Video Transcript
In this video, we will learn how to
solve problems about the equilibrium of a particle under the action of three
coplanar forces using Lami’s theorem.
When there are three coplanar,
concurrent, noncollinear forces acting to keep an object in static equilibrium,
Lami’s theorem may be used. This states that 𝐴 over sin 𝛼 is
equal to 𝐵 over sin 𝛽, which is equal to 𝐶 over sin 𝛾. We will begin by considering the
derivation of this theorem. Let’s consider three concurrent
forces 𝐴, 𝐵, and 𝐶 which means they act at the same point as shown. We will let the angle between force
𝐵 and force 𝐶 be 𝛼, the angle between force 𝐴 and force 𝐶 be 𝛽, and the angle
between force 𝐴 and force 𝐵 be 𝛾. We know that these three angles sum
to 360 degrees.
Using geometric properties, we can
redraw this in the form of a triangle such that the sum of the forces is equal to
the vector addition 𝐁 plus 𝐀 plus 𝐂. We know that for any triangle, the
sine rule states that side length 𝑎 over sin of capital 𝐴 is equal to side length
𝑏 over the sin of capital 𝐵 which is equal to side length 𝑐 over the sin of angle
capital 𝐶. Substituting in the values from our
diagram, we have 𝐴 over sin of 180 minus 𝛼 is equal to 𝐵 over sin 180 minus 𝛽,
which is equal to 𝐶 over sin of 180 degrees minus 𝛾.
We know that the sin of any angle
𝐴 is equal to the sin of 180 degrees minus angle 𝐴. Our equation simplifies to 𝐴 over
sin 𝛼 is equal to 𝐵 over sin 𝛽, which is equal to 𝐶 over sin 𝛾. This is known as Lami’s theorem,
where 𝐴, 𝐵, and 𝐶 are the magnitudes of the three coplanar, concurrent, and
noncollinear vectors 𝐀, 𝐁, and 𝐂 which keep the object in static equilibrium. And 𝛼, 𝛽, and 𝛾 are the angles
directly opposite to the vectors. We will now look at some examples
where we can apply Lami’s theorem.
In the given figure, particle 𝐴 is
in equilibrium under the effect of the forces shown which are in newtons. Find the force 𝐹.
One way of solving this problem is
using Lami’s theorem. This states that when we have three
coplanar, concurrent, noncollinear forces which keep an object in static
equilibrium, then 𝐴 over sin 𝛼 is equal to 𝐵 over sin 𝛽, which is equal to 𝐶
over sin 𝛾, where 𝐴, 𝐵, and 𝐶 are the magnitudes of the forces and 𝛼, 𝛽, and
𝛾 are the angles directly opposite them. In our diagram, we’re given two of
the angles. They’re both equal to 150
degrees. As angles at a point sum to 360
degrees, the missing angle is 60 degrees. Substituting our values into Lami’s
theorem gives us 31 over sin 150 degrees is equal to 31 over sin of 150 degrees,
which is equal to 𝐹 over sin of 60 degrees.
If we consider the second and third
parts of the equation, we can calculate the value 𝐹. We know that the sin of 60 degrees
is equal to root three over two, and the sin of 150 degrees is equal to
one-half. This means that 𝐹 over root three
over two is equal to 31 over a half. Multiplying both sides of the
equation by one-half gives us 𝐹 over root three is equal to 31. We can then multiply both sides of
this equation by root three. 𝐹 is equal to 31 root three. We are told in the question that
the forces are in newtons. Therefore, the force 𝐹 is equal to
31 root three newtons.
In our next question, it will be
helpful to draw a diagram before applying Lami’s theorem.
A weight of 90 gram-weight is
suspended by two inextensible strings. The first is inclined at an angle
𝜃 the vertical, and the second is at 30 degrees to the vertical. If the magnitude of the tension in
the first string is 45 gram-weight, find 𝜃 and the magnitude of the tension 𝑇 in
the second string.
We will begin here by sketching a
diagram. We are told that a 90 gram-weight
is suspended by two inextensible strings as shown. They’re inclined at angles 𝜃 and
30 degrees to the vertical. The magnitude of the tension in the
first string is 45 gram-weight. And we need to calculate the
magnitude of the tension 𝑇 in the second string. As angles on a straight line sum to
180 degrees, the angle between the tension 𝑇 and the 90 gram-weight is 150
degrees. In the same way, the angle between
the 45 gram-weight tension and the 90 gram-weight is 180 minus 𝜃 degrees.
