Lesson Explainer: Resultant of Parallel Coplanar Forces Mathematics

In this explainer, we will learn how to find the resultant of a system of parallel coplanar forces and how to locate its point of action.

Multiple forces that act at a point sum to a resultant force that acts at that point. The line of action of the resultant force intersects the lines of action of its components at the point where the forces act.

Consider, however, the forces shown in the following figure.

The forces shown do not act at the same point, and so, it is not immediately obvious what would be meant by the resultant of these forces. If a resultant can be defined for these forces, it is obvious that it will act in the same direction as either of the forces. What is not obvious is the position of the point that the resultant force will act at.

Let us suppose that the dashed line shown in the figure, which is perpendicular to the tails of the arrows representing the forces, is actually a light, thin rod. The forces would act on the rod to generate a net moment on the rod about points on the rod. The moment, 𝑀, due to a force, 𝐹, about a point equals the product of the magnitude of the force and the perpendicular distance, 𝑑, from the line of action of the force to the point about which moments are taken. This can be expressed as 𝑀=𝐹𝑑.

Let us assume that the forces shown in the figure have equal magnitudes. If moments are taken about the midpoint of the rod, π‘š, the forces produce moments about π‘š of equal magnitudes. One of the moments will be counterclockwise, hence positive, while the other will be clockwise, hence negative. The sum of these moments will be given by 𝑀=𝑀+(βˆ’π‘€)=0.net

The rod will, therefore, be in rotational equilibrium.

Let us now consider a force, 𝐹, acting vertically downward on the rod with magnitude 2𝐹. Taking vertically upward as positive, the net force on the rod would be given by 𝐹+πΉβˆ’πΉ=2πΉβˆ’2𝐹=0.

𝐹 acts to exactly oppose the upward forces and so satisfies a necessary condition for being the negative equivalent of the resultant of the upward forces. This is, however, not the only condition that must be satisfied. The action of the upward forces on the rod produces zero moment about π‘š, and so, for 𝐹 to be considered the negative equivalent of the resultant of the upward forces, it must also produce a zero net moment about π‘š.

𝐹 must act at a particular point on the rod. Depending on the point at which 𝐹 acts, it may produce a nonzero moment about π‘š, and if 𝐹 produces a nonzero moment about π‘š, then 𝐹 cannot be considered to be the negative equivalent of the resultant of the upward forces. We can then ask: at what point, or points, on the rod can 𝐹 act that produces a zero net moment about π‘š?

If 𝐹 acts at π‘š, the moment about π‘š due to 𝐹 is necessarily zero. If 𝐹 acts at any point other than π‘š, it produces a nonzero moment about π‘š. We see, therefore, that π‘š is the only point at which a force can act that allows it to be considered the negative equivalent to the resultant of the upward forces. This tells us that π‘š satisfies all necessary conditions to be considered to be the point at which the resultant of the upward forces acts. The line of action of the resultant of the upward forces is parallel to the upward forces and intersects π‘š, as shown in the following figure.

If the upward forces on the rod were not equal, the line of action of the resultant upward force would not be at π‘š, but there would still be a unique point corresponding to the point at which the resultant upward force acted. This would correspond to the point at which the clockwise and counterclockwise moments due to the upward forces were equal.

Let us look at an example where the resultant of unequal parallel forces is determined.

Example 1: Finding the Resultant Force and Point of Action of Two Parallel Forces Acting in the Same Direction

Two parallel forces have magnitudes of 10 N and 20 N. The distance between their lines of action is 30 cm. If the two forces are acting in the same direction, find their resultant ⃑𝑅 and the distance π‘₯ between its line of action and point 𝐴.

