### Video Transcript

In this video, we will learn how to
solve problems on the equilibrium of a body on a rough inclined plane. We will begin by recalling some key
definitions.

In many problems in mathematics, it
is assumed that the surface is smooth. This means there will be no
frictional force, whereas when dealing with a rough surface, there will be some
frictional resistance. When the frictional force is at its
maximum, friction is said to be limiting. If the body at this point is still
stationary, it is said to be in limiting equilibrium. At this point, the sum of the
forces parallel to the plane and those perpendicular to the plane are equal to
zero.

The coefficient of friction,
denoted by the Greek letter π, can be calculated by dividing the frictional force
by the normal reaction force. This means that the frictional
force πΉ r is equal to π multiplied by the normal reaction force π
. The normal reaction force is
sometimes written as π or πΉ sub n. The value of the coefficient of
friction π lies between zero and one inclusive. We will now model what this will
look like in a free body diagram and resolve parallel and perpendicular to the
plane.

Letβs consider a body resting on an
inclined plane as shown in the diagram. The angle of inclination is πΌ. If the body is of mass π
kilograms, there will be a vertical downward force equal to its weight. This will be the mass multiplied by
gravity. In this video, we will assume that
π is equal to 9.8 meters per second squared. The normal reaction force π
will
be perpendicular to the plane as shown. If the body is in limiting
equilibrium and is on the point of sliding down the plane, there will be a
frictional force πΉ r parallel to the plane acting in an upward direction. When a body is in limiting
equilibrium, the sum of the net forces is equal to zero. We can therefore resolve parallel
and perpendicular to the plane in the knowledge that the sum of the forces will
equal zero. We may sometimes see the shorthand
notation for parallel and perpendicular.

As the weight force is acting
vertically downwards, we firstly need to find the components of this parallel and
perpendicular to the plane. We do this using our knowledge of
right angle trigonometry. We know that the sin of angle π is
equal to the opposite over the hypotenuse and the cos of angle π is equal to the
adjacent over the hypotenuse. We have labeled the side opposite
angle πΌ π₯, and the side adjacent is π¦. Therefore, sin πΌ is equal to π₯
over ππ and cos of πΌ is equal to π¦ over ππ. We can then multiply both sides of
both of these equations by ππ. This means that the component of
the weight force parallel to the plane is equal to ππ sin πΌ and that
perpendicular to the plane is equal to ππ cos πΌ.

Resolving parallel to the plane
with downwards being the positive direction gives us ππ sin πΌ minus πΉ r is equal
to zero. Resolving perpendicular to the
plane, the sum of the net forces is equal to π
minus ππ cos πΌ. This is also equal to zero. We also know that in limiting
equilibrium, the frictional force πΉ r is equal to π multiplied by the normal
reaction force. We will now look at some questions
where we need to resolve parallel and perpendicular to the plane as well as using
the equation πΉ r is equal to ππ
to calculate unknowns.

A body weighing 60 newtons rests on
a rough plane inclined to the horizontal at an angle whose sin is three-fifths. The body is pulled upward by a
force of 63 newtons acting parallel to the line of greatest slope. Given that the body is on the point
of moving up the plane, find the coefficient of friction between the body and the
plane.

We will begin by sketching a
diagram to model the problem. We are told that the body weighs 60
newtons. Therefore, there will be a downward
vertical force equal to 60 newtons. The plane is inclined to the
horizontal at an angle πΌ, where sin of πΌ is equal to three-fifths. We are told that the body is pulled
upwards with a force of 63 newtons acting parallel to the line of greatest
slope. There will be a normal reaction
force π
acting perpendicular to the plane. And as the body is on the point of
moving up the plane, there will be a frictional force parallel to the plane acting
downwards.

