Lesson Video: Resultant of Parallel Coplanar Forces Mathematics

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

19:42

Video Transcript

In this video, we will learn how to find the resultant of a system of parallel coplanar forces and how to locate its point of action. We recall that multiple forces that act at a point sum to a resultant force that acts at that point. And the line of action of the resultant force intersects the lines of action of its components at the point where the forces act. In this video, however, we will deal with forces that do not act at the same point, as shown in the diagram.

Let’s begin by considering the scenario where we have three parallel coplanar forces, 𝐹 sub one, 𝐹 sub two, and 𝐹 sub three, as shown in the diagram. Finding the resultant force together with the point at which it acts is slightly more complicated than when the forces act at one point. There is, however, a step-by-step process we can apply.

As forces are vectors, they have direction as well as magnitude. So our first step is to set up a sign convention. In this example, we will let the positive direction be vertically upwards. Our next step is to find the sum of the parallel forces. Since 𝐹 sub one and 𝐹 sub three act upwards, they will be positive, whereas 𝐹 sub two acts downwards so will therefore be negative. The resultant force is equal to 𝐹 sub one minus 𝐹 sub two plus 𝐹 sub three.

Let’s now assume that the three forces have magnitude two newtons, three newtons, and four newtons, respectively. The resultant force is therefore equal to two minus three plus four, which is equal to three newtons. As this is positive, we know that the resultant force acts vertically upwards with magnitude three newtons.

We now need to consider where along our line this resultant force acts. In order to do this, we model the line as a light thin rod and then take moments about any point on the line. This is the third step of our process. Typically, when taking moments, we consider counterclockwise moments to be positive and clockwise moments to be negative. Whilst we can take moments about any point, typically, we choose a point where one of the forces acts. In this case, we’ll take moments about the point where the force 𝐹 sub one acts.

As already mentioned, we do not know where the resultant three-newton force acts. However, we will add it to our diagram a perpendicular distance of π‘₯ meters from where the force 𝐹 sub one acts. We know that we can calculate the moment of a force by multiplying the magnitude of the force by the perpendicular distance from the force to the point at which we are taking moments. This is usually written as 𝑀 is equal to 𝐹 multiplied by 𝑑.

Returning to our diagram and the general notation 𝐹 sub one, 𝐹 sub two, and 𝐹 sub three, we can add the distances 𝑑 sub two and 𝑑 sub three. These are the perpendicular distances of 𝐹 sub two and 𝐹 sub three from the point at which we are taking moments. As the 𝐹 sub one force is acting at the point at which we are taking moments, this will have a moment equal to zero. The resultant force is acting in a counterclockwise direction about the point at which we are taking moments. It will therefore have a positive moment equal to 𝑅 multiplied by π‘₯. The 𝐹 two newton force acts in a clockwise direction, so it will have a negative moment. This is equal to negative 𝐹 sub two multiplied by 𝑑 sub two. Finally, the 𝐹 sub three force is acting in a counterclockwise direction so will have a positive moment. This is equal to 𝐹 sub three multiplied by 𝑑 sub three.

If 𝑑 sub two was equal to two meters and 𝑑 sub three four meters, we can substitute all these values into our equation. This gives us three π‘₯ is equal to negative three multiplied by two plus four multiplied by four. The right-hand side simplifies to negative six plus 16, and this is equal to 10. We can then divide through by three such that π‘₯ is equal to 10 over three or ten-thirds. The resultant force 𝑅 acts a perpendicular distance of 10 over three meters from the 𝐹 sub one force.

This leads us to a general rule that we can use to calculate the perpendicular distance of the resultant from a point at which we are taking moments. The distance π‘₯ is equal to the sum of the moments divided by the resultant force. We will now look at some examples where we need to follow this four-step process.

Two parallel forces have magnitudes of 10 newtons and 20 newtons. The distance between their lines of action is 30 centimeters. 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.

We begin by noticing from the question that we are dealing with two parallel coplanar forces. Both of these forces act in the positive 𝑦-direction and have magnitudes of 10 and 20 newtons. The distance between the two forces is 30 centimeters. We are asked to find the resultant 𝑅 and the distance π‘₯ between the resultant’s line of action and point 𝐴.