Answer

The magnitude of the resultant force is the sum of the magnitudes of the upward forces; hence, ⃑𝑅=10+20=30.N

The point at which ⃑𝑅 acts is a point 𝑃 that is an element of 𝐴𝐡. At 𝑃, the moments due to the forces at 𝐴 and 𝐡 about 𝑃 are of equal magnitudes. The distance between 𝐴 and 𝐡 is 30 cm, and the distance between 𝐴 and 𝑃 is π‘₯ cm. At 𝑃, it is the case that the moments due to the forces at 𝐴 and 𝐡 are given by 𝑀+(βˆ’π‘€)=0;

hence, 𝑀=π‘€οŒ οŒ‘ and, therefore, 𝐹𝑑=𝐹𝑑, where 𝐹 and 𝐹 are the magnitudes of the forces and π‘‘οŒ  and π‘‘οŒ‘ are the distances of the forces from the point of action of the resultant of the forces. The length of 𝐴𝐡 is 30 cm, so we have that 10π‘₯=20(30βˆ’π‘₯)10π‘₯=600βˆ’20π‘₯30π‘₯=600π‘₯=20.cm

Let us now consider an example in which parallel forces act in opposite directions.

Example 2: Finding the Resultant Force and Point of Action of Two Parallel Forces Acting in Opposite Directions

Two parallel forces have magnitudes of 24 N and 60 N as shown in the figure. The distance between their lines of action is 90 cm. Given that the two forces are acting in opposite directions, determine their resultant ⃑𝑅 and the distance π‘₯ between its line of action and point 𝐴.

Answer

The magnitude of the resultant force is the sum of the magnitudes of the forces, where upward is taken as positive; hence, ⃑𝑅=24βˆ’60=βˆ’36.N

At any point along 𝐴𝐡, the moment due to either of the forces is either zero or clockwise. This tells us that there is no point along 𝐴𝐡 that can intersect the line of action of the resultant of the forces. This does not mean that the resultant of these forces cannot be defined, just that the line of action of the resultant is at a point that is not part of 𝐴𝐡.

The point at which ⃑𝑅 acts is a point 𝑃. At 𝑃, the moments due to the forces at 𝐴 and 𝐡 about 𝑃 are equal. The distance between 𝐴 and 𝐡 is 90 cm, and the distance between 𝐴 and 𝑃 is π‘₯; hence, at 𝑃, 𝐹𝑑=𝐹𝑑, where 𝐹 and 𝐹 are the magnitudes of the forces (with upward force taken as positive) and π‘‘οŒ  and π‘‘οŒ‘ are the distances of the forces from the point of action of the resultant of the forces. The length of 𝐴𝐡 is 90 cm, so we have that 24π‘₯=βˆ’60(90βˆ’π‘₯)24π‘₯=60π‘₯βˆ’5400βˆ’36π‘₯=βˆ’5400π‘₯=150.cm

We see from this that the line of action of the resultant of two parallel forces is not necessarily located between the lines of action of the forces. This can be understood as meaning that, for a light, thin rod of length 150 cm, if the forces in the question acted on the rod and an upward force of 36 N acted at the opposite end of the rod from 𝐴, the rod would be in equilibrium. This is shown in the following figure.

Let us now look at an example in which the resultant of two parallel forces of unknown magnitudes is known.

Example 3: Finding the Magnitude of Two Parallel Forces given Their Resultant and Point of Action

In the figure below, 𝐹 and 𝐹 are two parallel forces measured in newtons, where ⃑𝑅 their resultant. If ⃑𝑅=30N, 𝐴𝐡=36cm, and 𝐡𝐢=24cm, determine the magnitude of 𝐹 and 𝐹.

Answer

The moment about 𝐴 due to ⃑𝑅 is zero, as ⃑𝑅 acts at 𝐴, and so the sum of the moments of 𝐹 and 𝐹 about 𝐴 must be zero.

The lines of action of 𝐹 and 𝐹 are not perpendicular to 𝐴𝐢, so the moments due to 𝐹 and 𝐹 are the moments due to their components perpendicular to 𝐴𝐢. The angle that the lines of action of 𝐹 and 𝐹 make with 𝐴𝐢 can be denoted by πœƒ as shown in the following figure.

