What is the resultant force when two forces act in the opposite direction?

The SSC MTS Tier II Admit Card has been released. The paper II will be held on 6th November 2022. Earlier, the result for the Tier I was released. The candidates who are qualified in the SSC MTS Paper I are eligible for the Paper II. A total of 7709 vacancies are released, out of which 3854 vacancies are for MTS Group age 18-25 years, 252 vacancies are for MTS Group age 18-27 years and 3603 vacancies are for Havaldar in CBIC. 

Show

    What is the resultant force when two forces act in the opposite direction?

    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.

    1. The magnitude of a force is its size, which is measured in newtons (N). By using a directed line segment to represent the force ⃑𝐹, and drawing the line to a suitable scale, we can use the length of the line to denote the magnitude, ‖‖⃑𝐹‖‖.
    2. The direction of a force is the direction in which it acts. By using a directed line segment to represent the force ⃑𝐹, we can use the direction of the arrow to show the direction of the force.
    3. The point of action of a force is the point at which it is applied.
    4. The line of action of a force is a geometric way to represent how the force is applied. It is drawn as a line through the point of action in the same direction as ⃑𝐹.

    For example, the diagram below shows the force ⃑𝐹 represented by the directed line segment 𝐴𝐡.

    The magnitude of the force is determined by ‖‖𝐴𝐡‖‖. 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 𝐴𝐡 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, βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, 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+βƒ‘πΉοŠ§οŠ¨ can be represented in two ways, as illustrated in the following diagram.

    As βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, 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 βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, πœƒοŠ§ the angle between ⃑𝑅 and βƒ‘πΉοŠ§, and πœƒοŠ¨ the angle between ⃑𝑅 and βƒ‘πΉοŠ¨, as shown in the diagram below.

    The law of sines in this triangle gives us πΉπœƒ=πΉπœƒ=𝑅(180βˆ’π›Ό),∘sinsinsin where 𝐹, 𝐹, and 𝑅 are the magnitudes of βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, and ⃑𝑅 respectively.

    As sinsin(180βˆ’π‘₯)=π‘₯∘ for all π‘₯, we find the relationship given in the following box.

    We have πΉπœƒ=πΉπœƒ=𝑅𝛼,sinsinsin where 𝐹, 𝐹, and 𝑅 are the magnitudes of βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, and ⃑𝑅, respectively, 𝛼 is the angle between forces βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, πœƒοŠ§ is the angle between ⃑𝑅 and βƒ‘πΉοŠ§, and πœƒοŠ¨ is the angle between ⃑𝑅 and βƒ‘πΉοŠ¨.

    Applying the law of cosines in our triangle now, we find that 𝑅=𝐹+πΉβˆ’2𝐹𝐹(180βˆ’π›Ό).∘cos

    As coscos(180βˆ’π‘₯)=βˆ’π‘₯∘ for all π‘₯, 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 βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, and ⃑𝑅, respectively, and 𝛼 is the angle between forces βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨.

    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, βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, that act at a single point with an angle 𝛼 between them. Then, 𝑅=𝐹+𝐹+2πΉπΉπ›Όπœƒ=𝐹𝛼𝐹+𝐹𝛼,cosandtansincos where 𝐹, 𝐹, and 𝑅 are the magnitudes of βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, and ⃑𝑅, respectively, and πœƒ is the angle between ⃑𝑅 and βƒ‘πΉοŠ§.

    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 βƒ‘πΉοŠ§ and the other force βƒ‘πΉοŠ¨. The resultant of these forces, ⃑𝐹+βƒ‘πΉοŠ§οŠ¨, acts vertically as it is perpendicular to βƒ‘πΉοŠ§, as shown in the following figure.

    The force βƒ‘πΉοŠ¨ can be represented by an arrow with its tail at the head of βƒ‘πΉοŠ§ and its head at the head of ⃑𝐹+βƒ‘πΉοŠ§οŠ¨, as shown in the following figure.

    The resultant force ⃑𝑅 is given by ⃑𝑅=⃑𝐹+⃑𝐹.

    As βƒ‘πΉοŠ§ 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 β€–β€–βƒ‘πΉβ€–β€–οŠ§ and β€–β€–βƒ‘πΉβ€–β€–οŠ¨, we find that 35+‖‖⃑𝑅‖‖=91‖‖⃑𝑅‖‖=91βˆ’35=7056‖‖⃑𝑅‖‖=√7056=84.N

    Note that as the magnitude of a vector is always positive, βˆ’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 πœƒ with the 88 N force. Find the value of sinπœƒ.

    Answer

    It will be convenient to assume that one of the forces acts horizontally. Let us call this force βƒ‘πΉοŠ§ and the other force βƒ‘πΉοŠ¨, as shown in the following figure.

    By choosing to make βƒ‘πΉοŠ§ correspond to the line adjacent to πœƒ, we have chosen this force to be the 88-newton force. The magnitude of βƒ‘πΉοŠ¨ is 44 newtons; therefore, the magnitude of βƒ‘πΉοŠ¨ is half that of βƒ‘πΉοŠ§. The magnitude of the resultant of the forces, 𝑅, can be expressed as 𝑅=𝐹+𝐹=ο„ŸπΉ+𝐹2.

