### Video Transcript

A uniform beam π΄π΅ of weight 106
newtons is resting with its end π΄ on a rough horizontal ground and with its end π΅
against a rough vertical wall, where the coefficient of friction between the beam
and the wall is four times that between the beam and the ground. If the beam is about to move when
it is inclined to the wall at an angle whose tan is 14 over 45, determine the
reaction of the wall rounded to two decimal places.

Letβs begin with a diagram of the
scenario. We have the uniform beam π΄π΅ with
its end π΄ resting on the horizontal ground and its end π΅ resting against the
wall. The angle between the beam and the
wall is π whose tan is 14 over 45. Therefore, π is equal to the
inverse tan of 14 over 45. The beam has a weight of 106
newtons, which acts straight downwards from the beamβs center of mass, which in this
case, since the beam is uniform, is from the very center of the line π΄π΅. The coefficient of friction between
the beam and the ground is unknown. Letβs call this π. And the coefficient of friction
between the beam and the wall is four times this, so this is four π.

The question asks us to determine
the reaction of the wall. This is the force π
w which acts
on the end π΅ of the beam directly out from the wall and horizontally. The beam is given to be at
rest. So we can balance all of the
horizontal and vertical components of all of the forces acting on the beam.

First, we need to determine what
all of these forces are. In addition to the reaction force
from the wall, there will also be a reaction force from the ground, π
G, which acts
directly away from the ground vertically upwards. There is a frictional force πΉ w
acting on the end π΅ of the beam. This opposes the direction of
motion of the beam along the wall. Since the beam is about to slip,
the direction of motion would be directly down the wall. Therefore, πΉ w acts directly up
the wall. Likewise, there is another
frictional force πΉ G acting on the beam at the point π΄. Again, this frictional force
opposes the direction of motion of the beam. And since the beam is about to
slip, its motion would be in the rightward direction along the ground. So πΉ G acts in the leftward
direction.

All of these forces are either
directly horizontal or directly vertical. Therefore, they are perpendicular
or parallel to each other. And we can resolve their horizontal
and vertical components separately. Starting with the horizontal forces
then, to the left, we have the frictional force of the ground, πΉ G. And to the right, we have the
reaction force of the wall, π
w. So πΉ G is equal to π
w. And for the vertical forces, in the
upward direction, we have the frictional force of the wall πΉ w and the reaction
force of the ground π
G. And in the downward direction, we
have the weight of the beam, 106 newtons. So πΉ w plus π
G equals 106.

Now recall that the coefficient of
friction is defined by the equation πΉ equals ππ
, where the object is at the point
of slip, π
is the reaction force on the object, and πΉ is the frictional force. In this scenario then, πΉ G is
equal to the coefficient of friction of the ground, π, multiplied by the reaction
force of the ground, π
G. And πΉ w is equal to the
coefficient of friction of the wall, four π, multiplied by the reaction force of
the wall, π
w. We can therefore rewrite both of
these equations as ππ
G equals π
w and four ππ
w plus π
G equals 106. But we have a problem. We now have three unknowns in this
system: π, π
G, and π
w. But we only have two equations.

To solve a system of equations for
any of the variables, we need at least the same number of equations as there are
unknowns. We therefore need to find more
different equations for each of these unknowns in this scenario. In a system in equilibrium, it is
not only the forces that are balanced but also the moments of the forces about any
point in space. For instance, the moments on this
beam are balanced about the point π΄ and the point π΅. If we therefore calculate the
moments of all of the forces about the points π΄ and π΅, they should all be balanced
as well.

Letβs begin by calculating the
moments of the forces about the point π΄, defining a positive moment as being in the
counterclockwise direction. Recall that the moment of a force
π is equal to the magnitude of the force πΉ multiplied by the perpendicular
distance π between the line of action of the force and the pivot point π. The perpendicular distances between
the lines of action of π
G and πΉ G will clearly be equal to zero since they are
acting from the point π΄.

The perpendicular distance between
the line of action of the weight and the pivot point π΄ will be this distance here,
π one. This distance will be half the
length of the beam, which we will call π, multiplied by the sine of the angle
between the beam and the wall, π. The perpendicular distance between
the line of action of the reaction force of the wall π
w and the pivot point π΄
will be this distance here, π two. This distance will be the length of
the beam π multiplied by the cosine of the angle between the beam and the wall
π. And finally, the perpendicular
distance between the line of action of the frictional force of the wall πΉ w and the
pivot point π΄ will be given by this distance here, π three. This distance will be equal to the
length of the beam π multiplied by the sine of the angle between the beam and the
wall π.

Going through all the moments then,
starting with the moment of the weight of the beam, we have the magnitude of the
force, which is 106 newtons, multiplied by the perpendicular distance, π one, equal
to π over two sin π. Clearly this moment will be in the
anticlockwise direction, so this is a positive moment.

