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

# Lesson Explainer: Lami’s Theorem Mathematics

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 , , and . As , 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 , where is the length of the side opposite angle , is the length of the side opposite angle , and is the length of the side opposite angle , it is the case that

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 , , and in the first diagram of this explainer.

Let us, therefore, explore the link between these angles, , , and , and the angles , , and in the triangle of forces.

The external angles of triangle 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 is the angle between and , that is, angle . Hence, we have which can be rearranged to

In the same way, we find that

Since for all , we see that we can replace , , and by , , and 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 , , and acting on a particle in equilibrium, it is the case that where is the angle between and , is the angle between and , and is the angle between and .

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 is in equilibrium under the effect of the forces shown, which are in newtons. Find the force .

According to Lami’s theorem,

We can determine as follows:

We have then that

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 to the vertical, and the second is at to the vertical. If the magnitude of the tension in the first string is 45 g-wt, find and the magnitude of the tension in the second string.

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

Angle is given by

According to Lami’s theorem,

Only one of the terms in the expression does not contain an unknown, which is

We have then that

For this to be the case, it must be the case that

Given that we have that then

We can now see that

Rearranging to make the subject, we obtain

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.

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

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 . 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

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

Applying Lami’s theorem gives

Hence,

The magnitude of the reaction of each lamina is 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, , holds the body in equilibrium when the measure of the angle between the wall and the string is . Find , the tension in the string, and , the horizontal force.

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 is . Since is a right triangle at and angle is , we conclude that angle is . The angle between and the weight is therefore . It can be found as well by considering the alternate internal angles between the two vertical (and hence parallel) lines and the line of action of the weight.

Finally, the angle between and is .

Applying Lami’s theorem, we find that

Substituting in the values of and the sines, we find which gives

Hence, and

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 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 and , weighing 50 and 48 newtons, respectively, are attached to the other ends of the strings and that the system is in equilibrium, determine to the nearest minute and to the nearest two decimal places.

The system of forces acting on the block is shown in the following figure. The tensions in each string, and , are equal to the weights of and respectively.

The angle between the vector for the weight of the block and the vector for the weight of body is given by

Applying Lami’s theorem, we find that

Let us find first.

Rearranging we find that

is the angle in the second quadrant with the same sine value as .

Therefore, we have and as , we have

To the nearest minute, this is .

We see from this that

Applying Lami’s theorem, we have then that

Rearranging to make the subject, we obtain

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 , , and acting on a particle in equilibrium, it is the case that where is the angle between and , is the angle between and , and is the angle between and .