Video Transcript
In this video, we’re talking about
the motion of a body on a rough plane. We’re speaking then about objects
moving across surfaces, both inclined and level where friction is involved. To analyze and understand this
motion, we’ll be using coefficients of friction, free-body diagrams, and equations
of motion, including Newton’s second law.
We can start off by imagining an
object at rest on a rough surface, and we’ll see that this object has a mass 𝑚. If we draw the forces currently
acting on this body, we see that there are two. First, there’s the weight force
acting straight downward with a magnitude of 𝑚 times 𝑔, where 𝑔 is the
acceleration due to gravity. And there’s also an equal and
opposite normal or reaction force. Now, at this point, we can recall
Newton’s second law of motion, which tells us that the net force acting on an object
is equal to that object’s mass times its acceleration.
For our object here, the forces
acting on it cancel one another out, and so the net force is zero. And therefore, according to this
law, it has no acceleration. But now imagine that we apply a
force on this box pushing it from left to right. If the surface that our object was
at rest on was smooth, that is, if there was no friction between the surface and the
body, then 𝐹 sub 𝐴 would give a net force in the horizontal direction and our
object would start to accelerate. But in this case, because there is
friction between the two surfaces, an opposing frictional force is generated.
The interesting thing about this
frictional force is that it exists purely in response to the applied force. If there were no applied force,
there would be no force of friction. And even more interesting, as our
applied force increases in magnitude, say, for example, we were to push that much
harder on the box, then, up to a certain point, our friction force will increase to
match it. We say up to a certain point
because there’s a certain maximum value the force of friction can attain. That maximum value is given by this
equation. The Greek letter 𝜇 represents what
is called the coefficient of friction. And as we saw, 𝐹 sub 𝑁 is the
normal or reaction force. When it comes to the coefficient of
friction, this is a unit list number that is typically but not always between zero
and one.
There are two factors that
determine the value of 𝜇 in a given scenario. The first is what specific
materials are involved in the interaction, in other words, what material is our body
made of and what material is our surface made of. The second factor affecting 𝜇 is
whether our object is stationary or in motion. If our object is not moving, then
we use what’s called the coefficient of static friction, a 𝜇 with an 𝑠
subscript.
On the other hand, say that we
increase the applied force enough that our box does start moving despite the
opposition due to friction. In that case, there’s still a
frictional force opposing the box’s motion. But it’s of a different magnitude
because now we’re using what’s called the coefficient of kinetic friction, 𝜇 sub
𝑘. In general, these two coefficients
are not the same. Typically, the coefficient of
kinetic friction is less than that of static friction. We use one or the other in our
equation for the frictional force, depending on how our object is moving. One other point about our equation
for the friction force is that it’s important that we use the normal force here
rather than the weight force. This is especially true when our
body is on an inclined rough surface, as we’ll see later on in an example.
To get a sense for how the friction
force can change over time, let’s go back to our original condition where we only
had vertically acting forces on our body. If we were to make a plot of the
force of friction acting on our body against the applied force acting on it, at this
moment, we know that the applied force is zero and so is the force of friction. Then, say that we gradually
increase the applied force from zero up to one unit where we’ll leave the units
unspecified. And say further that as we do this,
we don’t exceed the maximum static frictional force that can act on our box. This tells us that in response to
this applied force, our frictional force kicks in, so to speak, and matches it.
All throughout this time then, our
object is still not in motion. It still has a zero net force
acting on it. Say then that we continue to
increase our applied force from one to two units. And that as we do this, our
frictional force is still able to keep up, so our box remains stationary. In that case, our graph would look
like this where the magnitude of frictional force is always equal to the magnitude
of the applied force. So far, the coefficient of friction
we would use in our friction equation is 𝜇 sub 𝑠, the coefficient of static
friction because our box is not in motion. But as we know from practical
experience, if we continue to push harder on the box, eventually it will start to
move. When that happens, we’ve overcome
the maximum force of static friction.
We might imagine that our graph
then would look like this where the force of friction remains at this level while
our applied force increases. Actually, though, it typically has
more of this shape. The maximum frictional force
achievable is 𝜇 sub 𝑠, the coefficient of static friction, times the normal force,
whereas the flat part of this curve represents 𝜇 sub 𝑘 times 𝐹 sub 𝑁. This is confirmation that indeed 𝜇
sub 𝑘 is typically less than 𝜇 sub 𝑠. Considering this vertical dashed
line on our graph, we can say that, to the left of this line, our box is stationary,
whereas to the right our box is in motion.
For each condition of motion, we’ll
want to be careful to use the correct coefficient of friction. These values, by the way, are
typically either given to us in a problem statement or we can look them up in a
table. Whether or not an object is moving,
Newton’s second law of motion can help us analyze the forces acting on it. When a body is in motion, we then
may be able to apply other equations of motion, sometimes called the kinematic
equations and sometimes called SUVAT equations. These names both refer to the same
thing, a set of four equations that help us describe the motion of objects in
motion. These equations involve initial and
final velocities, accelerations and displacements, and times.