Next, for the moment of the
reaction force of the wall π
w, we have the magnitude π
w itself multiplied by π
two, which is equal to π cos π. π
w will clearly have a clockwise
turning force about the point π΄. So this is a negative moment. And finally, we have the moment of
the frictional force of the wall, πΉ w, which is equal to the magnitude, πΉ w
itself, multiplied by the distance π three, which is π sin π. Once again, this will be a
clockwise turning force about the point π΄. So this is a negative moment. Since the system is at rest, the
sum of the moments about any point including the point π΄ will be equal to zero.

We can now simplify this
equation. We begin by noting that we have a
common factor of π in all three turns on the left-hand side. And since π is nonzero, we can
divide by π on both sides to cancel it out completely. Since the beam is about to slip, πΉ
w is also equal to four ππ
w. And finally, in the first term, we
can simplify by taking 106 multiplied by the one-half to give 53. So this gives us 53 sin π minus π
w cos π minus four ππ
w sin π is equal to zero.

But we can simplify this further to
make this even better. π is clearly not equal to 90
degrees since its tangent is well defined. Therefore, we can divide both sides
by cos π since cos π will be nonzero. The cos π will cancel with the
second term. And for the other two terms, sin π
over cos π is identically equal to tan π. And we are given in the question
that tan π is equal to 14 over 45. This gives us 53 multiplied by 14
over 45, which is equal to 742 over 45, minus π
w minus four times 14 over 45,
which is equal to negative 56 over 45, multiplied by ππ
w is equal to zero. We can simplify this one more time
by multiplying by 45 to give us 742 minus 45π
w minus 56ππ
w is equal to
zero. We could also factorize the π
w,
but it will be more useful to keep these terms separate in just a moment.

We now have three equations for the
three unknowns, which can be solved to give values for each of them. We can go about solving these
equations by eliminating the variables in any order that we choose. It turns out that solving for the
coefficient of friction π is slightly easier than other ways. So letβs try this way.

If we eliminate π
G from the
second and third equation, we will then be able to solve for both π and what we
want, π
w. So we begin by rearranging the
first equation to give π
G equals π
w over π. We can substitute this into the
second equation here, where π
G appears. And then we have two equations in
π and π
w. So for the second equation, we now
have four ππ
w plus π
w over π equals 106. We can rearrange the second
equation by multiplying by π on both sides, which gives four π squared π
w plus
π
w equals 106π. We have a common term of π
w on
the left-hand side. So we can factorize this to give π
w multiplied by four π squared plus one is equal to 106π. Dividing both sides by four π
squared plus one gives us π
w equals 106π divided by four π squared plus one.

We now have an explicit expression
for π
w in terms of the coefficient of friction π. So we will be able to come back to
this later after solving for π to calculate π
w easily. For now, we can substitute this
expression for π
w into the final equation, which will give us an equation only in
π, which can be solved. So this gives us 742 minus 45 times
106π over four π squared plus one minus 56π times 106π over four π squared plus
one is equal to zero.

Now, we can simplify this somewhat
if we like by noting that 742 is 106 times seven. Therefore, we have a common factor
of 106 in all three terms. Dividing through by 106 will give
us seven for the first term, will cancel out the 106 in the second and third
terms. This gives us seven minus 45π over
four π squared plus one minus 56π squared over four π squared plus one is equal
to zero. We can multiply both sides by four
π squared plus one, which will cancel out the denominator on the second and third
terms. This gives us seven times four π
squared plus one minus 45π minus 56π squared is equal to zero.

Expanding the parentheses and
collecting like terms in π gives us negative 28π squared minus 45π plus seven is
equal to zero. This is a quadratic in π, which
can be solved using the quadratic formula. The quadratic formula is given by
π is equal to negative π plus or minus the square root of π squared minus four
ππ all over two π, where π is the coefficient of the π squared term, so in this
case negative 28. π is the coefficient of the
π-term, so negative 45. And π is the constant coefficient,
which is seven. So this gives us π equals 45 plus
or minus the square root of negative 45 squared minus four times negative 28 times
seven all over two times negative 28.

The term under the square root
comes to 2809, and the denominator comes to negative 56. The square root of 2809 is plus or
minus 53. The coefficient of friction π must
be greater than or equal to zero. So clearly, we need to make the
numerator negative so that π is overall positive. Therefore, we will be taking the
negative square root. This gives us negative eight over
negative 56, so π is equal to one-seventh. We can now substitute this value of
π into our equation for π
w, and we will have our answer. So we have π
w equals 106
multiplied by one over seven all divided by four times one over seven squared plus
one. Performing this calculation gives
us 14 exactly, so we have our final answer. The reaction of the wall rounded to
two decimal places, π
w, equals 14.00 and the unit is newtons.