As we have three coplanar,
concurrent, and noncollinear forces acting to keep the object in static equilibrium,
we can use Lami’s theorem. This states that 𝐴 over sin 𝛼 is
equal to 𝐵 over sin 𝛽, which is equal to 𝐶 over sin 𝛾, where 𝐴, 𝐵, and 𝐶 are
the magnitude of the forces and 𝛼, 𝛽, and 𝛾 are the angles opposite them. In this question, the three forces
will be 𝑇, 90, and 45. The angles will be 180 minus 𝜃, 𝜃
plus 30, and 150. Substituting in these values, we
have 𝑇 over sin of 180 minus 𝜃 is equal to 90 over sin of 𝜃 plus 30 which is
equal to 45 over sin of 150.
Let’s consider the second and third
parts to this equation. sin of 150 degrees is equal to
one-half. So, our equation becomes 90 over
sin of 𝜃 plus 30 is equal to 45 over a half. 45 divided by a half is equal to
90. And we can then multiply both sides
by the sin of 𝜃 plus 30. Dividing both sides of this new
equation by 90 gives us one is equal to the sin of 𝜃 plus 30. We can then take the inverse sin of
both sides of this equation. The inverse sin of one is equal to
90 degrees. So, 90 is equal to 𝜃 plus 30. Subtracting 30 from both sides of
this equation gives us 𝜃 is equal to 60. The first part of our answer is 𝜃
is equal to 60 degrees.
We can now substitute this value
back into our equation to help calculate 𝑇. Using the first two parts of our
equation, we have 𝑇 over sin of 120 degrees is equal to 90 over sin of 90
degrees. sin of 90 degrees is equal to one,
and we can then multiply both sides by the sin of 120 degrees. 𝑇 is equal to 90 multiplied by the
sin of 120 degrees. The sin of 120 degrees is equal to
root three over two. Multiplying this by 90, we get 𝑇
is equal to 45 root three. The tension in the second string is
equal to 45 root three gram-weight. The two answers to this question
are 𝜃 equals 60 degrees and 𝑇 is equal to 45 root three gram-weight.
In our final question, we will
again study the equilibrium of a particle under three forces.
A body weighing 12 newtons is
attached to one end of a light inextensible string. The other end of the string is
fixed to a vertical wall. A horizontal force 𝐹 holds the
body in equilibrium when the measure of the angle between the wall and the string is
30 degrees. Find 𝑇, the tension in the string,
and 𝐹, the horizontal force.
As we have three coplanar,
concurrent, noncollinear forces acting to keep the body in static equilibrium, we
can use Lami’s theorem. This states that 𝐴 over sin 𝛼 is
equal to 𝐵 over sin 𝛽, which is equal to 𝐶 over sin 𝛾, where 𝐴, 𝐵, and 𝐶 are
the magnitude of the forces, in this case, the tension, the force 𝐹, and the weight
12 newtons. The angles 𝛼, 𝛽, and 𝛾 are
opposite the corresponding forces. We can see from the diagram that
the angle between the force 𝐹 and the weight is 90 degrees.
Using the fact that angles in a
triangle sum to 180 degrees, the tension force is 60 degrees above the
horizontal. This means that the angle between
the tension force and the weight force is 150 degrees. Angles at a point sum to 360
degrees. Therefore, the angle between the
tension force and the force 𝐹 is 120 degrees. Substituting our values into Lami’s
theorem gives us 𝐹 over sin of 150 degrees is equal to 𝑇 over sin of 90 degrees,
which is equal to 12 over sin of 120 degrees.
The sin of 150 degrees is equal to
one-half, the sin of 90 degrees is equal to one, and the sin of 120 degrees is equal
to root three over two. This means that our first term is
equal to 𝐹 over one-half. This simplifies to two 𝐹. The second term is equal to 𝑇 over
one which simplifies to 𝑇. The third term is equal to 12 over
root three over two. This simplifies to 24 over root
three.
We can rationalize the denominator
by multiplying the numerator and denominator of the fraction by root three. This gives us 24 root three over
three, which in turn simplifies to eight root three. Two 𝐹 is equal to 𝑇 which is
equal to eight root three. This means that the tension in the
string 𝑇 is equal to eight root three newtons. As this is equal to two 𝐹, the
horizontal force 𝐹 will be half of this. 𝐹 is equal to four root three
newtons. In order for the body to remain in
equilibrium, 𝑇 is equal to eight root three newtons and 𝐹 is equal to four root
three newtons.
We will now summarize the key
points from this video. When there are three coplanar,
concurrent, noncollinear forces acting to keep an object in static equilibrium,
Lami’s theorem may be used. This states that 𝐴 over sin 𝛼 is
equal to 𝐵 over sin 𝛽, which is equal to 𝐶 over sin 𝛾, where 𝐴, 𝐵, and 𝐶 are
the magnitudes of the three forces and 𝛼, 𝛽, and 𝛾 are the angles directly
opposite them.
