WEBVTT
00:00:01.400 --> 00:00:06.350
In this video, we’re going to learn about forces and Newton’s second law of motion.
00:00:06.670 --> 00:00:12.910
We’ll learn what the law says, what it means, and how to use the second law practically.
00:00:13.610 --> 00:00:21.020
To start out, imagine that your great aunt Thelma has a beautiful and very fragile antique crystal bowl.
00:00:21.450 --> 00:00:28.970
And one day, to help celebrate your 12-year-old nephew’s birthday party, she fills the bowl to overflowing with candy.
00:00:29.550 --> 00:00:37.990
Now, this was a questionable decision because soon enough the bowl is overrun by 12-year-olds hungry for the candy inside.
00:00:38.440 --> 00:00:48.630
As they jostle for the candy, each one exerts a force on the bowl until this antique fragile crystal bowl is under considerable stress.
00:00:49.230 --> 00:00:56.490
The question is, what is the net, or overall, force exerted on the bowl, and is it big enough to make the bowl crack?
00:00:57.290 --> 00:01:01.180
Newton’s second law of motion helps us understand this question better.
00:01:01.840 --> 00:01:07.940
To start talking about Newton’s second law, let’s consider that we have a box at rest in front of us.
00:01:08.350 --> 00:01:17.130
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.
00:01:17.730 --> 00:01:28.530
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.
00:01:29.050 --> 00:01:31.020
If it was at rest, it stays at rest.
00:01:31.020 --> 00:01:34.730
If it was moving at a constant speed, it continues on at that speed.
00:01:35.310 --> 00:01:40.140
But now, let’s imagine that we add a force to the forces acting on this box.
00:01:40.550 --> 00:01:47.630
When the forces on an object don’t balance out, there is a net force which causes the object’s motion to change.
00:01:47.890 --> 00:01:50.400
Specifically, the object accelerates.
00:01:50.940 --> 00:01:58.770
As Isaac Newton was looking into these quantities of net force and acceleration of an object, he found that they’re related.
00:01:59.350 --> 00:02:06.230
He saw that as net force goes up, by that same factor acceleration would go up and vice versa.
00:02:06.850 --> 00:02:13.070
There’s a third quantity we can connect to net force and acceleration, and that’s an object’s mass.
00:02:13.420 --> 00:02:19.660
Let’s consider a thought experiment where we have a very small mass and a very large mass sitting on the same surface.
00:02:20.060 --> 00:02:25.910
Let’s say that we apply a force 𝐹 sub 𝑝 horizontally to this very small mass.
00:02:26.240 --> 00:02:32.860
With the mass being so very small, even for a moderately sized force, we expect it to accelerate rapidly.
00:02:33.040 --> 00:02:36.420
It will have a large acceleration for the small mass.
00:02:36.850 --> 00:02:43.710
Now, let’s take the very same force 𝐹 sub 𝑝 and apply it to the very large mass capital 𝑀.
00:02:44.120 --> 00:02:52.850
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.
00:02:52.850 --> 00:02:54.860
Its acceleration will be quite small.
00:02:55.440 --> 00:03:05.420
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.
00:03:06.080 --> 00:03:13.680
And for the same force magnitude, if our mass is quite large, we expect our acceleration to be quite small.
00:03:14.270 --> 00:03:20.270
Overall, then, we might say that mass and acceleration are inversely proportional.
00:03:20.710 --> 00:03:25.550
That is, when one gets large, the other gets smaller, for a given force.
00:03:26.220 --> 00:03:44.540
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.
00:03:45.090 --> 00:03:48.510
This is the form of Newton’s second law of motion.
00:03:48.900 --> 00:03:51.290
We often see it in a slightly different arrangement.
00:03:51.800 --> 00:03:55.930
If we multiply both sides by the mass 𝑚, we see this form.
00:03:55.970 --> 00:04:00.420
The net force is equal to an object’s mass multiplied by its acceleration.
00:04:00.870 --> 00:04:04.610
Notice that this equation doesn’t apply for any old force.
00:04:04.880 --> 00:04:10.220
Specifically, it’s the net force acting on an object which is equal to 𝑚 times 𝑎.
00:04:10.740 --> 00:04:14.560
A second thing we might notice is that this is a vector equation.
00:04:14.790 --> 00:04:18.780
Both force and acceleration have a magnitude and direction.
00:04:19.040 --> 00:04:22.960
And the direction of acceleration equals the direction of the net force.
00:04:23.540 --> 00:04:27.570
Let’s get some practice using this second law in a couple of examples.
00:04:28.280 --> 00:04:35.830
A sprinter with a mass of 57.0 kilograms accelerates at 3.922 meters per second squared.
00:04:36.160 --> 00:04:39.740
What is the magnitude of the net external force on her?
00:04:40.360 --> 00:04:46.330
We can call this magnitude of the net external force acting on the sprinter capital 𝐹.
00:04:46.900 --> 00:04:58.580
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.
00:04:59.350 --> 00:05:08.180
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.
00:05:08.600 --> 00:05:11.440
It’s that force magnitude we want to solve for.
