WEBVTT
00:00:01.440 --> 00:00:04.740
In this video, we’re going to talk about resultant force.
00:00:05.000 --> 00:00:08.530
We’ll learn what it is, how to calculate resultant force.
00:00:08.670 --> 00:00:11.870
And we’ll also get some practice using it in a few examples.
00:00:12.410 --> 00:00:16.310
To start out, imagine that you have a tree stump in your backyard.
00:00:16.440 --> 00:00:18.520
And you’d like to pull this stump out of the ground.
00:00:19.070 --> 00:00:30.240
Being a mechanically minded person, you decide to rig up a system of pulleys over which you put a rope attached to the stump that you’ll pull the other end of to pull the stump up.
00:00:30.860 --> 00:00:37.320
And in effort to pull up the stump with dramatic speed, you leap onto the rope and pull with all your weight.
00:00:37.900 --> 00:00:40.110
Unfortunately, the stump stays put.
00:00:40.780 --> 00:00:42.670
You see that more weight is needed.
00:00:42.700 --> 00:00:47.970
So you call a few friends over, who clamber up the rope and pull themselves with their full weight.
00:00:48.560 --> 00:00:56.040
With all four people hanging on the rope at the same time, you’re curious to know what is the total force acting to pull up the stump.
00:00:56.690 --> 00:01:01.440
To better understand this force, we want to learn a bit about resultant force.
00:01:02.150 --> 00:01:09.990
We can define resultant force as the single net force acting on an object that’s created by a combination of forces.
00:01:10.470 --> 00:01:18.180
Any object with more than one force acting on it will have a resultant force, that is, the net force acting on that object.
00:01:18.810 --> 00:01:23.130
Imagine, for example, a skydiver who’s falling and still speeding up.
00:01:23.650 --> 00:01:30.340
On the skydiver, there’s the weight force acting down and the frictional force, the air resistance force, acting up.
00:01:30.860 --> 00:01:36.580
The combination of these two forces yields the resultant force, 𝐹 sub 𝑅, which points down.
00:01:37.240 --> 00:01:42.320
Or imagine another example where a team of oxen pull along a sled behind them.
00:01:42.790 --> 00:01:47.360
Each pair of oxen exerts a force forward pulling on the sled.
00:01:47.900 --> 00:01:52.880
And the resultant force on the sled is the combination or some of those forces.
00:01:53.700 --> 00:01:58.120
Calculating resultant force for an object is a two-step process.
00:01:58.540 --> 00:02:02.830
In the first step, we find out all the forces that are acting on that object.
00:02:03.470 --> 00:02:07.570
This step may involve drawing the forces in on a free body diagram.
00:02:07.890 --> 00:02:12.200
Or it may involve listing the forces out algebraically by their components.
00:02:12.830 --> 00:02:16.570
Step two is then to add the forces together as vectors.
00:02:16.910 --> 00:02:20.160
The result is the resultant force on that object.
00:02:20.720 --> 00:02:25.350
From this process, we can see that another name for resultant force is net force.
00:02:25.990 --> 00:02:30.300
Now, let’s get some practice through a couple of examples calculating resultant force.
00:02:31.420 --> 00:02:37.060
A particle is accelerated when acted on by two forces, as shown in the accompanying diagram.
00:02:37.460 --> 00:02:41.930
The force 𝐹 sub 𝐵 has twice the magnitude of force 𝐹 sub 𝐴.
00:02:42.360 --> 00:02:50.670
Find the direction in which the particle’s net acceleration occurs in terms of the angle below the negative 𝑥-direction from the position of the particle.
00:02:51.420 --> 00:03:01.820
So given the two forces acting on the particle, 𝐹 sub 𝐵 and 𝐹 sub 𝐴, we want to solve for a direction, the direction in which the particle’s net acceleration occurs.
00:03:02.130 --> 00:03:09.860
We’ll call this direction 𝜃, where 𝜃 is an angle measured from the negative 𝑥-direction from the position of the particle.
00:03:10.440 --> 00:03:17.610
In order to solve for the direction of the particle’s acceleration, we’ll want to solve for the resultant force acting on the mass 𝑚.
00:03:18.100 --> 00:03:23.500
If we call that resultant force 𝐹 sub 𝑅, we know that force will have two components.
00:03:23.740 --> 00:03:28.930
One we can call 𝐹 sub 𝑥 in the 𝑖-direction, that is, along the 𝑥-axis.
00:03:29.300 --> 00:03:34.820
And the second component we can call 𝐹 sub 𝑦 in the 𝑗-direction along the 𝑦-axis.
00:03:35.330 --> 00:03:43.600
To solve for 𝐹 sub 𝑅, the resultant force, we’ll add together 𝐹 sub 𝐵 and 𝐹 sub 𝐴, the two forces acting on the mass 𝑚.