We note that the resultant force, ⃑𝑅, is given by ⃑𝑅=πΉβˆ’πΉ.

The moments due to the resultant force and due to 𝐹 act clockwise, while the moment due to 𝐹 acts counterclockwise; hence, the moment about 𝐴 due to 𝐹 is given by 𝑀=36πΉπœƒ=36(30+𝐹)πœƒπ‘€=(1080+36𝐹)πœƒ.sinsinsin

The length of 𝐴𝐢 is given by 𝐴𝐢=36+24=60,cm and so, the moment about 𝐴 due to 𝐹 is given by 𝑀=(60𝐹)πœƒ.sin

The moments about 𝐴 sum to zero, where π‘€οŠ§ is positive and π‘€οŠ¨ is negative, so we can see that π‘€βˆ’π‘€=0𝑀=𝑀(1080+36𝐹)πœƒ=(60𝐹)πœƒ.sinsin

The factor of sinπœƒ can be eliminated by dividing both sides of the equation by sinπœƒ, leaving 1080+36𝐹=60𝐹24𝐹=1080𝐹=108024=45.N

The magnitude of the resultant force is the sum of the magnitudes of the forces, where upward is taken as positive; hence, ⃑𝑅=30=πΉβˆ’πΉ.

This expression can be rearranged to give an expression for 𝐹 in terms of 𝐹: 𝐹=30+𝐹𝐹=30+45=75.N

Let us now look at an example in which the resultant of an arbitrary number of parallel forces is determined.

Example 4: Finding the Resultant and Point of Action of Five Parallel Forces

Points 𝐴, 𝐡, 𝐢, 𝐷, and 𝐸 are lying in the same straight line such that 𝐴𝐡=8cm, 𝐡𝐢=18cm, 𝐢𝐷=12cm, and 𝐷𝐸=11cm. Five forces of magnitudes 40, 25, 20, 45, and 50 newtons are acting as shown in the figure. Determine their resultant ⃑𝑅 and the distance π‘₯ between its line of action and point 𝐴.

Answer

The magnitude of ⃑𝑅 is given by the sum of the forces acting. Taking upward as positive, this is given by ⃑𝑅=40+25βˆ’20βˆ’45+50=50.N

We can define a parameter π‘₯ that is equal to the distance in the positive π‘₯-direction between the line of action of the 40-newton force and the line of action of ⃑𝑅. The moment due to the 40-newton force is therefore given by 𝑀=40π‘₯.

The distance between the lines of action of the other forces and the line of action of ⃑𝑅 can also be expressed in terms of π‘₯. For example, the line of action of the 25-newton force has a distance from 𝐴 of 8 cm in the positive π‘₯-direction. The distance between the line of action of ⃑𝑅 and the line of action of the 25-newton force is given by 𝑑=(π‘₯βˆ’8).

The moment due to the 25-newton force is, therefore, given by 𝑀=25(π‘₯βˆ’8).

Expressing the moments due to all of the forces this way, we obtain 𝑀=40π‘₯+25(π‘₯βˆ’8)βˆ’20(π‘₯βˆ’26)βˆ’45(π‘₯βˆ’38)+50(π‘₯βˆ’49).

The sum of the moments due to the forces is zero. If we distribute the parentheses throughout, then separate the integer terms from the terms that are multiples of π‘₯, we find that 40π‘₯+25π‘₯βˆ’20π‘₯βˆ’45π‘₯+50π‘₯=200βˆ’520βˆ’1710+245050π‘₯=420π‘₯=42050=8.4.cm

Let us now look at an example in which two parallel forces are represented by their perpendicular components.

Example 5: Finding the Resultant and Point of Action of Two Parallel Forces Given in Vector Form

Given that the two parallel forces ⃑𝐹=2⃑𝑖+βƒ‘π‘—οŠ§ and ⃑𝐹=βˆ’4βƒ‘π‘–βˆ’2βƒ‘π‘—οŠ¨ are acting at 𝐴(βˆ’3,βˆ’5) and 𝐡(5,3), respectively, determine their resultant ⃑𝑅, and find its point of action.