Knowing all this about how objects
move when friction is involved, let’s get a bit of practice through an example.
Using the information in the
figure, calculate the coefficient of kinetic friction, rounding the result to the
nearest two decimal places. Given that the mass of the body is
28 kilograms and that the acceleration due to gravity, 𝑔, equals 9.8 meters per
second squared.
Taking a look at our figure, we see
our body which is on some surface, moving along with an acceleration to the right of
2.2 meters per second squared. This acceleration is due to a force
of 155 newtons that’s being applied at an angle of 45 degrees to the horizontal. Knowing this, along with the fact
that the mass of our body — we’ll call it 𝑚 — is 28 kilograms, we want to solve for
the coefficient of kinetic friction between the body and the surface.
We can begin solving for this
coefficient by first sketching in all the forces that are acting on our body. This is called a free-body
diagram. For one thing, we know that our
body is subject to the force of gravity. We also know that there’s an
applied force, we’ll call it 𝐹 sub 𝐴, which is given as 155 newtons, acting at a
45-degree angle on the body. Furthermore, there’s a normal or
reaction force acting straight up from the surface. And lastly, there’s a frictional
force; we’ll call it 𝐹 sub 𝑓. And this opposes the motion of our
object. That’s always true of friction,
which is how we knew that it acted to the left.
Now that we know all the forces
acting on our object, we can recall Newton’s second law of motion. This tells us that the net force
that acts on somebody is equal to that body’s mass times its acceleration. Now because motion in the vertical
direction is independent of motion in the horizontal direction, we can apply
Newton’s second law to either one independently. In other words, we could say that
the net force in the horizontal plane acting on our body is equal to the object’s
mass multiplied by its horizontal acceleration. And indeed, this is the dimension
we’ll focus on because we want to solve for the coefficient of kinetic friction,
which is bound up in the force of friction.
When we say that, we just mean that
the force of friction acting on an object that’s in motion is equal to the
coefficient of kinetic friction multiplied by the normal force acting on the
object. So, here’s what we’ll do. Focusing first on this horizontal
direction, let’s clear a bit of space on screen and apply Newton’s second law in
this horizontal plane. The first thing we’ll do is
establish sign conventions, positive and negative.
Note that the acceleration given to
us in our problem statement is a positive value and that acceleration is to the
right. We’ll say then that any force or
motion in that direction is positive, and any force or motion in the other direction
is negative. As we study the forces acting
horizontally in our free-body diagram, we see that there’s a component of the
applied force that acts horizontally. And then there’s the friction force
which acts entirely in this direction.
Given a 45-degree right triangle
where the hypotenuse is our applied force magnitude, 155 newtons, we can solve for
this horizontal leg of that triangle by multiplying 𝐹 sub 𝐴 by the cos of 45
degrees. That then accounts for the positive
horizontal force acting on our body. We then subtract from that the
frictional force. That force recall is equal to the
coefficient of kinetic friction, what we want to solve for, times the normal
force. This accounts for all of the
horizontal forces acting on our mass. And therefore, by Newton’s second
law, this sum is equal to mass times acceleration in the horizontal direction.
We’re given 𝐹 sub 𝐴 as well as
the mass and acceleration of our body, but we don’t know 𝐹 sub 𝑁 to let us solve
for 𝜇 sub 𝑘. To solve for it, we’re actually
going to need to apply Newton’s second law again, but this time in the vertical
dimension. As we do this, we can set up the
conventions that motion up is in the positive direction and motion down is in what
we’ll call the negative direction. So, let’s consider the vertically
acting forces on our body. First, there’s the normal force, 𝐹
sub 𝑁. Added to that is the vertical
component of our applied force, 𝐹 sub 𝐴.
Returning to our triangle, that
side length is equal to 𝐹 sub 𝐴 times the sin of 45 degrees. And then lastly in the vertical
direction, we have the weight force, 𝑚 times 𝑔, acting downward. By our convention, this force is
negative. And by Newton’s second law, all of
this is equal to the mass of our body times its acceleration in the vertical
direction, what we’ll call 𝑎 sub 𝑣. We then realize that our object
isn’t accelerating vertically, so this value is zero. That means these forces add up to
zero. So, if we subtract this term from
both sides of the equation and add this term to both sides, then we find an
expression for the normal force, which is in terms of the weight force minus the
vertical component of the applied force.
We can then substitute this whole
right-hand side in for 𝐹 sub 𝑁 here. And now, we have an expression we
can use to solve for 𝜇 sub 𝑘, the coefficient of kinetic friction. This is because we know the
numerical values of 𝐹 sub 𝐴, 𝑚, 𝑔, and 𝑎, as well as the sin and cos of 45
degrees. Before we plug in those values
though, let’s rearrange this expression so that 𝜇 sub 𝑘 is the subject. First, we’ll subtract 𝐹 sub 𝐴,
cos of 45 degrees, from both sides. Then, we’ll divide both sides of
the equation by this expression in parentheses. And lastly, we’ll multiply both
sides by negative one. We end up with this expression. And now, we’re ready to substitute
in our values.
