In this video, we’re going to learn about forces and Newton’s second law of motion. We’ll learn what the law says, what it means, and how to use the second law practically.
To start out, imagine that your great aunt Thelma has a beautiful and very fragile antique crystal bowl. And one day, to help celebrate your 12-year-old nephew’s birthday party, she fills the bowl to overflowing with candy. Now, this was a questionable decision because soon enough the bowl is overrun by 12-year-olds hungry for the candy inside. As they jostle for the candy, each one exerts a force on the bowl until this antique fragile crystal bowl is under considerable stress. The question is, what is the net, or overall, force exerted on the bowl, and is it big enough to make the bowl crack?
Newton’s second law of motion helps us understand this question better. To start talking about Newton’s second law, let’s consider that we have a box at rest in front of us. Let’s imagine further that on this box there are a series of forces which act in various directions and, overall, cancel one another out. When the forces on an object like this box all balance out, then the object experiences no net force, that net force is zero, and its motion doesn’t change. If it was at rest, it stays at rest. If it was moving at a constant speed, it continues on at that speed.
But now, let’s imagine that we add a force to the forces acting on this box. When the forces on an object don’t balance out, there is a net force which causes the object’s motion to change. Specifically, the object accelerates. As Isaac Newton was looking into these quantities of net force and acceleration of an object, he found that they’re related. He saw that as net force goes up, by that same factor acceleration would go up and vice versa.
There’s a third quantity we can connect to net force and acceleration, and that’s an object’s mass. Let’s consider a thought experiment where we have a very small mass and a very large mass sitting on the same surface. Let’s say that we apply a force 𝐹 sub 𝑝 horizontally to this very small mass. With the mass being so very small, even for a moderately sized force, we expect it to accelerate rapidly. It will have a large acceleration for the small mass.
Now, let’s take the very same force 𝐹 sub 𝑝 and apply it to the very large mass capital 𝑀. So long as the floor doesn’t provide any frictional resistance, we do expect the mass to accelerate, but we imagine it will barely move. Its acceleration will be quite small. Summing up the results of this thought experiment, we see that when our mass is very small, for a given force, we expect our acceleration then to be quite large. And for the same force magnitude, if our mass is quite large, we expect our acceleration to be quite small.
Overall, then, we might say that mass and acceleration are inversely proportional. That is, when one gets large, the other gets smaller, for a given force. If we bring the two proportionalities we’ve talked about together, they combine to give us an expression that says the net force that acts on an object divided by the mass of that object is equal to the acceleration the object experiences as result of that net force. This is the form of Newton’s second law of motion.
We often see it in a slightly different arrangement. If we multiply both sides by the mass 𝑚, we see this form. The net force is equal to an object’s mass multiplied by its acceleration. Notice that this equation doesn’t apply for any old force. Specifically, it’s the net force acting on an object which is equal to 𝑚 times 𝑎. A second thing we might notice is that this is a vector equation. Both force and acceleration have a magnitude and direction. And the direction of acceleration equals the direction of the net force. Let’s get some practice using this second law in a couple of examples.
A sprinter with a mass of 57.0 kilograms accelerates at 3.922 meters per second squared. What is the magnitude of the net external force on her?
We can call this magnitude of the net external force acting on the sprinter capital 𝐹. We’re told the sprinter’s mass, 57.0 kilograms, which we’ll label 𝑚 and also her acceleration, 3.922 meters per second squared, which we’ll name 𝑎. As the sprinter runs, it’s the frictional force of the track on her feet that push her forward and give her a net forward force. It’s that force magnitude we want to solve for. And to do it, we’ll recall Newton’s second law of motion.
Newton’s second law says that the net force acting on an object equals the object’s mass times its acceleration. In the case of our sprinter, we’re working with magnitudes rather than vectors. And we’re given values for the sprinter’s mass as well as her acceleration 𝑎. When we plug in for those two values and calculate this product, we find it’s equal to 224 newtons. That’s the net force magnitude causing the sprinter’s acceleration. Now, let’s look at an example where the applied force and the object’s resulting acceleration are not in the same direction.
