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In this explainer, we will learn how to find the resultant of two forces acting on one point and how to find the direction of the resultant. We start by defining a force and exploring its properties. Force is defined as the effect of one natural body on another. Each force is described in terms of its magnitude (size), direction, point of action, and line of action. We often represent a force by using the notation ⃑𝐹.
For example, the diagram below shows the force ⃑𝐹 represented by the directed line segment 𝐴𝐵.
The magnitude of the force is determined by ‖‖𝐴𝐵‖‖. The direction of the arrow corresponds to the direction of ⃑𝐹. The point of action is 𝐴. The line of action is indicated by extending 𝐴𝐵 in the same direction (as shown by the dotted line).
A force acting on a body is represented by vector ⃑𝐹. When two forces act on a body, we call their resultant the force that describes their combined effect.
When two forces, ⃑𝐹 and ⃑𝐹, act on a body at the same point, the combined effect of these two forces is the same as the effect of a single force, called the resultant force.
The resultant force, ⃑𝑅, is given by ⃑𝑅=⃑𝐹+⃑𝐹.
The vector equality ⃑𝑅=⃑𝐹+⃑𝐹 can be represented in two ways, as illustrated in the following diagram.
As ⃑𝐹, ⃑𝐹, and ⃑𝑅 are three sides of a triangle, we can use either the law of sines or the law of cosines in the triangle to find the resultant of the two forces, the angles between the resultant and the forces, or any other unknown.
Let 𝛼 be the angle between forces ⃑𝐹 and ⃑𝐹, 𝜃 the angle between ⃑𝑅 and ⃑𝐹, and 𝜃 the angle between ⃑𝑅 and ⃑𝐹, as shown in the diagram below.
The law of sines in this triangle gives us 𝐹𝜃=𝐹𝜃=𝑅(180−𝛼),∘sinsinsin where 𝐹, 𝐹, and 𝑅 are the magnitudes of ⃑𝐹, ⃑𝐹, and ⃑𝑅 respectively.
As sinsin(180−𝑥)=𝑥∘ for all 𝑥, we find the relationship given in the following box.
We have 𝐹𝜃=𝐹𝜃=𝑅𝛼,sinsinsin where 𝐹, 𝐹, and 𝑅 are the magnitudes of ⃑𝐹, ⃑𝐹, and ⃑𝑅, respectively, 𝛼 is the angle between forces ⃑𝐹 and ⃑𝐹, 𝜃 is the angle between ⃑𝑅 and ⃑𝐹, and 𝜃 is the angle between ⃑𝑅 and ⃑𝐹.
Applying the law of cosines in our triangle now, we find that 𝑅=𝐹+𝐹−2𝐹𝐹(180−𝛼).∘cos
As coscos(180−𝑥)=−𝑥∘ for all 𝑥, we find the relationship given in the following box.
We have 𝑅=𝐹+𝐹+2𝐹𝐹𝛼,cos where 𝐹, 𝐹, and 𝑅 are the magnitudes of ⃑𝐹, ⃑𝐹, and ⃑𝑅, respectively, and 𝛼 is the angle between forces ⃑𝐹 and ⃑𝐹.
By taking the square root of both sides of the above equality and recalling that the magnitude of a vector is positive, we can obtain an explicit formula for 𝑅, the magnitude of ⃑𝑅. It is also straightforward to derive an accompanying formula for the direction of ⃑𝑅. We state these results below.
Let ⃑𝑅 be the resultant force of two forces, ⃑𝐹 and ⃑𝐹, that act at a single point with an angle 𝛼 between them. Then, 𝑅=𝐹+𝐹+2𝐹𝐹𝛼𝜃=𝐹𝛼𝐹+𝐹𝛼,cosandtansincos where 𝐹, 𝐹, and 𝑅 are the magnitudes of ⃑𝐹, ⃑𝐹, and ⃑𝑅, respectively, and 𝜃 is the angle between ⃑𝑅 and ⃑𝐹.
Let us start with an example in which the magnitude of the resultant of two forces acting at a point is determined.
Two forces of magnitudes 35 N and 91 N are acting at a particle. Given that the resultant is perpendicular to the first force, find the magnitude of the resultant.
Answer
It will be convenient to assume that the first force acts horizontally. Let us call this force ⃑𝐹 and the other force ⃑𝐹. The resultant of these forces, ⃑𝐹+⃑𝐹, acts vertically as it is perpendicular to ⃑𝐹, as shown in the following figure.
The force ⃑𝐹 can be represented by an arrow with its tail at the head of ⃑𝐹 and its head at the head of ⃑𝐹+⃑𝐹, as shown in the following figure.
