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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 βπΉ. For example, the diagram below shows the force βπΉ represented by the directed line segment ο«π΄π΅. The magnitude of the force is determined by ββο π΄π΅ββ. The direction of the arrow corresponds to the direction of βπΉ. The point of action is π΄. The line of action is indicated by extending ο«π΄π΅ in the same direction (as shown by the dotted line). A force acting on a body is represented by vector βπΉ. When two forces act on a body, we call their resultant the force that describes their combined effect. When two forces, βπΉο§ and βπΉο¨, 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, βπ , is given by βπ =βπΉ+βπΉ.ο§ο¨ The vector equality βπ =βπΉ+βπΉο§ο¨ can be represented in two ways, as illustrated in the following diagram. As βπΉο§, βπΉο¨, and βπ 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 πΌ be the angle between forces βπΉο§ and βπΉο¨, πο§ the angle between βπ and βπΉο§, and πο¨ the angle between βπ and βπΉο¨, as shown in the diagram below. The law of sines in this triangle gives us πΉπ=πΉπ=π (180βπΌ),ο§ο¨ο¨ο§βsinsinsin where πΉο§, πΉο¨, and π are the magnitudes of βπΉο§, βπΉο¨, and βπ respectively. As sinsin(180βπ₯)=π₯β for all π₯, we find the relationship given in the following box. We have πΉπ=πΉπ=π πΌ,ο§ο¨ο¨ο§sinsinsin where πΉο§, πΉο¨, and π are the magnitudes of βπΉο§, βπΉο¨, and βπ , respectively, πΌ is the angle between forces βπΉο§ and βπΉο¨, πο§ is the angle between βπ and βπΉο§, and πο¨ is the angle between βπ and βπΉο¨. Applying the law of cosines in our triangle now, we find that π =πΉ+πΉβ2πΉπΉ(180βπΌ).ο¨ο§ο¨ο¨ο¨ο§ο¨βcos As coscos(180βπ₯)=βπ₯β for all π₯, we find the relationship given in the following box. We have π =πΉ+πΉ+2πΉπΉπΌ,ο¨ο§ο¨ο¨ο¨ο§ο¨cos where πΉο§, πΉο¨, and π are the magnitudes of βπΉο§, βπΉο¨, and βπ , respectively, and πΌ is the angle between forces βπΉο§ and βπΉο¨. 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 π , the magnitude of βπ . It is also straightforward to derive an accompanying formula for the direction of βπ . We state these results below. Let βπ be the resultant force of two forces, βπΉο§ and βπΉο¨, that act at a single point with an angle πΌ between them. Then, π =οπΉ+πΉ+2πΉπΉπΌπ=πΉπΌπΉ+πΉπΌ,ο¨ο§ο¨ο¨ο§ο¨ο¨ο§ο¨cosandtansincos where πΉο§, πΉο¨, and π are the magnitudes of βπΉο§, βπΉο¨, and βπ , respectively, and π is the angle between βπ and βπΉο§. 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. AnswerIt will be convenient to assume that the first force acts horizontally. Let us call this force βπΉο§ and the other force βπΉο¨. The resultant of these forces, βπΉ+βπΉο§ο¨, acts vertically as it is perpendicular to βπΉο§, as shown in the following figure. The force βπΉο¨ can be represented by an arrow with its tail at the head of βπΉο§ and its head at the head of βπΉ+βπΉο§ο¨, as shown in the following figure. The resultant force βπ is given by βπ =βπΉ+βπΉ.ο§ο¨ As βπΉο§ and βπ are perpendicular, we see that the two forces and their resultant form a right triangle. Therefore, applying the Pythagorean theorem gives βββπΉββ+βββπ ββ=βββπΉββ.ο§ο¨ο¨ο¨ο¨ It is worth noting that the Pythagorean theorem is just a special case of the law of cosines. Substituting in the values of βββπΉββο§ and βββπΉββο¨, we find that 35+βββπ ββ=91βββπ ββ=91β35=7056βββπ ββ=β7056=84.