    We can see from this that 𝑅=𝐹+𝐹2=𝐹+𝐹4=ο€Ό54𝐹.

    Taking square roots, we have that 𝑅=√5𝐹2.

    Applying the law of sines in the triangle gives sinsinπœƒπΉ=90𝑅.∘

    As sin90=1∘, we have sinsinπœƒ=πΉπ‘…πœƒ==1√5=√55.οŒ₯√οŒ₯

    We have, therefore, that sinπœƒ=√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, βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, act at a point. Their resultant, ⃑𝑅, has magnitude 188 N and makes an angle of 60∘ with βƒ‘πΉοŠ§. Find the magnitudes of βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨.

    Answer

    The perpendicular forces, βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, and their resultant are shown in the following figure.

    We see that βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ are perpendicular and the resultant ⃑𝐹+βƒ‘πΉοŠ§οŠ¨ makes an angle of 60∘ with βƒ‘πΉοŠ§. As we have a right triangle, we have coscosN60=‖‖⃑𝐹‖‖188‖‖⃑𝐹‖‖=18860=12Γ—188=94∘∘ and sinsinN60=‖‖⃑𝐹‖‖188‖‖⃑𝐹‖‖=18860=√32Γ—188=94√3.∘∘

    βƒ‘πΉοŠ§ has a magnitude of 94 N, and βƒ‘πΉοŠ¨ has a magnitude of 94√3 N.

    Let us now look at an example involving two nonperpendicular forces.

    The angle between forces βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ is 112∘, and the measure of the angle between their resultant and βƒ‘πΉοŠ¨ is 56∘. If the magnitude of βƒ‘πΉοŠ§ is 28 N, what is the magnitude of βƒ‘πΉοŠ¨?

    Answer

    The following figure shows the forces βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ and their resultant ⃑𝐹+βƒ‘πΉοŠ§οŠ¨. The forces act at a point 𝑃.

    The resultant forces βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ form a parallelogram whose diagonal through 𝑃 is the resultant.

    The angle, πœƒ, between βƒ‘πΉοŠ§ and the resultant of βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ is given by πœƒ=112βˆ’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 βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, and ⃑𝐹+βƒ‘πΉοŠ§οŠ¨, we find that ‖‖⃑𝐹‖‖56=‖‖⃑𝐹‖‖56,∘∘sinsin that is, ‖‖⃑𝐹‖‖=‖‖⃑𝐹‖‖.

    The magnitude of βƒ‘πΉοŠ§ is given as 28 N, so the magnitude of βƒ‘πΉοŠ¨ 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, βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, the resultant is the diagonal of the parallelogram formed by βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, with its tail being the point of application of βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨. 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=β€–β€–βƒ‘πΉβ€–β€–οŠ§οŠ§, 𝐹=β€–β€–βƒ‘πΉβ€–β€–οŠ¨οŠ¨, 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 βˆ’βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ will be 180βˆ’π›Όβˆ˜.

    The magnitude of βˆ’βƒ‘πΉοŠ§ is the same as the magnitude of βƒ‘πΉοŠ§, 𝐹.

    Applying the law of cosines in the triangle formed by βˆ’βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, and their resultant gives us 𝑅′=2𝐹+2𝐹(180βˆ’π›Ό),∘cos that is, 𝑅′=2πΉβˆ’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=2πΉβˆ’2𝐹𝛼=90.coscos

    This is true only if cos𝛼=0, that is, if 𝛼=90∘. Forces βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ are, thus, perpendicular.

    Hence, we have 2𝐹=90𝐹=ο„ž902𝐹=90ο„ž12𝐹=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, βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, acting on a body at the same point is a single force that is given by ⃑𝑅=⃑𝐹+⃑𝐹.
    • The combined effect of βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ is the same as the effect of only ⃑𝑅.
    • βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, 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 βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ and their resultant, ⃑𝑅, gives πΉπœƒ=πΉπœƒ=𝑅𝛼,sinsinsin where 𝐹, 𝐹, and 𝑅 are the magnitudes of βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, and ⃑𝑅, respectively, 𝛼 is the angle between forces βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, πœƒοŠ§ is the angle between ⃑𝑅 and βƒ‘πΉοŠ§, and πœƒοŠ¨ is the angle between ⃑𝑅 and βƒ‘πΉοŠ¨.
    • Applying the law of cosines in the triangle formed by two forces βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ and their resultant, ⃑𝑅, gives 𝑅=𝐹+𝐹+2𝐹𝐹𝛼,cos where 𝐹, 𝐹, and 𝑅 are the magnitudes of βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, and ⃑𝑅, respectively, and 𝛼 is the angle between forces βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨.
    • Let ⃑𝑅 be the resultant force of two forces, βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨, that act at a single point with an angle 𝛼 between them. Then, 𝑅=𝐹+𝐹+2πΉπΉπ›Όπœƒ=𝐹𝛼𝐹+𝐹𝛼,cosandtansincos where 𝐹, 𝐹, and 𝑅 are the magnitudes of βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, and ⃑𝑅, respectively, and πœƒ is the angle between ⃑𝑅 and βƒ‘πΉοŠ§.