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

Lesson Explainer: Lami’s Theorem Mathematics • Second Year of Secondary School

In this explainer, 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 a particle is in equilibrium, Newton’s first law tells us that the sum of the forces that act on the particle is zero. If the particle in equilibrium is subjected to only three coplanar, concurrent, and noncollinear forces, then the vectors representing the three forces are in a particular geometry, described by Lami’s theorem.

To understand Lami’s theorem, we need first to rearrange the three vectors representing the three forces βƒ‘πΉοŒΊ, βƒ‘πΉοŒ», and βƒ‘πΉοŒΌ. As ⃑𝐹+⃑𝐹+⃑𝐹=0, the three vectors representing the three forces form a triangle when arranged tail to head, called a triangle of forces.

Let us now recall the law of sines.

Definition: The Law of Sines

For a triangle 𝐴𝐡𝐢, where π‘Ž is the length of the side opposite angle 𝐴, 𝑏 is the length of the side opposite angle 𝐡, and 𝑐 is the length of the side opposite angle 𝐢, it is the case that π‘Žπ΄=𝑏𝐡=𝑐𝐢.sinsinsin

We see that the law of sines gives us a relationship between the magnitudes of the three forces and the angles in the triangle of forces. However, we often know the angles between the vectors representing the forces, that is, when the forces’ tails meet at the particle. These angles are denoted 𝛼, 𝛽, and 𝛾 in the first diagram of this explainer.

Let us, therefore, explore the link between these angles, 𝛼, 𝛽, and 𝛾, and the angles 𝐴, 𝐡, and 𝐢 in the triangle of forces.

The external angles of triangle 𝐴𝐡𝐢 are shown in the following figure. Each of them is a supplementary angle of the adjacent internal angle of the triangle.

We see that each external angle is the angle between the two vectors that join at the corresponding vertex. So, for instance, the external angle at 𝐴 is the angle between βƒ‘πΉοŒ» and βƒ‘πΉοŒΌ, that is, angle 𝛼. Hence, we have 𝛼=180βˆ’π΄,∘ which can be rearranged to 𝐴=180βˆ’π›Ό.∘

In the same way, we find that 𝐡=180βˆ’π›½,𝐢=180βˆ’π›Ύ.∘∘

Since sinsin(180βˆ’πœƒ)=πœƒβˆ˜ for all πœƒ, we see that we can replace sin𝐴, sin𝐡, and sin𝐢 by sin𝛼, sin𝛽, and sin𝛾 when applying the law of sines in the triangle of forces. It leads to Lami’s theorem.

Definition: Lami’s Theorem

For three coplanar, concurrent, and noncollinear forces βƒ‘πΉοŒΊ, βƒ‘πΉοŒ», and βƒ‘πΉοŒΌ acting on a particle in equilibrium, it is the case that ‖‖⃑𝐹‖‖𝛼=‖‖⃑𝐹‖‖𝛽=‖‖⃑𝐹‖‖𝛾,sinsinsin where 𝛼 is the angle between βƒ‘πΉοŒ» and βƒ‘πΉοŒΌ, 𝛽 is the angle between βƒ‘πΉοŒΊ and βƒ‘πΉοŒΌ, and 𝛾 is the angle between βƒ‘πΉοŒΊ and βƒ‘πΉοŒ».

Let us now look at an example in which Lami’s theorem is applied.

Example 1: Finding a Missing Force Using 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.

Answer

According to Lami’s theorem, 𝐹(𝐴)=31(150).sinsin

We can determine 𝐴 as follows: 𝐴=360βˆ’2(150)=60.∘

We have then that 𝐹(60)=31(150)𝐹=31ο€½(60)(150)𝐹=31βŽ›βŽœβŽœβŽο€½ο‰βŽžβŽŸβŽŸβŽ =31√3.sinsinsinsinN√

Let us now look at an application of Lami’s theorem involving an unknown force and an unknown angle.

Example 2: Finding a Missing Force and Angle Using Lami’s Theorem

A weight of 90 g-wt is suspended by two inextensible strings. The first is inclined at an angle πœƒ to the vertical, and the second is at 30∘ to the vertical. If the magnitude of the tension in the first string is 45 g-wt, find πœƒ and the magnitude of the tension 𝑇 in the second string.

Answer

The system of forces acting on the body is shown in the following figure.

Angle 𝛼 is given by 𝛼=180βˆ’30=150.∘

We see also that πœ™=180βˆ’πœƒ.∘

According to Lami’s theorem, 45(150)=90(30+πœƒ)=𝐹(180βˆ’πœƒ).sinsinsin

Only one of the terms in the expression does not contain an unknown, which is 45(150)=45=90.sin

We have then that 90=90(30+πœƒ).sin

For this to be the case, it must be the case that sin(30+πœƒ)=1.

Given that sin(90)=1, we have that 30+πœƒ=90; then πœƒ=60.∘

We can now see that 90=𝐹(180βˆ’60).sin

Rearranging to make 𝐹 the subject, we obtain 𝐹=90(120)=90Γ—βˆš32=45√3.sinkg-wt

Let us now look at applying Lami’s theorem to the forces on a spherical body.