00:05:11.640 --> 00:05:14.730
And to do it, we’ll recall Newton’s second law of motion.
00:05:15.180 --> 00:05:21.860
Newton’s second law says that the net force acting on an object equals the object’s mass times its acceleration.
00:05:22.290 --> 00:05:27.130
In the case of our sprinter, we’re working with magnitudes rather than vectors.
00:05:27.370 --> 00:05:31.580
And we’re given values for the sprinter’s mass as well as her acceleration 𝑎.
00:05:32.160 --> 00:05:40.250
When we plug in for those two values and calculate this product, we find it’s equal to 224 newtons.
00:05:40.730 --> 00:05:44.680
That’s the net force magnitude causing the sprinter’s acceleration.
00:05:45.440 --> 00:05:52.310
Now, let’s look at an example where the applied force and the object’s resulting acceleration are not in the same direction.
00:05:53.090 --> 00:05:57.590
A cart with a mass of 18 kilograms is at rest on a level floor.
00:05:57.900 --> 00:06:04.560
A constant 15-newton force is applied to the cart at an angle 52 degrees below the horizontal.
00:06:04.990 --> 00:06:12.470
If friction is negligible, what is the speed of the cart when the force is applied over a distance 6.7 meters?
00:06:12.990 --> 00:06:19.230
In this exercise, we want to solve for the speed of the cart after the force has been applied over some distance.
00:06:19.470 --> 00:06:21.560
We’ll call that speed 𝑣 sub 𝑓.
00:06:22.010 --> 00:06:31.700
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.
00:06:32.260 --> 00:06:43.790
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.
00:06:43.980 --> 00:06:45.810
It’s 6.7 meters.
00:06:45.950 --> 00:06:47.770
We’ll label that distance 𝑑.
00:06:48.280 --> 00:06:51.870
Let’s start on our solution by drawing a diagram of the scenario.
00:06:52.360 --> 00:06:58.660
We start off in this example with our cart of mass 𝑚 at rest on a flat frictionless surface.
00:06:59.190 --> 00:07:05.620
A force 𝐹 is applied to the cart in the direction 𝜃 degrees below the horizontal.
00:07:06.170 --> 00:07:11.790
This force continues to act on the cart over a distance, 𝑑, of 6.7 meters.
00:07:12.570 --> 00:07:18.650
After being under the influence of the force over that span, we want to know what is the final speed of the cart.
00:07:19.170 --> 00:07:22.800
To find out, let’s recall Newton’s second law of motion.
00:07:23.270 --> 00:07:30.290
The second law tells us that the net force that acts on an object is equal to that object’s acceleration times its mass.
00:07:30.870 --> 00:07:39.470
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.
00:07:40.000 --> 00:07:45.700
The vertical component of that force will contribute to the normal force that the mass experiences.
00:07:46.090 --> 00:07:50.910
Only the horizontal component of 𝐹 will contribute to its acceleration to the right.
00:07:51.520 --> 00:08:02.410
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.
00:08:02.830 --> 00:08:08.210
By the second law, that’s equal to the mass of the case multiplied by its acceleration.
00:08:08.670 --> 00:08:22.900
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.
00:08:23.390 --> 00:08:30.540
Since all these terms are constant, acceleration 𝑎 is also constant over the span of the distance 𝑑.
00:08:31.090 --> 00:08:39.290
When the acceleration of an object is constant, that’s a sign that the kinematic equations apply to describe the motion of that object.
00:08:39.850 --> 00:08:50.880
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.
00:08:51.520 --> 00:08:55.510
The second equation we’ve written is a match for these conditions.
00:08:55.900 --> 00:09:08.720
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.
00:09:09.130 --> 00:09:15.760
We’re told in the problem statement that our case starts at rest, which means 𝑣 sub zero is equal to zero.
00:09:16.200 --> 00:09:24.700
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.
00:09:25.130 --> 00:09:29.830
And 𝑎, the acceleration, is something we’ve solved for symbolically earlier.
00:09:30.410 --> 00:09:41.870
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.
00:09:42.180 --> 00:09:44.730
We’re ready to plug in and solve for 𝑣 sub 𝑓.
00:09:45.220 --> 00:09:54.630
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.
00:09:55.100 --> 00:10:00.040
That’s our object’s final velocity after being accelerated over the distance 𝑑.
00:10:00.900 --> 00:10:05.900
Let’s summarize what we’ve learnt so far about forces and Newton’s second law of motion.
00:10:06.650 --> 00:10:13.840
We’ve seen that Newton’s second law of motion describes a mass’s acceleration when a net force acts on it.
00:10:14.470 --> 00:10:19.420
Mathematically, we would write that the net force is equal to mass times acceleration.
00:10:19.930 --> 00:10:27.580
We noted that this relationship doesn’t hold for any force, but only for the net force that acts on an object.
00:10:28.260 --> 00:10:37.110
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.
00:10:37.660 --> 00:10:41.480
Newton’s second law is a helpful way of understanding object motion.
00:10:41.700 --> 00:10:47.810
And by recalling these points, we’ll be that much better equipped to answer questions involving this law.