The body is in limiting
equilibrium, so the sum of the net forces will equal zero. There is no acceleration. We will need to resolve parallel
and perpendicular to the plane. Before doing this, we will need to
find the component of the weight force acting parallel and perpendicular to the
plane. We will do this using our knowledge
of right angle trigonometry. We know that the sine of an angle
is equal to the opposite over the hypotenuse and the cosine is equal to the adjacent
over the hypotenuse. This means that the force acting
parallel to the plane is equal to 60 sin πΌ, and the force acting perpendicular to
the plane is equal to 60 cos πΌ.

By considering a three, four, five
Pythagorean triple, we know that if sin πΌ is equal to three-fifths, then cos πΌ is
equal to four-fifths. Resolving parallel to the plane, we
have 63 minus 60 sin πΌ minus the frictional force is equal to zero. Resolving perpendicular to the
plane, we have π
minus 60 cos πΌ equals zero. We can now solve these two
equations to calculate the frictional force and the normal reaction force. Adding πΉ r to both sides of
equation one and replacing sin πΌ with three-fifths, we get 63 minus 60 multiplied
by three-fifths is equal to πΉ r. The left-hand side simplifies to 63
minus 12 multiplied by three, which is equal to 27. The frictional force is equal to 27
newtons. Substituting in our value for cos
πΌ, equation two becomes π
is equal to 60 multiplied by four-fifths. This gives us π
is equal to
48. The normal reaction force is 48
newtons.

We are asked to calculate the
coefficient of friction. And we know that the frictional
force is equal to π, the coefficient of friction, multiplied by the normal reaction
force. Substituting in our values gives us
27 is equal to π multiplied by 48. Dividing both sides of this
equation by 48 gives us π is equal to 27 over 48. As both the numerator and
denominator are divisible by three, this simplifies to nine over 16 or nine
sixteenths. This is the coefficient of friction
between the body and the plane.

In our next question, we will need
to calculate the tension in a string that is attached to a body on an inclined
plane.

A body weighing 56 newtons rests on
a rough plane inclined at an angle of 30 degrees to the horizontal. The coefficient of friction between
the body and the plane is root three over six. The body is pulled upward by a
string making an angle of 30 degrees to the line of greatest slope of the plane. Determine the maximum tension in
the string required to cause the body to be on the point of moving up the plane.

We will begin by drawing a free
body diagram. The angle of the plane is 30
degrees, and the body has weight 56 newtons. Therefore, this force will act
vertically downwards. There will be a normal reaction
force π
acting perpendicular to the plane and a frictional force πΉ r acting
parallel to the plane in a downward direction, as the body is on the point of moving
up the plane. We are also told that the body is
pulled upward by a string acting at an angle of 30 degrees to the line of greatest
slope. As the body is on the point of
moving, we know that the sum of the net forces is equal to zero. There is no acceleration, and we
will need to resolve parallel and perpendicular to the plane.

Both the tension and weight force
will have components in these directions. We will need to use our knowledge
of right angle trigonometry to calculate these. As the sine of an angle is equal to
the opposite over the hypotenuse and the cosine of an angle is equal to the adjacent
over the hypotenuse, the four forces will be as shown. We have three forces acting
parallel to the plane. If we consider up the plane to be
the positive direction, we have π cos 30 minus 56 sin 30 minus πΉ r is equal to
zero. There are also three forces acting
perpendicular to the plane, giving us the equation π
plus π sin 30 minus 56 cos 30
is equal to zero.

We can now simplify these two
equations using the fact that sin of 30 degrees is one-half and cos of 30 degrees is
root three over two. Equation one becomes root three
over two π minus 28 minus πΉ r is equal to zero. And equation two becomes π
plus a
half π minus 28 root three is equal to zero. We also know that the frictional
force πΉ r is equal to π, the coefficient of friction, multiplied by the normal
reaction force π
. The coefficient of friction is root
three over six, so πΉ r is equal to root three over six multiplied by π
. We will now clear some space so we
can solve the equations to calculate the maximum tension π.

