# Lesson Video: Newton’s First Law of Motion Mathematics

In this video, we will learn how to solve problems using Newton’s first law.

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### 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.

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