Lesson Video: Lami’s Theorem | Nagwa Lesson Video: Lami’s Theorem | Nagwa

# Lesson Video: Lami’s Theorem Mathematics

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.

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### 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.