 Welcome to the second video in the Mechanics series. Here we will explore the relationship between force, mass, and acceleration. In the first video, we saw that to the extent we can neglect air resistance, near Earth's surface, all objects fall with the same constant acceleration, approximately 9.8 meters per second squared. If x is distance-fallen, v is velocity, and a is acceleration, then a equals g, v equals gt, and x equals one-half gt squared. A is a constant, v varies linearly with time, and x varies quadratically with time. We also developed the technique of differential calculus. If we plot the curve showing an object's position versus time, for a given time t, this tells us the object's position, x, and the slope of the curve at that point tells us how fast position is changing with respect to time, which is the object's velocity. To determine the velocity, we consider position at times t and t plus delta t. We develop a position at the later time, x plus delta x. In delta t seconds, position changes by delta x meters, so the average velocity over this interval is the slope of the orange triangle's hypotenuse, delta x over delta t. The instantaneous velocity at time t, which we write as dx over dt, is the limit as the interval delta t shrinks to zero of the average velocity delta x over delta t. We say that velocity is the derivative of position, and acceleration is the derivative of velocity. Acceleration tells us how fast velocity is changing, and velocity tells us how fast position is changing. Let's consider why things fall. The usual explanation is that something called gravity exerts a downward force on objects. Here are some intuitive observations about this gravitational force. If you hold a weight in your hand, you need to exert a force to keep it from falling. That force appears to be the same if you move the weight either parallel to the ground or up and down. A reasonable conclusion is that gravitational force does not depend on position. Of course, this statement is necessarily limited to our experience near Earth's surface. If I had a second weight, I need to exert more force to keep the combined weight from falling. A reasonable conclusion is that gravitational force is proportional to the mass, the amount of matter. We write f is proportional to m, or f over m equals the constant k. Now if all objects experience the same gravitational acceleration g, and the ratio of force to mass is the same constant for all objects, then since it's trivially true that two constants are proportional to each other, we can write that gravitational force divided by mass is proportional to the gravitational acceleration. Let's define the gravitational force on an object of mass m, what we call its weight, to be f equals m times g. The unit of force is called the Newton. One Newton of force equals one kilogram meter per second squared. Now we have a relation between force, mass, and acceleration. In Earth's gravitational field, force equals mass times acceleration. Let's postulate that this is true for any object subject to any type of force. This is what Isaac Newton did in his classic work, Mathematical Principles of Natural Philosophy, published in 1687. The original was in Latin and is sometimes called the Principia after its Latin title. We'll be presenting excerpts from the English translation. Newton presents three laws of motion as postulates, statements to be initially accepted without proof, their validity rests on the correctness of the physical predictions they lead to. His second law reads, the change of motion is proportional to the motive force impressed. Here motion is defined as follows, the quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjointly. In modern notation we write this statement as p equals m times v, where p is momentum, m is mass, what Newton calls quantity of matter, and v is velocity. Newton uses the term motion for what we now call momentum, and by change of motion he means what we now refer to as rate of change of momentum. So we write Newton's second law as force f equals rate of change of momentum, mv. Assuming mass doesn't change with time, we factor this out to get m times rate of change of velocity, which is mass times acceleration. F equals ma. As to what a force is, Newton gives the following definition, an impressed force is an action exerted upon a body in order to change its state of motion. But this all seems a bit circular, Newton tells us that the change of a body state of motion is proportional to impressed force, but impressed force is defined as something that changes a body state of motion. As it stands, f equals ma is merely a definition of force. If we observe the acceleration a of a mass m, then the force f acting on that mass is m times a. F equals ma becomes useful when we can specify force independently of acceleration from a force law such as Newton's law of gravitation. Then we can determine the acceleration as force divided by mass. And since acceleration is the rate of change of velocity and velocity is the rate of change of position, knowing the initial velocity and position of the mass, we can solve for its velocity and position any time in the future. In this way, f equals ma allows us to make predictions about the motion of bodies. Let's see how these ideas work in a system with a non-gravitational force. Specifically let's consider a mass suspended by a spring. Let x be the downward displacement of the mass. Gravity exerts a constant downward force. The spring exerts an upward force that increases as the spring is stretched. The sum of these is the