We know that when dealing with parallel coplanar forces, the resultant force 𝑅 is equal to the sum of the other forces. Since both forces are acting in the positive 𝑦-direction, we have 𝑅 is equal to 10 plus 20. This is equal to 30. The resultant force 𝑅 is equal to 30 newtons and acts vertically upwards.

We don’t know at present where this resultant force acts. Let’s assume it acts a perpendicular distance of π‘₯ centimeters from 𝐴. By considering moments, we recall that this distance π‘₯ is equal to the sum of the moments divided by the resultant force 𝑅. We know that the moment of a force is equal to the magnitude of the force multiplied by the perpendicular distance to the point at which we are taking moments. We will consider moments in the counterclockwise direction to be positive and take moments about point 𝐴.

The moment of the 10-newton force will therefore be equal to zero, as this acts at that point. The only other force we need to consider is the 20-newton force. This will have a moment of 20 newtons, its magnitude, multiplied by 30 centimeters, its distance from point 𝐴. 20 multiplied by 30 is equal to 600. Therefore, the sum of the moments is equal to 600 newton centimeters. Substituting this into our formula, we have π‘₯ is equal to 600 divided by 30. This is the same as 60 divided by three. π‘₯ is therefore equal to 20 centimeters. The distance between the line of action of the resultant force and point 𝐴 is therefore equal to 20 centimeters.

We will now consider an example where the parallel forces act in opposite directions.

Two parallel forces have magnitudes of 24 newtons and 60 newtons as shown in the figure. The distance between their lines of action is 90 centimeters. 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.

In this question, we have two parallel coplanar forces acting in opposite directions. They have magnitudes 24 and 60 newtons, where the 24-newton force acts in the positive direction and the 60-newton force acts in the negative direction. We are also told that the distance between their lines of action is 90 centimeters. We have been asked to calculate the resultant force 𝑅 together with the distance π‘₯ between its line of action and point 𝐴.

We know that the resultant force is equal to the sum of the other forces. Since the positive direction is vertically upwards, this is equal to 24 plus negative 60, which is equal to negative 36. The resultant force 𝑅 is equal to negative 36 newtons. This means that it acts in the downward direction with magnitude 36.

We don’t know at present where this resultant force acts. But we can calculate the distance π‘₯ between its line of action and point 𝐴 using the formula π‘₯ is equal to the sum of the moments divided by the resultant force 𝑅, where the moment of any force can be calculated by multiplying the magnitude of that force by the perpendicular distance to the point at which we are taking moments.

In this question, we’ll take moments about point 𝐴, where moments acting in the counterclockwise direction are positive. The 24-newton force acts at point 𝐴, which means it is a distance of zero centimeters from point 𝐴. It will therefore have a moment equal to zero. This means that the only moment we need to consider is that of the 60-newton force. This acts in a clockwise direction about point 𝐴. Therefore, the moment is equal to negative 60 multiplied by 90, where 90 centimeters is the perpendicular distance of this force from point 𝐴. 60 multiplied by 90 is 5400. Therefore, negative 60 multiplied by 90 is negative 5400.

We can therefore calculate π‘₯ by dividing negative 5400 by negative 36. Dividing a negative number by a negative number gives a positive answer. Therefore, π‘₯ equals 150. The distance between the line of action of the resultant force and point 𝐴 is 150 centimeters.

We can see from this answer an interesting point about this question. The line of action of the resultant force does not lie between the line of action of our other forces. In practice, we could consider a light thin rod 𝐴𝐢 of length 150 centimeters. If forces of magnitude 24 newtons and 60 newtons are acting upon it as shown, then we would require a vertical force of magnitude 36 newtons at point 𝐢 in order to maintain the equilibrium.

We will now look at one final example where the parallel coplanar forces are joined by a line that is not perpendicular to their lines of action.