The resultant force ⃑𝑅 is given by ⃑𝑅=⃑𝐹+⃑𝐹.
As ⃑𝐹 and ⃑𝑅 are perpendicular, we see that the two forces and their resultant form a right triangle. Therefore, applying the Pythagorean theorem gives ‖‖⃑𝐹‖‖+‖‖⃑𝑅‖‖=‖‖⃑𝐹‖‖.
It is worth noting that the Pythagorean theorem is just a special case of the law of cosines.
Substituting in the values of ‖‖⃑𝐹‖‖ and ‖‖⃑𝐹‖‖, we find that 35+‖‖⃑𝑅‖‖=91‖‖⃑𝑅‖‖=91−35=7056‖‖⃑𝑅‖‖=√7056=84.N
Note that as the magnitude of a vector is always positive, −84 N is not a valid solution.
The magnitude of the resultant of the forces is 84 N.
Let us now look at an example in which the direction of the line of action of the resultant of two forces acting at a point is determined.
Two perpendicular forces of magnitudes 88 N and 44 N act at a point. Their resultant makes an angle 𝜃 with the 88 N force. Find the value of sin𝜃.
Answer
It will be convenient to assume that one of the forces acts horizontally. Let us call this force ⃑𝐹 and the other force ⃑𝐹, as shown in the following figure.
By choosing to make ⃑𝐹 correspond to the line adjacent to 𝜃, we have chosen this force to be the 88-newton force. The magnitude of ⃑𝐹 is 44 newtons; therefore, the magnitude of ⃑𝐹 is half that of ⃑𝐹. The magnitude of the resultant of the forces, 𝑅, can be expressed as 𝑅=𝐹+𝐹=𝐹+𝐹2.
We can see from this that 𝑅=𝐹+𝐹2=𝐹+𝐹4=54𝐹.
Taking square roots, we have that 𝑅=√5𝐹2.
Applying the law of sines in the triangle gives sinsin𝜃𝐹=90𝑅.∘
As sin90=1∘, we have sinsin𝜃=𝐹𝑅𝜃==1√5=√55.√
We have, therefore, that sin𝜃=√55.
Let us now look at an example in which the magnitude and direction of the line of action of the resultant of two perpendicular forces are known and the magnitudes of the forces must be determined.
Two perpendicular forces, ⃑𝐹 and ⃑𝐹, act at a point. Their resultant, ⃑𝑅, has magnitude 188 N and makes an angle of 60∘ with ⃑𝐹. Find the magnitudes of ⃑𝐹 and ⃑𝐹.
Answer
The perpendicular forces, ⃑𝐹 and ⃑𝐹, and their resultant are shown in the following figure.
We see that ⃑𝐹 and ⃑𝐹 are perpendicular and the resultant ⃑𝐹+⃑𝐹 makes an angle of 60∘ with ⃑𝐹. As we have a right triangle, we have coscosN60=‖‖⃑𝐹‖‖188‖‖⃑𝐹‖‖=18860=12×188=94∘∘ and sinsinN60=‖‖⃑𝐹‖‖188‖‖⃑𝐹‖‖=18860=√32×188=94√3.∘∘
⃑𝐹 has a magnitude of 94 N, and ⃑𝐹 has a magnitude of 94√3 N.
Let us now look at an example involving two nonperpendicular forces.
The angle between forces ⃑𝐹 and ⃑𝐹 is 112∘, and the measure of the angle between their resultant and ⃑𝐹 is 56∘. If the magnitude of ⃑𝐹 is 28 N, what is the magnitude of ⃑𝐹?
Answer
The following figure shows the forces ⃑𝐹 and ⃑𝐹 and their resultant ⃑𝐹+⃑𝐹. The forces act at a point 𝑃.
The resultant forces ⃑𝐹 and ⃑𝐹 form a parallelogram whose diagonal through 𝑃 is the resultant.
The angle, 𝜃, between ⃑𝐹 and the resultant of ⃑𝐹 and ⃑𝐹 is given by 𝜃=112−56=56.∘
We can now add this angle and its alternate interior angle in our diagram as shown.
Applying the law of sines in the triangle formed by ⃑𝐹, ⃑𝐹, and ⃑𝐹+⃑𝐹, we find that ‖‖⃑𝐹‖‖56=‖‖⃑𝐹‖‖56,∘∘sinsin that is, ‖‖⃑𝐹‖‖=‖‖⃑𝐹‖‖.
The magnitude of ⃑𝐹 is given as 28 N, so the magnitude of ⃑𝐹 is also 28 N.
Let us look at our last example where the direction of one of the forces is reversed.