net force, let's call it f of x. At some position the net force will be zero. This is the equilibrium position where the mass can remain suspended at rest. And we'll use this as our x equals zero reference. For non-zero x values, the net force acts to accelerate the mass back toward the equilibrium position. We are going to observe the motion of a mass on a spring and see if f equals ma is valid for this system. Our first task is to determine the force law, f of x. The force law for a spring is called Hooke's law after Robert Hooke, who first stated it in 1676. In these three photos we show a spring fixed at its top. In the right frame it has a single 2.5 pound weight at its bottom. In the middle frame two of these weights and in the left frame three weights. In all cases the weights are held at rest in equilibrium. The gravitational force on each weight is about 11 newtons downward. In equilibrium the spring exerts an equal force upward. In each case we measure the displacement, x of the spring's end, marked with a red dot in the photos. Now we plot spring force in newtons versus spring length in millimeters as red dots. We see that the relation between force and displacement is well represented by the blue line. We calculate the slope of this line to be 47 newtons per meter. This is our spring constant. Finally, we write Hooke's law. F of x equals minus k times x. The minus sign signifies that the force acts opposite to the displacement. K is the spring constant and we assume x equals 0 is the displacement when no weight is attached to the spring, that is when the force is 0. Now let's consider the net force due to the spring and gravity. The spring force is minus kx and the force of gravity is mg. We can write this as minus k times the quantity x minus mg over k. If we now replace x by x plus mg over k, that is we shift the x coordinate so that x equals 0 corresponds to the position where the net force is 0, then the net force could be written simply as minus kx. Now we attach a 5 pound mass to the spring, raise it above the equilibrium position and let it drop. Taking a video of the mass's oscillation, for any video frame we can determine the time and the corresponding x coordinate of a reference point on the mass. Here are our observed x coordinates versus time for roughly one half an oscillation period. Note that the position coordinate increases as we go from top to bottom. X is the downward displacement of the mass. Now we fit a function x equals x0 cosine omega t to these data by adjusting the values of the oscillation amplitude x0 and the frequency omega until the difference between the data and the curve is minimized. The resulting fit is very good and we use this curve to represent the mass's position as a function of time. V is the derivative of position, the slope of the position curve. The derivative of x0 cosine omega t is minus x0 omega sine omega t. Here we plot this in green, but without the constant factor omega, so that the scale is the same as the position curve. Initially the velocity is 0. As time goes on the mass accelerates downward and the velocity increases. When x reaches 0 the net force vanishes, acceleration stops and the velocity curve flattens out. After that the net force and acceleration are upward and the velocity decreases reaching 0 at the bottom of the mass's oscillation. Acceleration is the derivative of velocity, the slope of the velocity curve. The derivative of minus x0 omega sine omega t is minus x0 omega squared cosine omega t. Here we plot this in red, but without the constant factor omega squared, so that the scale is the same as the position and velocity curves. The red curve is the mirror image of the blue curve. This tells us that acceleration is proportional to the negative of position. Red force minus kx is also proportional to the negative of position, so acceleration and force are proportional to each other. Let's work out the constants of proportionality needed to satisfy f equals ma. F is minus kx, which is minus kx0 cosine omega t. Ma is m times minus x0 omega squared cosine omega t. Canceling a common factor of minus x0 cosine omega t, we're left with k equals m omega squared. Solving for the frequency omega, we get omega equals square root of k over m. But omega is also 2 pi over the oscillation period t. So we can write t as 2 pi square root of m over k. Plugging in the mass, 2.27 kilograms, which is 5 pounds, and the spring constant, 47 Newtons per meter, we find a period of 1.38 seconds. The half period is 0.69 seconds, or 690 milliseconds. When we look at our plots of position versus time, we see that, indeed, one half period of the sinusoidal oscillation corresponds to something close to 690 milliseconds. Better still, here we plot minus the spring constant k times our position data as red dots. We also plot mass times our acceleration curve. The agreement is excellent. F equals ma is true at all times for the mass spring system to a high level of accuracy. We started with the observation that F equals ma was almost trivially true for free fall motion. We guessed that it might be true more generally, and we found that it is true for the more complicated mass spring system. For more than two centuries after Newton proposed it as a law of nature, it was applied to an amazingly broad range of systems and forces and found to give an accurate description of physical phenomena in all cases. Entire sub-disciplines of science and engineering have been developed primarily from this single equation, F equals ma, undoubtedly one of the most important equations in all of physics.