00:03:44.120 --> 00:03:52.510
We can start by writing the force vector 𝐹 sub 𝐴 as the magnitude of that force multiplied by its direction in the 𝑖-direction.
00:03:53.030 --> 00:04:12.100
Likewise, we can write the force vector 𝐹 sub 𝐵 as negative the magnitude of that force multiplied by the cos of 45 degrees — that’s the 𝑖-component of 𝐹 sub 𝐵 — minus the magnitude 𝐹 sub 𝐵 times the sin of 45 degrees — that’s the 𝑗-component of that force.
00:04:12.670 --> 00:04:16.850
In the exercise statement, we’re told something about the magnitude 𝐹 sub 𝐵.
00:04:17.400 --> 00:04:21.100
We’re told that it’s equal to twice the magnitude of 𝐹 sub 𝐴.
00:04:21.600 --> 00:04:30.440
We substitute in two 𝐹 sub 𝐴 for 𝐹 sub 𝐵 and factor it out along with the minus sign from this expression for the force vector 𝐹 sub 𝐵.
00:04:31.100 --> 00:04:41.430
We can further simplify this expression for 𝐹 sub 𝐵 by realizing that the cos of 45 degrees and the sin of 45 degrees are both the square root of two over two.
00:04:42.020 --> 00:04:53.120
And we see that now the factors of two cancel out, so that 𝐹 sub 𝐵, the vector, is equal to negative root two the magnitude of 𝐹 sub 𝐴 in the 𝑖- and the 𝑗-direction.
00:04:53.800 --> 00:04:58.340
Our next step is to add these two forces, 𝐹 sub 𝐴 and 𝐹 sub 𝐵, together.
00:04:58.780 --> 00:05:09.410
We find that their sum, which is the resultant force 𝐹 sub 𝑅, is equal to the magnitude of 𝐹 sub 𝐴 times one minus the square root of two 𝑖 minus the square root of two 𝑗.
00:05:09.980 --> 00:05:14.230
We’ve now solved for the resultant or net force acting on our particle.
00:05:14.540 --> 00:05:26.180
And if we were to sketch in approximately the direction of this force on our diagram, it might point in a direction like this, slightly farther towards the negative 𝑦-direction and 𝐹 sub 𝐵.
00:05:26.710 --> 00:05:38.650
The particular direction this force and therefore the particle’s acceleration is directed is given by the angle 𝜃 measured from the negative 𝑥-axis with respect to the particle’s position.
00:05:39.220 --> 00:05:50.290
If we were to draw in the 𝑥- and 𝑦-components of that resultant force, 𝐹 sub 𝑅, on our diagram, we can see that the angle 𝜃 is part of that right triangle.
00:05:50.880 --> 00:05:59.960
And we can see that the tangent of that angle 𝜃 is equal to the 𝑦-component of the resultant force 𝐹 sub 𝑦 divided by the 𝑥-component 𝐹 sub 𝑥.
00:06:00.470 --> 00:06:09.940
Looking at the resultant force we solved for earlier, we see that the 𝑦-component of that force is the magnitude 𝐹 sub 𝐴 times negative root two.
00:06:10.750 --> 00:06:18.260
And we see further that the 𝑥-component of that force is the magnitude of 𝐹 sub 𝐴 multiplied by one minus the square root of two.
00:06:18.750 --> 00:06:23.580
Looking at this fraction, we see that the magnitudes 𝐹 sub 𝐴 cancel out.
00:06:24.070 --> 00:06:35.420
And if we then take the inverse tangent of both sides of the equation, we find that 𝜃 is equal to the inverse tangent of negative the square root of two divided by one minus the square root of two.
00:06:35.970 --> 00:06:42.550
Entering this expression on our calculator, we find that, to two significant figures, 𝜃 is 74 degrees.
00:06:42.890 --> 00:06:50.320
That’s the direction of the particle’s acceleration as measured from the negative 𝑥-axis with respect to the particle’s position.
00:06:52.120 --> 00:06:55.830
Now, let’s try a second example involving resultant force.
00:06:57.010 --> 00:06:58.760
A planet orbits a star.
00:06:58.790 --> 00:07:00.690
And the planet is orbited by a moon.
00:07:01.010 --> 00:07:11.140
At a particular time in the moon’s orbit around the planet, the gravitational force on the moon from the planet acts perpendicularly to the gravitational force on the moon from the star.
00:07:11.570 --> 00:07:17.830
The force on the moon from the planet, 𝐹 sub one, equals 1.21 times 10 to the 19th newtons.
00:07:18.110 --> 00:07:24.680
And the force on the moon from the star, 𝐹 sub two, equals 5.00 times 10 to the 19th newtons.
