Video: Resultant Force

In this video we learn about resultant or net force acting on an object, and we calculate this force through examples.

11:02

Video Transcript

In this video, we’re going to talk about resultant force. We’ll learn what it is, how to calculate resultant force. And we’ll also get some practice using it in a few examples. To start out, imagine that you have a tree stump in your backyard. And you’d like to pull this stump out of the ground. 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. And in effort to pull up the stump with dramatic speed, you leap onto the rope and pull with all your weight. Unfortunately, the stump stays put.

You see that more weight is needed. So you call a few friends over, who clamber up the rope and pull themselves with their full weight. 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. To better understand this force, we want to learn a bit about resultant force. We can define resultant force as the single net force acting on an object that’s created by a combination of forces. Any object with more than one force acting on it will have a resultant force, that is, the net force acting on that object. Imagine, for example, a skydiver who’s falling and still speeding up. On the skydiver, there’s the weight force acting down and the frictional force, the air resistance force, acting up. The combination of these two forces yields the resultant force, 𝐹 sub 𝑅, which points down.

Or imagine another example where a team of oxen pull along a sled behind them. Each pair of oxen exerts a force forward pulling on the sled. And the resultant force on the sled is the combination or some of those forces. Calculating resultant force for an object is a two-step process. In the first step, we find out all the forces that are acting on that object. This step may involve drawing the forces in on a free body diagram. Or it may involve listing the forces out algebraically by their components. Step two is then to add the forces together as vectors. The result is the resultant force on that object. From this process, we can see that another name for resultant force is net force. Now, let’s get some practice through a couple of examples calculating resultant force.

A particle is accelerated when acted on by two forces, as shown in the accompanying diagram. The force 𝐹 sub 𝐵 has twice the magnitude of force 𝐹 sub 𝐴. 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.

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. We’ll call this direction 𝜃, where 𝜃 is an angle measured from the negative 𝑥-direction from the position of the particle. 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. If we call that resultant force 𝐹 sub 𝑅, we know that force will have two components. One we can call 𝐹 sub 𝑥 in the 𝑖-direction, that is, along the 𝑥-axis. And the second component we can call 𝐹 sub 𝑦 in the 𝑗-direction along the 𝑦-axis.

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. 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. 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.

In the exercise statement, we’re told something about the magnitude 𝐹 sub 𝐵. We’re told that it’s equal to twice the magnitude of 𝐹 sub 𝐴. 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. 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. 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.

Our next step is to add these two forces, 𝐹 sub 𝐴 and 𝐹 sub 𝐵, together. 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. We’ve now solved for the resultant or net force acting on our particle. 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. 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. 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. 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.

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. 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. Looking at this fraction, we see that the magnitudes 𝐹 sub 𝐴 cancel out. 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. Entering this expression on our calculator, we find that, to two significant figures, 𝜃 is 74 degrees. That’s the direction of the particle’s acceleration as measured from the negative 𝑥-axis with respect to the particle’s position.

Now, let’s try a second example involving resultant force.

A planet orbits a star. And the planet is orbited by a moon. 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. The force on the moon from the planet, 𝐹 sub one, equals 1.21 times 10 to the 19th newtons. And the force on the moon from the star, 𝐹 sub two, equals 5.00 times 10 to the 19th newtons. The moon’s mass is 1.13 times 10 to the 22nd kilograms. What is the magnitude of the moon’s resultant acceleration?

Let’s start on our solution by drawing a diagram of this situation. In this scenario, we have a star. And orbiting around that star is a planet. And orbiting around the planet is a moon. And we’re told that there comes a particular time in the orbit of the moon. 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. 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. And the force of the star on the moon is 𝐹 sub two. In the problem statement, we’re told the magnitude of each one of those forces, 𝐹 sub one and 𝐹 sub two. And we’re also told the mass of the moon, which we can call 𝑚 sub 𝑚.

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. We can call that acceleration 𝑎. And to get started solving for it, let’s consider again our two forces and what the resultant force on the moon is. Finding the resultant force acting on the moon due to 𝐹 one and 𝐹 two involves adding those two forces together. 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. That means that, to solve for the magnitude of the resultant force on the moon, we can use the Pythagorean theorem.

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. 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. We’re given the magnitudes of 𝐹 sub one and 𝐹 sub two and can plug those into this expression now. Before entering this value on our calculator, we can recall that we wanna solve for the moon’s acceleration magnitude 𝑎. By Newton’s second law of motion, this acceleration 𝑎 is equal to the net force acting on an object divided by its mass. So in our case, we take our resultant force magnitude, divide it by the mass of the moon whose value we’re given. And that fraction will be equal to our acceleration magnitude 𝑎. 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. That’s the moon’s acceleration magnitude under the influence of the planet and the star.

Now, let’s summarize what we’ve learnt about resultant force. 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. Net force is another expression for resultant force. When we’re calculating resultant force, we want to be sure to add forces together by their vector components. And finally, we’ve seen that resultant force is another way of saying net force. These two terms mean the same thing. When we calculate the resultant force on an object, we’re finding out which way that object will move or accelerate. As such, it’s a useful skill to have.

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