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 formulae. Friction is the resistance an
object encounters in moving over another. For example, it is easier to drag
an object over glass than sandpaper. The reason for this is that the
sandpaper exerts more frictional resistance. In many problems, it is assumed
that a surface is smooth, which means it does not exert any frictional force. In this video, however, we will
deal with a rough surface. This is one which will offer some
frictional resistance.
We know that a system is said to be
in equilibrium if the object is not moving. When the frictional force is at its
maximum possible value, friction is said to be limiting. If this friction is limiting, yet
the object is still stationary, it is said to be in limiting equilibrium. Let’s consider how we can model
this on a horizontal surface. Let’s consider an object at rest on
a rough horizontal surface. It will have a downward force equal
to its mass multiplied by gravity. There will be a normal reaction
force in the opposite direction, going vertically upwards. Let’s assume there is a horizontal
force 𝐹 trying to move the body to the right. The frictional force 𝐹 r will
therefore act horizontally to the left.
We will assume that the body is in
limiting equilibrium. This means that it is on the point
of moving but is still stationary. Newton’s second law states that the
sum of the net forces is equal to the mass multiplied by the acceleration. As the body is stationary, we know
that the acceleration is equal to zero. This means that the sum of the
vertical and horizontal forces equals zero. Resolving vertically, the forces
upwards must be equal to the forces downwards. Therefore, 𝑅 is equal to 𝑚𝑔. Resolving horizontally, the force
going to the right must be equal to the frictional force going to the left. The coefficient of friction 𝜇 is a
number which represents the friction between two surfaces. When a body is in limiting
equilibrium, the maximum frictional force is equal to the coefficient of friction
multiplied by the normal reaction force. This means that the frictional
force is equal to 𝜇𝑅.
We will now use this equation as
well as resolving vertically and horizontally to solve problems involving the
equilibrium of a body on a rough horizontal plane.
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 can begin by sketching a
diagram. We don’t know anything about the
mass of the body. Therefore, the downward force will
be equal to 𝑚 multiplied by 𝑔. There is a vertical normal reaction
force labeled 𝑁. Whilst we would often label this
𝑅, in this case, we will use a different letter as 𝑅 is the resultant of the
friction force and this normal reaction force. As the plane is rough, there will
be a frictional horizontal force. We will also have a horizontal
force in the opposite direction labeled 𝐹. We recall the formula that links
the frictional force and the normal reaction force. Friction is equal to 𝜇 multiplied
by 𝑅. In this question, however, as we
defined the normal reaction force to be 𝑁, we have 𝐹 r is equal to 𝜇 multiplied
by 𝑁.
We are told that the coefficient of
friction is equal to 0.2. Therefore, 𝐹 r is equal to 0.2
multiplied by 𝑁. We are also told that the limiting
friction force is equal to 80 newtons. Therefore, 80 is equal to
0.2𝑁. Dividing both sides of this
equation by 0.2 gives us 𝑁 is equal to 400. The normal reaction force is
therefore equal to 400 newtons. This also tells us that the body is
on the point of moving. Therefore, the forward force 𝐹
must be equal to the frictional force.
To find the resultant force 𝑅, we
will now use a force triangle. The vertical force is equal to 400
newtons, and the horizontal force is equal to 80 newtons. Using the Pythagorean theorem, we
see that 𝑅 squared is equal to 400 squared plus 80 squared. The right-hand side of this
equation is equal to 166,400. Square rooting both sides of this
equation gives us 𝑅 is equal to 80 root 26. The resultant of the force of
friction and the normal reaction force is therefore equal to 80 root 26 newtons. As a decimal answer, this is equal
to 407.92 newtons to two decimal places.
In our next question, we need to
calculate the frictional force.