ο¨ο¨ο¨ο¨ο¨ο¨N Note that as the magnitude of a vector is always positive, β84 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 with the 88 N force. Find the value of sinπ. AnswerIt will be convenient to assume that one of the forces acts horizontally. Let us call this force βπΉο§ and the other force βπΉο¨, as shown in the following figure. By choosing to make βπΉο§ correspond to the line adjacent to π, we have chosen this force to be the 88-newton force. The magnitude of βπΉο¨ is 44 newtons; therefore, the magnitude of βπΉο¨ is half that of βπΉο§. The magnitude of the resultant of the forces, π , can be expressed as π =οπΉ+πΉ=οπΉ+ο½πΉ2ο.ο§ο¨ο¨ο¨ο§ο¨ο§ο¨ We can see from this that π =πΉ+ο½πΉ2ο=πΉ+πΉ4=οΌ54οπΉ.ο¨ο§ο¨ο§ο¨ο§ο¨ο§ο¨ο§ο¨ Taking square roots, we have that π =β5πΉ2.ο§ Applying the law of sines in the triangle gives sinsinππΉ=90π .ο¨β As sin90=1β, we have sinsinπ=πΉπ π==1β5=β55.ο¨ο₯ο¨βο«ο₯ο¨ο ο We have, therefore, that sinπ=β55. 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, βπΉο§ and βπΉο¨, act at a point. Their resultant, βπ , has magnitude 188 N and makes an angle of 60β with βπΉο§. Find the magnitudes of βπΉο§ and βπΉο¨. AnswerThe perpendicular forces, βπΉο§ and βπΉο¨, and their resultant are shown in the following figure. We see that βπΉο§ and βπΉο¨ are perpendicular and the resultant βπΉ+βπΉο§ο¨ makes an angle of 60β with βπΉο§. As we have a right triangle, we have coscosN60=βββπΉββ188βββπΉββ=18860=12Γ188=94βο§ο§β and sinsinN60=βββπΉββ188βββπΉββ=18860=β32Γ188=94β3.βο¨ο¨β βπΉο§ has a magnitude of 94 N, and βπΉο¨ has a magnitude of 94β3 N. Let us now look at an example involving two nonperpendicular forces. The angle between forces βπΉο§ and βπΉο¨ is 112β, and the measure of the angle between their resultant and βπΉο¨ is 56β. If the magnitude of βπΉο§ is 28 N, what is the magnitude of βπΉο¨? AnswerThe following figure shows the forces βπΉο§ and βπΉο¨ and their resultant βπΉ+βπΉο§ο¨. The forces act at a point π. The resultant forces βπΉο§ and βπΉο¨ form a parallelogram whose diagonal through π is the resultant. The angle, π, between βπΉο§ and the resultant of βπΉο§ and βπΉο¨ is given by π=112β56=56.β 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 βπΉο§, βπΉο¨, and βπΉ+βπΉο§ο¨, we find that βββπΉββ56=βββπΉββ56,ο§βο¨βsinsin that is, βββπΉββ=βββπΉββ.ο§ο¨ The magnitude of βπΉο§ is given as 28 N, so the magnitude of βπΉο¨ 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 πΉ 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 πΉ. AnswerLet us represent the first situation. When we add two forces, βπΉο§ and βπΉο¨, the resultant is the diagonal of the parallelogram formed by βπΉο§ and βπΉο¨, with its tail being the point of application of βπΉο§ and βπΉο¨. 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 π =πΉ+πΉ+2πΉπΉπΌ,ο¨ο§ο¨ο¨ο¨ο§ο¨cos with πΉ=βββπΉββο§ο§, πΉ=βββπΉββο¨ο¨, and π =βββπ ββ. Since πΉ=πΉ=πΉο§ο¨, we have π =2πΉ+2πΉπΌ.ο¨ο¨ο¨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 ββπΉο§ and βπΉο¨ will be 180βπΌβ. The magnitude of ββπΉο§ is the same as the magnitude of βπΉο§, πΉ. Applying the law of cosines in the triangle formed by ββπΉο§, βπΉο¨, and their resultant gives us π β²=2πΉ+2πΉ(180βπΌ),ο¨ο¨ο¨βcos that is, π β²=2πΉβ2πΉπΌ.ο¨ο¨ο¨cos We are told that the magnitude of the resultant is the same in both cases, 90 N. Hence, we have π =π β²=90,N which means that 2πΉ+2πΉπΌ=2πΉβ2πΉπΌ=90.ο¨ο¨ο¨ο¨ο¨coscos This is true only if cosπΌ=0, that is, if πΌ=90β. Forces βπΉο§ and βπΉο¨ are, thus, perpendicular. Hence, we have 2πΉ=90πΉ=ο902πΉ=90ο12πΉ=45β2.ο¨ο¨ο¨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.
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