Example 3: Finding the Reactions on a Sphere Resting in Equilibrium between Two Inclined Laminae

A sphere is resting on two laminae. The distance between the two points of contact is equal to the sphere’s radius. Determine the reaction of each lamina on the sphere, given that the weight of the sphere is 261 N.

Answer

From the figure, we see that the line of action of the weight of the sphere passes through the point where the two laminae meet. We know that if two tangents on a circle intersect, then the distances from the point where they touch the circle to the point where they cross are the same. It means that the line of action of the weight is a line of symmetry of our system. Therefore, we already know that 𝑅=𝑅.

Furthermore, the center of the circle and the two points of contact with the laminae are equidistant, therefore forming an equilateral triangle with internal angles of 60∘. As the line of action of the weight is a line of symmetry, the angles between the weight and the reaction force of each lamina are therefore equal and given by 12(360βˆ’60)=150.∘∘∘

The system of forces acting on the body is shown in the following figure.

Applying Lami’s theorem gives π‘Š60=𝑅150=𝑅150.sinsinsin∘∘∘

Hence, 𝑅=𝑅=π‘Š60β‹…150=261β‹…=261√3=261√33=87√3.∘∘√sinsinN

The magnitude of the reaction of each lamina is 87√3 N.

Let us now look at an example in which Lami’s theorem is applied to a system of forces that correspond to a right triangle.

Example 4: Finding Missing Forces Using Lami’s Theorem in a Real-World Problem

A body weighing 12 N 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∘. Find 𝑇, the tension in the string, and 𝐹, the horizontal force.

Answer

We can solve this question by applying Lami’s theorem. For this, we need to find the angles between the forces. We already know that the angle between the weight and ⃑𝐹 is 90∘. Since 𝐴𝐡𝐢 is a right triangle at 𝐡 and angle 𝐴 is 30∘, we conclude that angle 𝐢 is 60∘. The angle between ⃑𝑇 and the weight is therefore 90+60=150∘. It can be found as well by considering the alternate internal angles between the two vertical (and hence parallel) lines ⃖⃗𝐴𝐡 and the line of action of the weight.

Finally, the angle between ⃑𝐹 and ⃑𝑇 is 360βˆ’90βˆ’150=120∘.

Applying Lami’s theorem, we find that π‘Š120=𝑇90=𝐹150.sinsinsin∘∘∘

Substituting in the values of π‘Š and the sines, we find 12=𝑇=𝐹,√ which gives 24√3=𝑇=2𝐹.

Hence, 𝑇=2𝐹=24√33=8√3,N and 𝐹=4√3.N

Let us now look at an example involving applying Lami’s theorem to a system in equilibrium.

Example 5: Finding a Missing Force and Angle Using Lami’s Theorem in a Real-World Problem

As shown in the figure, a block weighing π‘Š N is suspended from a string connected to two other pieces of string, each of which passes over a smooth pulley. Given that two bodies 𝐴 and 𝐡, weighing 50 and 48 newtons, respectively, are attached to the other ends of the strings and that the system is in equilibrium, determine πœƒ to the nearest minute and π‘Š to the nearest two decimal places.

Answer

The system of forces acting on the block is shown in the following figure. The tensions in each string, π‘‡οŠ§ and π‘‡οŠ¨, are equal to the weights of 𝐴 and 𝐡 respectively.

The angle between the vector for the weight of the block and the vector for the weight of body 𝐴 is given by 45+90=135.∘

Applying Lami’s theorem, we find that 48135=π‘Šπœ™=50(90+πœƒ).sinsinsin

Let us find πœƒ first.

Rearranging 48135=50(90+πœƒ),sinsin we find that sinsinsinsin(90+πœƒ)=5048135ο€Ό5048135οˆβ‰ˆ47.44.∘

90+πœƒ is the angle in the second quadrant with the same sine value as 47.44∘.

Therefore, we have 90+πœƒβ‰ˆ180βˆ’47.44β‰ˆ132.56,∘ and as 90+πœƒβ‰ˆ132.56∘, we have πœƒβ‰ˆ42.56.∘

To the nearest minute, this is 4234β€²βˆ˜.

We see from this that πœ™β‰ˆ180βˆ’45βˆ’42.56β‰ˆ92.44.∘

Applying Lami’s theorem, we have then that 48135β‰ˆπ‘Š92.44.sinsin

Rearranging to make π‘Š the subject, we obtain π‘Šβ‰ˆ48Γ—92.44135.sinsin

To two decimal places, this is 67.82 N.

Let us now summarize what we have learned in these examples.

Key Points

  • For three coplanar, concurrent, and noncollinear forces βƒ‘πΉοŒΊ, βƒ‘πΉοŒ», and βƒ‘πΉοŒΌ acting on a particle in equilibrium, it is the case that ‖‖⃑𝐹‖‖𝛼=‖‖⃑𝐹‖‖𝛽=‖‖⃑𝐹‖‖𝛾,sinsinsin where 𝛼 is the angle between βƒ‘πΉοŒ» and βƒ‘πΉοŒΌ, 𝛽 is the angle between βƒ‘πΉοŒΊ and βƒ‘πΉοŒΌ, and 𝛾 is the angle between βƒ‘πΉοŒΊ and βƒ‘πΉοŒ».

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