Video Transcript
In this video, we will learn how to
solve problems involving the equilibrium of a body on a rough horizontal plane. We will begin by looking at some
key definitions and then consider how we can resolve vertically and horizontally to
calculate unknowns. These will include the normal
reaction force, the frictional force between the body and the plane, and the
coefficient of friction 𝜇. We will also consider the angle of
friction.
Let’s begin by considering a body
resting on a rough horizontal surface as shown. If the body is in equilibrium, it
has zero resultant force acting on it. Two forces act on the body:
firstly, its weight, which is equal to the mass of the body multiplied by the
acceleration due to gravity. Secondly, as a result of Newton’s
third law, we have the normal reaction force 𝐍. For a body on a horizontal surface,
this acts vertically upwards. As there is zero resultant force,
these two forces must be equal such that 𝐍 is equal to 𝑊, which we can also write
as 𝐍 is equal to 𝑚𝑔.
If the body was on a smooth
surface, any net force applied horizontally on the body would accelerate the body
horizontally. However, on a rough surface, the
body will only accelerate if the horizontal force applied to it has magnitude
greater than the frictional force between the body and the surface. In this video, we will label this
frictional force 𝐅𝑟. This frictional force acts in the
opposite direction to the applied force 𝐅.
In this video, we will consider
problems where the body is on the point of moving and, as such, the frictional force
will be at its maximum. At this stage, the body will still
be in equilibrium. And when we resolve horizontally,
the applied force 𝐅 will be equal to the frictional force 𝐅𝑟. The maximum frictional force is
also called the limiting friction. And this limiting friction 𝐅𝑟
between a body at rest and the surface it rests on satisfies the formula 𝐅𝑟 is
equal to 𝜇𝐍. And since 𝐍, the normal reaction
force, is equal to 𝑚𝑔, the frictional force 𝐅𝑟 is equal to 𝜇𝑚𝑔.
We will now look at an example
where the maximum applied force on a body that remains in equilibrium is to be
determined.
A body weighing 25.5 newtons rests
on a rough horizontal plane. A horizontal force acts on the
body, causing it to be on the point of moving. Given that the coefficient of
friction between the body and the plane is three seventeenths, determine the
magnitude of the force.
We can begin by sketching a diagram
to model the situation. We know that any body resting on a
horizontal plane will have a downward force equal to its weight. And this weight 𝑊 will be equal to
the mass of the body multiplied by the acceleration due to gravity. In this question, we are told the
body weighs 25.5 newtons. So there will be a force acting
vertically downwards equal to 25.5 newtons. Newton’s third law tells us that
there will be a normal reaction force acting in the opposite direction to this. Since the body is in equilibrium,
we know that these two forces are equal. The normal reaction force 𝐍 is
equal to 25.5 newtons.
We are told that a horizontal force
𝐅 acts on the body. And as the plane is rough, there
will be a frictional force acting between the body and the plane. The body is on the point of moving,
which means that the frictional force will be at its maximum. This is known as the limiting
friction such that 𝐅𝑟 is equal to 𝜇 multiplied by 𝐍, where 𝜇 is the coefficient
of friction between the body and the plane. In this question, we are told this
is equal to three seventeenths. The frictional force 𝐅𝑟 is
therefore equal to three seventeenths multiplied by 25.5, which is equal to 4.5. The maximum frictional force is
equal to 4.5 newtons.
Since the body is on the point of
moving and is still in equilibrium, the horizontal forces must also be equal to one
another. The applied force 𝐅 must be equal
to the frictional force 𝐅𝑟. We can therefore conclude that the
magnitude of the horizontal force acting on the body is 4.5 newtons.
In our next example, we need to
calculate a resultant force of which the limiting friction is a component.
A body is resting on a rough
horizontal plane. The coefficient of friction between
the body and the plane is 0.2, and the limiting friction force that is acting on the
body is 80 newtons. Given that 𝑅 is the resultant of
the force of friction and the normal reaction force, find the magnitude of 𝑅.
We will begin by sketching a
diagram to model the situation. If the body is in limiting
friction, it will have four forces acting on it. Firstly, acting vertically
downwards, we have the weight force, which is equal to the mass of the body
multiplied by the acceleration due to gravity. Acting vertically upwards, we have
the normal reaction force 𝐍. We will have an applied force 𝐅
that is causing the body to be on the point of moving, and finally a frictional
force 𝐅𝑟 between the body and the plane acting in the opposite direction to
this.
Whilst the weight and applied force
are not mentioned in this question, we know that for a body at rest on a surface,
the normal reaction force 𝐍 must be equal to the weight of the body. The limiting friction is the
maximum frictional force that the surface exerts on the body before the body starts
to move. And we therefore know that the
applied force 𝐅 must be equal to the frictional force 𝐅𝑟. This limiting friction is equal to
𝜇 multiplied by 𝐍, where 𝜇 is the coefficient of static friction between the
surface and the body.
