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.