A cart with a mass of 18 kilograms is at rest on a level floor. A constant 15-newton force is applied to the cart at an angle 52 degrees below the horizontal. If friction is negligible, what is the speed of the cart when the force is applied over a distance 6.7 meters?
In this exercise, we want to solve for the speed of the cart after the force has been applied over some distance. We’ll call that speed 𝑣 sub 𝑓. We’re told the cart’s mass, 18 kilograms, we’ll label that 𝑚, the force magnitude acting on the cart 15 newtons, which we’ll call 𝐹. The direction relative to the horizontal that that force acts 52 degrees, we’ll name that angle 𝜃, and finally, we’re told the distance over which that force 𝐹 is applied. It’s 6.7 meters. We’ll label that distance 𝑑.
Let’s start on our solution by drawing a diagram of the scenario. We start off in this example with our cart of mass 𝑚 at rest on a flat frictionless surface. A force 𝐹 is applied to the cart in the direction 𝜃 degrees below the horizontal. This force continues to act on the cart over a distance, 𝑑, of 6.7 meters. After being under the influence of the force over that span, we want to know what is the final speed of the cart.
To find out, let’s recall Newton’s second law of motion. The second law tells us that the net force that acts on an object is equal to that object’s acceleration times its mass. In our case, as we look at the force applied to our mass, we see that only a component of the force will contribute to its acceleration to the right. The vertical component of that force will contribute to the normal force that the mass experiences. Only the horizontal component of 𝐹 will contribute to its acceleration to the right.
Looking at our diagram, we see that it’s the cos of that angle 𝜃 multiplied by the magnitude of the force 𝐹 which is equal to the net force to the right on our case. By the second law, that’s equal to the mass of the case multiplied by its acceleration. If we divide both sides of this equation by mass 𝑚, canceling that out on the right-hand side, we see that the acceleration of our object is equal to 𝐹 times the cos of 𝜃 all divided by the object’s mass, 𝑚.
Since all these terms are constant, acceleration 𝑎 is also constant over the span of the distance 𝑑. When the acceleration of an object is constant, that’s a sign that the kinematic equations apply to describe the motion of that object. Looking over these four equations, we seek to find one which matches what we want to solve for, a final velocity, as well as the information we’re given in the problem statement.
The second equation we’ve written is a match for these conditions. It tells us that the final speed of the cart squared is equal to its initial speed squared plus two times its acceleration, again which is constant, multiplied by the distance over which that acceleration occurs. We’re told in the problem statement that our case starts at rest, which means 𝑣 sub zero is equal to zero. If we then take the square root of both sides of this equation, we see that 𝑣 sub 𝑓 is equal to the square root of two times 𝑎 times 𝑑. And 𝑎, the acceleration, is something we’ve solved for symbolically earlier.
If we plug in this expression for 𝑎, we now have an expression for what we want to solve for, 𝑣 sub 𝑓, in terms of known values 𝐹, 𝜃, 𝑚, and 𝑑. We’re ready to plug in and solve for 𝑣 sub 𝑓. With these values inserted, when we calculate the result of the square root, we find that, to two significant figures, it equals 2.6 meters per second. That’s our object’s final velocity after being accelerated over the distance 𝑑.
Let’s summarize what we’ve learnt so far about forces and Newton’s second law of motion. We’ve seen that Newton’s second law of motion describes a mass’s acceleration when a net force acts on it. Mathematically, we would write that the net force is equal to mass times acceleration. We noted that this relationship doesn’t hold for any force, but only for the net force that acts on an object.
We’ve also seen that the direction of the net force is the same as the object’s acceleration direction, that these two quantities are vectors. Newton’s second law is a helpful way of understanding object motion. And by recalling these points, we’ll be that much better equipped to answer questions involving this law.