Let’s now substitute the values we
are given in this question. We are told that 𝜇 is equal to
0.2. We are also told that the limiting
frictional force is equal to 80 newtons. Substituting these into 𝐅𝑟 is
equal to 𝜇𝐍, we have 80 is equal to 0.2 multiplied by 𝐍. We can then divide through by
0.2. 𝐍 is equal to 80 divided by 0.2,
which equals 400. The normal reaction force acting on
the body is 400 newtons. Whilst it is not required in this
question, we could then calculate the weight force and the applied force 𝐅. These are equal to 400 newtons and
80 newtons, respectively.
We need to calculate the magnitude
of 𝑅, where 𝑅 is the resultant of the force of friction and the normal reaction
force. This will act in a direction shown
on the diagram. We can calculate the magnitude by
creating a right triangle. And using the Pythagorean theorem,
we have 𝑅 squared is equal to 400 squared plus 80 squared. The right-hand side is equal to
166,400. We can then take the square root of
both sides of this equation. And since the resultant force must
be positive, we have 𝑅 is equal to 80 root 26. The magnitude of the resultant
force is equal to 80 root 26 newtons. As a decimal answer, this is equal
to 407.92 newtons to two decimal places.
This question leads us onto a key
fact about any body on a rough surface. A body on a rough surface has an
angle of friction. This is the angle between the
normal reaction force and the resultant of the normal reaction force and the
limiting frictional force. If we consider the following body
on a rough horizontal surface, then if the body is in limiting friction and on the
point of moving, there will be four forces acting on it in the horizontal and
vertical directions. We have the normal reaction force
𝐍, the weight force 𝑊 which is equal to 𝑚𝑔, the applied force 𝐅, and the
limiting frictional force 𝐅𝑟.
We recall that this limiting
frictional force 𝐅𝑟 is equal to 𝜇 multiplied by 𝐍. And whilst it is not required here,
it is worth recalling that when resolving vertically and horizontally, our forces
are equal. Drawing on the force 𝑅, which is
the resultant of the normal reaction force and the frictional force and not an
additional force, we see that the angle of friction 𝜃 is as shown. We can sketch a right triangle with
this resultant force 𝑅 together with the normal reaction force and the limiting
frictional force. Replacing 𝐅𝑟 with 𝜇𝐍, we can
use the tangent ratio to calculate 𝜃.
We know that in any right triangle,
the tan of angle 𝜃 is equal to the opposite over the adjacent. In our diagram, tan 𝜃 is equal to
𝜇𝐍 over 𝐍. And dividing through by 𝐍, we have
tan 𝜃 is equal to 𝜇. The angle of friction for a body on
a rough horizontal surface can therefore be determined by tan 𝜃 equals 𝜇.
Let’s now look at a specific
example where we need to calculate the angle of friction.
Given that the coefficient of
friction between a body and a plane is root three over four, what is the angle of
friction? Round your answer to the nearest
minute if necessary.
We will begin by recalling what we
mean by the angle of friction. The angle of friction is the angle
between the normal reaction force on a body and the resultant of the normal reaction
force and the limiting frictional force on the body. Since the limiting frictional force
𝐅𝑟 is equal to 𝜇𝐍, and using our knowledge of right angle trigonometry, we know
that the angle of friction 𝜃 satisfies the equation tan 𝜃 is equal to 𝜇𝐍 over
𝐍, which is equal to 𝜇, where 𝜇 is the coefficient of friction and in this
question is equal to root three over four. The tan of angle 𝜃 is therefore
equal to root three over four.
Taking the inverse tangent of both
sides of this equation gives us 𝜃 is equal to the inverse tan or arctan of root
three over four. Ensuring that our calculator is in
degree mode, typing in the right-hand side gives us 23.4132 and so on. We are asked to give our answer to
the nearest minute. We can do this by using the
degrees, minutes, and seconds button on the calculator or by multiplying the decimal
part of our answer by 60. This is equal to 24.79 and so on
minutes, giving us an answer to the nearest minute of 23 degrees and 25 minutes. This is the angle of friction when
the coefficient of friction between a body and a plane is root three over four.
We will now summarize the key
points from this video. We saw in this video that the
normal reaction force for a body on a horizontal surface has a magnitude equal to
the weight of the body. The frictional force between a body
and a rough surface acts in the opposite direction to the net force on the body and
parallel to the plane. The maximum frictional force
between a body and a rough surface is given by 𝐅𝑟 is equal to 𝜇𝐍, where 𝐍 is
the normal reaction force on the body and 𝜇 is the coefficient of static friction
between the body and the surface. Finally, we saw that the angle of
friction for a body on a rough horizontal surface is given by 𝜃, which is equal to
the inverse tan of 𝜇, where once again 𝜇 is the coefficient of static friction
between the body and the surface.