We can rearrange equation two so
that π
is equal to 28 root three minus a half π. The frictional force in equation
one is equal to root three over six multiplied by this. This gives us root three over two
π minus 28 minus root three over six multiplied by 28 root three minus a half π is
equal to zero. Distributing our parentheses gives
us negative 14 plus root three over 12 π. We can then collect like terms. This gives us seven root three over
12 π minus 42 is equal to zero. We can then add 42 to both sides of
this equation. Finally, we can divide both sides
by seven root three over 12 so that π is equal to 24 root three. The maximum tension in the string
required to cause the body to be on the point of moving up the plane is 24 root
three newtons.

In our last question, we will
calculate the maximum force required to maintain equilibrium.

A body weighing 75 newtons rests on
a rough plane inclined at an angle of 45 degrees to the horizontal under the action
of a horizontal force. The minimum horizontal force
required to maintain the body in a state of equilibrium is 45 newtons. Find the maximum horizontal force
that will also maintain the equilibrium.

In this question, we are looking at
two different scenarios: when the force is at its minimum and maximum. In both of these situations, the
body is in equilibrium. Therefore, the sum of the net
forces must equal zero. We have a vertical downward force
of 75 newtons equal to the weight. The plane is inclined at an angle
of 45 degrees. When we have the minimum horizontal
force equal to 45 newtons, the body will be on the point of sliding down the
plane. This means that the frictional
force will act up the plane.

We will resolve parallel and
perpendicular to the plane. Before doing this, we will need to
work out the components of the 75- and 45-newton forces in these directions. We do this using right angle
trigonometry, giving us the four forces shown. Parallel to the plane, we have
three forces. As the body is on the point of
sliding down the plane, we will use this as the positive direction. This gives us the equation 75 sin
45 minus 45 cos 45 minus πΉ r is equal to zero. The sin of 45 degrees and cos of 45
degrees are both equal to root two over two. Adding the frictional force to both
sides of this equation gives us πΉ r is equal to 15 root two.

There are also three forces acting
perpendicular to the plane. This gives us the equation π
minus
75 cos 45 minus 45 sin 45 is equal to zero. This simplifies to π
minus 60 root
two equal zero, which means that π
is equal to 60 root two. We now have the frictional force
and normal reaction force in newtons. And we know that πΉ r equals π,
the coefficient of friction, multiplied by π
. π is equal to 15 root two divided
by 60 root two. This simplifies to one-quarter.

We now need to consider the case
where the horizontal force is at its maximum. In this situation, most of our
forces remain the same. However, the body will be on the
point of moving up the plane, so the frictional force acts downwards. We need to calculate the unknown
force πΉ. Resolving parallel to the plane, we
have πΉ cos 45 minus 75 sin 45 minus πΉ r is equal to zero. Simplifying this and using the fact
that π equals one-quarter gives us root two over two πΉ minus 75 root two over two
minus one-quarter π
is equal to zero. In the perpendicular direction, we
have π
minus 75 cos 45 minus πΉ sin 45 is equal to zero. Simplifying this, we get π
is
equal to 75 root two over two plus root two over two πΉ.

We now have two simultaneous
equations that we can solve to calculate πΉ. We will replace the value of π
in
equation two into equation one. We can then divide through this
whole equation by root two over two. Distributing our parentheses gives
us πΉ minus 75 minus 75 over four minus a quarter πΉ is equal to zero. This in turn simplifies to
three-quarters πΉ is equal to 375 over four. Dividing both sides of the equation
by three-quarters gives us πΉ is equal to 125. The maximum force is equal to 125
newtons. This means that if the force is
between 45 and 125 newtons inclusive, the body will remain in equilibrium.

We will now summarize the key
points from this video. In all three questions, we saw that
when a body is in limiting equilibrium, the sum of the forces parallel and
perpendicular to the plane are equal to zero. We can resolve parallel and
perpendicular to the plane as well as using the equation πΉ r is equal to ππ
to
calculate any unknowns. π is the coefficient of friction
and lies between zero and one inclusive. We can also use right angle
trigonometry to calculate the relevant components.