Two forces, both of magnitude 𝐹 N, act at the same point. The magnitude of their resultant is 90 N. When the direction of one of the forces is reversed, the magnitude of their resultant is 90 N. Determine the value of 𝐹.
Answer
Let us represent the first situation.
When we add two forces, ⃑𝐹 and ⃑𝐹, the resultant is the diagonal of the parallelogram formed by ⃑𝐹 and ⃑𝐹, with its tail being the point of application of ⃑𝐹 and ⃑𝐹. If the two forces have the same magnitude, then the parallelogram is a rhombus, and the two forces and their resultant form an isosceles triangle, as shown in the following diagram.
Applying the law of cosines, we find that 𝑅=𝐹+𝐹+2𝐹𝐹𝛼,cos with 𝐹=‖‖⃑𝐹‖‖, 𝐹=‖‖⃑𝐹‖‖, and 𝑅=‖‖⃑𝑅‖‖.
Since 𝐹=𝐹=𝐹, we have 𝑅=2𝐹+2𝐹𝛼.cos
If we now reverse the direction of one of the forces (for symmetry reasons, it does not matter which force has its direction reversed; we will get the same result), the resultant will still be the diagonal of a rhombus congruent to the previous one, but it will be the other diagonal, and the angle between forces −⃑𝐹 and ⃑𝐹 will be 180−𝛼∘.
The magnitude of −⃑𝐹 is the same as the magnitude of ⃑𝐹, 𝐹.
Applying the law of cosines in the triangle formed by −⃑𝐹, ⃑𝐹, and their resultant gives us 𝑅′=2𝐹+2𝐹(180−𝛼),∘cos that is, 𝑅′=2𝐹−2𝐹𝛼.cos
We are told that the magnitude of the resultant is the same in both cases, 90 N. Hence, we have 𝑅=𝑅′=90,N which means that 2𝐹+2𝐹𝛼=2𝐹−2𝐹𝛼=90.coscos
This is true only if cos𝛼=0, that is, if 𝛼=90∘. Forces ⃑𝐹 and ⃑𝐹 are, thus, perpendicular.
Hence, we have 2𝐹=90𝐹=902𝐹=9012𝐹=45√2.N
It is worth noting that, in the previous example, we could have concluded that the two forces are perpendicular with simple geometric considerations: the diagonals in a rhombus have the same length only if the rhombus is a square.
Let us now summarize what has been learned in these examples.
- Force is defined as the effect of one natural body on another. Each force is described in terms of its magnitude (size), direction, point of action, and line of action. We often represent a force by using the notation ⃑𝐹.
- The resultant, ⃑𝑅, of two forces, ⃑𝐹 and ⃑𝐹, acting on a body at the same point is a single force that is given by ⃑𝑅=⃑𝐹+⃑𝐹.
- The combined effect of ⃑𝐹 and ⃑𝐹 is the same as the effect of only ⃑𝑅.
- ⃑𝐹, ⃑𝐹, and ⃑𝑅 are three sides of a triangle or two adjacent sides and a diagonal of a parallelogram.
- Applying the law of sines in the triangle formed by two forces ⃑𝐹 and ⃑𝐹 and their resultant, ⃑𝑅, gives 𝐹𝜃=𝐹𝜃=𝑅𝛼,sinsinsin where 𝐹, 𝐹, and 𝑅 are the magnitudes of ⃑𝐹, ⃑𝐹, and ⃑𝑅, respectively, 𝛼 is the angle between forces ⃑𝐹 and ⃑𝐹, 𝜃 is the angle between ⃑𝑅 and ⃑𝐹, and 𝜃 is the angle between ⃑𝑅 and ⃑𝐹.
- Applying the law of cosines in the triangle formed by two forces ⃑𝐹 and ⃑𝐹 and their resultant, ⃑𝑅, gives 𝑅=𝐹+𝐹+2𝐹𝐹𝛼,cos where 𝐹, 𝐹, and 𝑅 are the magnitudes of ⃑𝐹, ⃑𝐹, and ⃑𝑅, respectively, and 𝛼 is the angle between forces ⃑𝐹 and ⃑𝐹.
- Let ⃑𝑅 be the resultant force of two forces, ⃑𝐹 and ⃑𝐹, that act at a single point with an angle 𝛼 between them. Then, 𝑅=𝐹+𝐹+2𝐹𝐹𝛼𝜃=𝐹𝛼𝐹+𝐹𝛼,cosandtansincos where 𝐹, 𝐹, and 𝑅 are the magnitudes of ⃑𝐹, ⃑𝐹, and ⃑𝑅, respectively, and 𝜃 is the angle between ⃑𝑅 and ⃑𝐹.