Answer

The magnitudes of the components’ resultant force is the sum of the magnitudes of the components of the forces, where upward is taken as positive; hence, ⃑𝑅=(2βˆ’4)⃑𝑖+(1βˆ’2)⃑𝑗=βˆ’2βƒ‘π‘–βˆ’βƒ‘π‘—.

The following figure shows the forces acting at 𝐴 and 𝐡.

There is an angle of magnitude πœƒ between the line of action of either force and a line that intersects the points of action of both forces. It is possible to determine πœƒ from the information in the question, but this is actually not necessary. Determining the point at which the resultant acts will require determining the ratios of the moments due to the forces about this point. Multiplying both forces’ magnitudes by sinπœƒ will not change this ratio.

The magnitude of βƒ‘πΉοŠ§ is given by 𝐹=√2+1=√5.

The magnitude of βƒ‘πΉοŠ¨ is given by 𝐹=(βˆ’4)+(βˆ’2)=√20.

The forces do not act perpendicularly to 𝐴𝐡, so the components of the forces perpendicular to 𝐴𝐡 are given by 𝐹=√5πœƒοŠ§βŸ‚sin and 𝐹=√20πœƒ.οŠ¨βŸ‚sin

The force components acting perpendicularly to 𝐴𝐡 can be shown with 𝐴𝐡 drawn horizontally, as shown in the following figure, which also shows the distance of 𝐴 from 𝐡.

We can see from the figure that the lines from 𝐴 and 𝐡 that are parallel to the π‘₯- and 𝑦-axes, respectively, both have a length of 8. These lines are perpendicular and so form two legs of a right triangle that has a hypotenuse with a length given by β„Ž=√8+8=√128β„Ž=8ο„ž1288=8√2.

From the figure, we can see that π‘₯, the distance along 𝐴𝐡 from 𝐴, to the point at which the resultant acts, is given by equating the moments due to the forces at 𝐴 and at 𝐡: √5π‘₯πœƒ=βˆ’βˆš20πœƒο€»8√2βˆ’π‘₯.sinsin

The factor of sinπœƒ can be eliminated by dividing both sides of the equation by sinπœƒ, leaving √5π‘₯=βˆ’βˆš20ο€»8√2βˆ’π‘₯ο‡βˆš5π‘₯=√20π‘₯βˆ’8√20√2ο€»βˆš5βˆ’βˆš20π‘₯=βˆ’8√20√2π‘₯=βˆ’8√20√2ο€»βˆš5βˆ’βˆš20.

This expression for π‘₯ can be simplified as follows: π‘₯=βˆ’8√40ο€½βˆš5βˆ’2=βˆ’8√40ο€»βˆš5βˆ’2√5π‘₯=8√8=8ο€Ώ2ο„ž82=16√2.

The length of 𝐴𝐡 is 8√2, so the resultant acts at a point that is at a distance 2𝐴𝐡 from 𝐴 in the direction of 𝐴𝐡.

The distance of 𝐡 from 𝐴 is 8 in the π‘₯-direction and is also 8 in the 𝑦-direction; hence, the point of action of ⃑𝑅 is at a distance of 2(8)=16 in both the π‘₯- and 𝑦-directions from 𝐴. The coordinates of the point of action of ⃑𝑅 are therefore given by (βˆ’3+16,βˆ’5+16)=(13,11).

Let us summarize what we have learned in these examples.

Key Points

  • The point of action of the resultant of multiple parallel forces is the point at which the moments about the point due to the forces sum to zero.
  • If the perpendicular distance between the lines of action of parallel forces is a line segment 𝐴𝐡, the point of action of the resultant of the forces is not necessarily an element of 𝐴𝐡.
  • The position of the point of action of the resultant of parallel forces is not affected by the angle between the lines of action of the forces and the line that intersects the points of action of the forces.

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