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. We know that the body weighs 25.5
newtons. There will be a normal reaction
force going vertically upwards of 𝑅 newtons. We will call the horizontal force
that is applied to the body 𝐹. As the horizontal plane is rough,
there will be a frictional force 𝐹 r newtons acting in the opposite direction to
this. We are told that the body is on the
point of moving, which means it is in limiting equilibrium. This tells us that the sum of the
vertical and horizontal forces equals zero. Resolving vertically, the forces up
are equal to the forces down. This means that 𝑅 is equal to
25.5. The normal reaction force is equal
to 25.5 newtons. Resolving horizontally, the force
𝐹 is equal to the frictional force.
We are also told that the
coefficient of friction is equal to three seventeenths. In limiting equilibrium, the
frictional force is equal to the coefficient of friction multiplied by the reaction
force. We can therefore calculate the
frictional force by multiplying three seventeenths by 25.5. This is equal to 4.5 newtons. As the horizontal force acting on
the body is equal to this, we know that this is equal to 4.5 newtons.
In our final question, we will deal
with a system on a rough horizontal plane that includes smooth pulleys.
A body of weight 79 newtons rests
on a rough horizontal table. It is attached by a light
inextensible string passing over a smooth pulley fixed at the end of the table to a
weight of 41 newtons hanging freely below the pulley. Under these conditions, the system
is on the point of moving. The body is then attached by a
second inextensible string passing over a second pulley fixed at the opposite end of
the table to a second weight of 𝑊 newtons hanging freely vertically below the
pulley. Determine the weight 𝑊 which will
cause the body to be on the point of moving.
Let’s begin by modeling the initial
situation. We have a body of weight 79 newtons
resting on a rough horizontal table. There will be a normal reaction
force 𝑅 acting vertically upwards. As the table is rough, there will
be a frictional force acting away from the pulley. As the string is light and
inextensible and passes over a smooth pulley, the tension in the string will be
constant. We will call this 𝑇 one. We know that the freely hanging
body has a weight of 41 newtons. The system is on the point of
moving, which means it is in limiting equilibrium. When we resolve vertically and
horizontally, the sum of our forces will equal zero.
Resolving vertically for the body
at rest on the table, we have 𝑅 is equal to 79 newtons. The forces up are equal to the
forces down. Resolving horizontally, we see that
𝑇 one, the tension, is equal to the frictional force 𝐹 r. Resolving vertically for the freely
hanging body, we see that 𝑇 one is equal to 41 newtons. This in turns tells us that the
frictional force on the table is also 41 newtons. Whilst it is not essential for this
question, we could then use the equation that the frictional force is equal to 𝜇
multiplied by the normal reaction force to calculate the coefficient of friction
𝜇. In this question, this will be
equal to 41 over 79.
Let’s now clear some space and
concentrate on the second part of this question. We now have another body of weight
𝑊 newtons hanging freely below the other end of the table. Again, we have an inextensible
string passing over a smooth pulley. We will let the tension in this
string be equal to 𝑇 two. We need to calculate the value of
𝑊 so that the system is on the point of moving. Body A will be about to move to the
left. This means that the frictional
force will be moving to the right. Resolving vertically on body C, we
see that the tension 𝑇 two is equal to 𝑊. When resolving horizontally on body
A, we see that 𝑇 two is equal to 𝑇 one plus the frictional force 𝐹 r.
We know from the first part of the
question that both of these are equal to 41 newtons. 41 plus 41 is equal to 82. Therefore, the tension in the
second string is equal to 82 newtons. As this tension is equal to the
weight of the body, 𝑊 is equal to 82 newtons. This is the weight of body C that
will mean that the system is on the point of moving.
We will now summarize the key
points from this video. We saw in this video that a system
is in limiting equilibrium if it is on the point of moving. The limiting frictional force 𝐹 r
is equal to the coefficient of friction 𝜇 multiplied by the normal reaction force
𝑅. The greater the value of 𝜇, the
greater the frictional force. This value of 𝜇 will lie between
zero and one inclusive. When dealing with problems in
equilibrium, the sum of the vertical and horizontal forces on an object equals
zero. This means that when resolving
vertically, the forces up are equal to the forces down, and when resolving
horizontally, the forces to the right are equal to the forces to the left.