00:07:25.150 --> 00:07:29.590
The moon’s mass is 1.13 times 10 to the 22nd kilograms.
00:07:29.950 --> 00:07:33.070
What is the magnitude of the moon’s resultant acceleration?
00:07:33.740 --> 00:07:37.470
Let’s start on our solution by drawing a diagram of this situation.
00:07:37.900 --> 00:07:40.030
In this scenario, we have a star.
00:07:40.480 --> 00:07:43.030
And orbiting around that star is a planet.
00:07:43.540 --> 00:07:46.570
And orbiting around the planet is a moon.
00:07:47.290 --> 00:07:51.110
And we’re told that there comes a particular time in the orbit of the moon.
00:07:51.490 --> 00:08:04.050
We’re told that there’s a particular time in the moon’s orbit when the force on the moon from the star and the force on the moon from the planet are at right angles or perpendicular to one another.
00:08:04.640 --> 00:08:14.450
If we draw out an expanded sketch of the forces acting on the moon, we can say that the force of the planet acting on the moon is 𝐹 sub one.
00:08:14.970 --> 00:08:18.940
And the force of the star on the moon is 𝐹 sub two.
00:08:19.640 --> 00:08:25.160
In the problem statement, we’re told the magnitude of each one of those forces, 𝐹 sub one and 𝐹 sub two.
00:08:25.400 --> 00:08:29.080
And we’re also told the mass of the moon, which we can call 𝑚 sub 𝑚.
00:08:29.510 --> 00:08:39.670
Given all this information, we want to solve for the magnitude of the acceleration the moon will experience when 𝐹 sub one and 𝐹 sub two are perpendicular to one another.
00:08:40.160 --> 00:08:42.320
We can call that acceleration 𝑎.
00:08:42.600 --> 00:08:50.150
And to get started solving for it, let’s consider again our two forces and what the resultant force on the moon is.
00:08:50.510 --> 00:08:56.950
Finding the resultant force acting on the moon due to 𝐹 one and 𝐹 two involves adding those two forces together.
00:08:57.490 --> 00:09:08.200
Adding these vectors graphically, we see that the resultant force, 𝐹 sub 𝑅, is the hypotenuse of a right triangle whose sides are 𝐹 sub one and 𝐹 sub two in magnitude.
00:09:08.660 --> 00:09:14.460
That means that, to solve for the magnitude of the resultant force on the moon, we can use the Pythagorean theorem.
00:09:14.830 --> 00:09:23.030
This theorem tells us that the magnitude of 𝐹 sub 𝑅 squared equals the magnitude of 𝐹 sub one squared plus the magnitude of 𝐹 sub two squared.
00:09:23.450 --> 00:09:30.250
And if we take the square root of both sides of this equation, we see we now have an expression for the magnitude of 𝐹 sub 𝑅 by itself.
00:09:30.940 --> 00:09:36.130
We’re given the magnitudes of 𝐹 sub one and 𝐹 sub two and can plug those into this expression now.
00:09:36.510 --> 00:09:43.290
Before entering this value on our calculator, we can recall that we wanna solve for the moon’s acceleration magnitude 𝑎.
00:09:43.720 --> 00:09:52.010
By Newton’s second law of motion, this acceleration 𝑎 is equal to the net force acting on an object divided by its mass.
00:09:52.390 --> 00:09:59.330
So in our case, we take our resultant force magnitude, divide it by the mass of the moon whose value we’re given.
00:09:59.620 --> 00:10:03.330
And that fraction will be equal to our acceleration magnitude 𝑎.
00:10:03.840 --> 00:10:14.800
When we enter this entire expression on our calculator, we find that 𝑎, to three significant figures, is 4.55 times 10 to the negative third meters per second squared.
00:10:15.150 --> 00:10:20.090
That’s the moon’s acceleration magnitude under the influence of the planet and the star.
00:10:21.180 --> 00:10:24.480
Now, let’s summarize what we’ve learnt about resultant force.
00:10:25.420 --> 00:10:33.870
We’ve seen that the resultant force on an object is the single net force created by the combination of forces acting on the object.
00:10:34.240 --> 00:10:37.470
Net force is another expression for resultant force.
00:10:38.220 --> 00:10:44.340
When we’re calculating resultant force, we want to be sure to add forces together by their vector components.
00:10:44.990 --> 00:10:49.920
And finally, we’ve seen that resultant force is another way of saying net force.
00:10:50.240 --> 00:10:52.290
These two terms mean the same thing.
00:10:53.060 --> 00:10:59.530
When we calculate the resultant force on an object, we’re finding out which way that object will move or accelerate.
00:10:59.860 --> 00:11:02.150
As such, it’s a useful skill to have.