cubed. Now, that’s equivalent to saying 𝑤
over 27 times 19 cubed. Now, we’re not going to work this
out and we’ll see why in a moment. We can, therefore, substitute this
back into our earlier equation and say 𝑅 is 𝑤𝑣 cubed over 27 times 19 cubed.

Now, we’re told that at some point
the soldier reaches a maximum speed. At this point, the speed will
remain unchanged. Now, for the velocity to remain
uniform, Newton’s first law of motion says that the sum of all of the forces in this
direction must be equal to zero. Now, let’s take downwards to be the
positive direction. We can say that the weight minus
the reaction force is the sum of these forces. So, 𝑤 minus 𝑅 is equal to
zero. Then, adding 𝑅 to both sides, we
get 𝑤 is equal to 𝑅.

But let’s replace 𝑅 with the
earlier expression in terms of 𝑤 and 𝑣. And since 𝑤 is not equal to zero,
we’re able to divide both sides of this equation by 𝑤. So, one is 𝑣 cubed over 27 times
19 cubed. We’re then going to multiply both
sides by 27 times 19 cubed. And finally, we’re going to find
the cube root of both sides. The cube root of 27 is three, and
the cube root of 19 cubed is 19. So, we find three times 19 is equal
to 𝑣, but three times 19 is 57. So, 𝑣 is equal to 57 and that’s
kilometers per hour.

Let’s recap the key points from
this lesson. In this video, we saw that we’re
able to consider bodies as points with no size or rotation. And this means we can ignore
extraneous factors such as friction or air resistance. Newton’s first law states that if
there’s no net force, essentially the forces acting on the body cancel one another
out, the object will maintain a constant velocity. Now, of course, if that velocity is
zero, then the object must remain at rest. In this case, we say that the
system of forces are in equilibrium; their sum is equal to zero. Finally, we saw that when we’re
working with forces acting in a direction other than horizontal or vertical, we can
use right-angled trigonometry to break them down into the components we require.