In the figure below, 𝐹 sub one and 𝐹 sub two are two parallel forces measured in newtons, where 𝑅 is their resultant. If 𝑅 is equal to 30 newtons, 𝐴𝐡 equals 36 centimeters, and 𝐡𝐢 equals 24 centimeters, determine the magnitude of 𝐹 sub one and 𝐹 sub two.

In this question, we have two parallel coplanar forces, 𝐹 sub one and 𝐹 sub two, acting in opposite directions. We are also given the resultant force 𝑅, which is equal to 30 newtons. The distance from point 𝐴 to 𝐡 is 36 centimeters, and the distance from 𝐡 to 𝐢 is 24 centimeters. The lines of action of our forces 𝐹 sub one, 𝐹 sub two, and 𝑅 are not perpendicular to the line segment 𝐴𝐢. However, as our three forces are parallel, we can add the angle πœƒ to our diagram as shown.

We can calculate the perpendicular components of these forces using our knowledge of right angle trigonometry. These are equal to 𝐹 sub one sin πœƒ, 𝐹 sub two sin πœƒ, and 𝑅 sin πœƒ. These will be useful when we come to take moments, as the moment of a force is equal to the magnitude of force multiplied by the perpendicular distance to the point at which we are taking moments. Whilst we can take moments about any point on our line, in this question, we will do so about point 𝐴.

We will consider moments acting in the counterclockwise direction to be positive and those acting in the clockwise direction to be negative. This means that the moment 𝑀 sub one of the force 𝐹 sub one acts in the positive direction and is equal to 𝐹 sub one sin πœƒ multiplied by 36. We can repeat this process for 𝑀 sub two, which is the moment of the force 𝐹 sub two. As this acts in the negative direction, this is equal to negative 𝐹 sub two sin πœƒ multiplied by 60.

Our expressions for 𝑀 sub one and 𝑀 sub two can be simplified as shown. We know that the distance π‘₯ from the line of action of the resultant force to the point at which we are taking moments is equal to the sum of the moments divided by 𝑅. In this case, we are taking moments about the point where the resultant acts. Therefore, π‘₯ is equal to zero. This means that the sum of our two moments must equal zero. 36 multiplied by 𝐹 sub one sin πœƒ plus negative 60 multiplied by 𝐹 sub two sin πœƒ equals zero. Since sin πœƒ cannot be equal to zero, we can divide through by this. We can also divide through by 12 such that three 𝐹 sub one minus five 𝐹 sub two equals zero. As there are two unknowns here, we will call this equation one.

We will now consider the resultant force and the fact that this is equal to the sum of the other forces. Going back to our initial diagram, if we let the positive direction be vertically upwards, we have 𝑅 is equal to 𝐹 sub one plus negative 𝐹 sub two. Since 𝑅 is equal to 30 newtons, we have 30 is equal to 𝐹 sub one minus 𝐹 sub two. Adding 𝐹 sub two to both sides of this equation, we have 𝐹 sub one is equal to 30 plus 𝐹 sub two. We will call this equation two. And we now have a pair of simultaneous equations that we can solve by substitution.

One way of doing this is to substitute the expression for 𝐹 sub one in equation two into equation one. This gives us three multiplied by 30 plus 𝐹 sub two minus five 𝐹 sub two is equal to zero. Distributing the parentheses gives us 90 plus three 𝐹 sub two. The left-hand side then simplifies to 90 minus two 𝐹 sub two. By adding two 𝐹 sub two to both sides and then dividing through by two, we have 𝐹 sub two is equal to 45. Substituting this value back into equation two gives us 𝐹 sub one is equal to 30 plus 45, which is equal to 75. The magnitude of the two forces 𝐹 sub one and 𝐹 sub two are 75 newtons and 45 newtons, respectively.

We will now summarize the key points from this video. We saw in this video that parallel coplanar forces can be combined into a resultant force 𝑅 such that 𝑅 is equal to the sum of 𝐹. The point of action of 𝑅 is the point at which the moments about the point due to the forces sum to zero. This means that we can calculate the distance π‘₯ from the line of action of the resultant force to any point on the line by finding the sum of the moments and dividing it by the resultant force 𝑅.

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