𝐹 sub 𝐴 is 155 newtons. Both the cos and sin of 45 degrees
is the square root of two over two. Our mass, 𝑚, is 28 kilograms. The acceleration, 𝑎, is 2.2 meters
per second squared. And lastly, the acceleration due to
gravity, 𝑔, is 9.8 meters per second squared. When we enter this expression on
our calculator, to two decimal places, we find the result of 0.29. This is the coefficient of kinetic
friction between our body and the surface.
Now, let’s look at an exercise
where our body is on an inclined rough surface.
A body of weight 24 newtons is
placed on a rough plane inclined at 30 degrees to the horizontal. The coefficient of static friction
between the body and the plane is one over six root three. And the coefficient of kinetic
friction is one over 12 root three. A force is pushing the body upward
in the direction of the line of greatest slope. Determine the magnitude of the
force that would make the body about to move.
Sketching in this scenario, it
could look like this where our body is at rest on an inclined plane of 30
degrees. Situated like this, our body is
subject to a static frictional force, which depends on the coefficient of static
friction given as one over six root three. We’re told that there’s a force,
we’ll call it 𝐹, that acts on the body in the direction of the line of greatest
slope. And we want to figure out how big
that force could be, such that the body would be about to move.
Let’s first give ourselves some
space on screen to work. And then let’s include in our
sketch all of the forces that are acting on our body. We’re told that there’s a weight
force of magnitude 24 newtons. And we also know that there’s a
normal force perpendicular to the surface of our slope. Along with all this, there is a
static frictional force. Now, if we had no applied force,
what we’ve called 𝐹, pushing our body up the incline, then our static frictional
force would point in that same direction. This would be so it would resist
the block sliding down the plane. But in our scenario, we want to
solve for the maximum possible value of 𝐹 such that the block is just barely not
moving up the incline. That means that, here, the static
frictional force is actually acting down the incline.
As we begin analyzing these forces
acting on our body, we can orient ourselves with a pair of orthogonal axes. We’ll say that forces in motion
acting up the incline are positive, and those acting the opposite way are
negative. Likewise, forces in motion
perpendicularly away from our surface are positive, and those in the other direction
are negative. Knowing this, we can apply Newton’s
second law of motion to both of what we’ll call our 𝑦-direction and our
𝑥-direction.
Looking first in the 𝑥-direction,
we see that our force 𝐹 acts in the positive 𝑥, while our frictional force acts
opposite this. These, however, aren’t the only
𝑥-direction forces. That’s because our weight force has
a component right here in the negative 𝑥-direction. Now, this angle right here in this
triangle is equal to 30 degrees. And since this intersection is a
right angle, we can see that this 𝑥-component of our weight force is 24 newtons
times the sin of 30 degrees. The sin of 30 degrees we can recall
is one-half, so we’ll substitute that in right away.
So these then are indeed all the
forces acting on what we’ve called our 𝑥-direction. By Newton’s second law, these add
up to our object’s mass times acceleration in this dimension. But remember that we’re solving for
𝐹 such that our object is not yet in motion. In other words, 𝑎 sub 𝑥 is
zero. That means our whole right side is
zero, which means we can rearrange and see that the maximum force we can apply to
our body without it yet being put in motion is equal to the static frictional force
plus 12 newtons. In general, the force of static
friction is equal to the coefficient of static friction times the normal force
acting on the relevant body.
In our scenario, we know 𝜇 sub
𝑠. But what about 𝐹 sub 𝑁? To figure that out, we’ll want to
look at Newton’s second law of motion applied in the 𝑦-direction. Here, we have a positive, normal
force but then a negative component of our weight force here on our sketch. This component is 24 newtons times
the cos of 30 degrees where the cos of that angle equals the square root of three
over two. Just like in the 𝑥-direction,
because there’s no acceleration in the 𝑦-direction, these forces add up to
zero. This tells us that the normal force
equals 12 times the square root of three newtons.
Now that we know 𝐹 sub 𝑁, we can
substitute it as well as 𝜇 sub 𝑠 into our equation for the frictional force. And multiplying these quantities
together, we see that a square root of three cancels out and that the maximum
frictional force, therefore, is two newtons. 𝐹 then is 14 newtons. This is the magnitude of the
largest force that could act directly up the incline without our body being put in
motion.
Let’s review now some key points
from what we’ve learned. In this lesson, we saw that a body
resting on a rough surface can experience a static friction force, while a body in
motion across a rough surface experiences a kinetic friction force. Along with this, we’ve learned
that, mathematically, the friction force is equal to the coefficient of friction
times the normal force acting on the relevant body. The coefficient of friction may be
represented 𝜇 sub 𝑠 if the body is stationary or 𝜇 sub 𝑘 if it’s in motion. Lastly, we saw that Newton’s second
law of motion and free-body diagrams are useful tools for understanding motion
involving friction.
