Lesson Video: Resultant of Coplanar Forces | Nagwa Lesson Video: Resultant of Coplanar Forces | Nagwa

Lesson Video: Resultant of Coplanar Forces Mathematics

In this video, we will learn how to find the resultant of a group of forces acting at a point.

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Video Transcript

In this video, we will learn how to find the resultant of a group of forces acting at a point.

Suppose that multiple forces act at a point, such as in the figure. We call the net force exerted by the combination of all the forces the resultant force. In this case, it is the vector 𝐑. To calculate the resultant of any number of forces, we can add the forces together tip to tail. That is, we move the starting point, or tail, of 𝐅 two to the finishing point, or tip, of 𝐅 one. We move the tail of 𝐅 three to the tip of 𝐅 two, and so on for however many forces we have. The resultant force is given by the vector from the starting point to the tip of the final force, 𝐅 three. 𝐑 is the vector sum of the forces 𝐅 one, 𝐅 two, and 𝐅 three.

Another way to find the resultant is to add the perpendicular components of each force together. Suppose we have a force 𝐅 one that can be decomposed into the horizontal and vertical components 𝐅 one 𝑥 and 𝐅 one 𝑦. Similarly, we have a force 𝐅 two that can be decomposed into 𝐅 two 𝑥 and 𝐅 two 𝑦. Then, we can find the resultant of these two forces by adding the components as shown. 𝐑 is the vector sum of 𝐅 one and 𝐅 two. And the horizontal component of 𝐑, 𝐑 𝑥, is the sum of the horizontal components 𝐅 one 𝑥 and 𝐅 two 𝑥. The vertical component of 𝐑, 𝐑 𝑦, is the sum of the vertical components 𝐅 one 𝑦 and 𝐅 two 𝑦.

The horizontal component, 𝐅 𝑥, and the vertical component, 𝐅 𝑦, of a force 𝐅 can be found from the inclination angle of 𝐅. The horizontal component 𝐅 𝑥 is given by the magnitude of the force 𝐅 multiplied by the cosine of the adjacent angle 𝜃 or by the sine of the opposite angle 𝜙. The vertical component 𝐅 𝑦 is given by 𝐅 multiplied by sin 𝜃 or 𝐅 multiplied by cos 𝜙. The magnitude of the force 𝐅 is given by the square root of the sum of the components squared. From this, we can also derive that the tan of 𝜃 is equal to the vertical component 𝐅 𝑦 over the horizontal component 𝐅 𝑥. And it follows that 𝜃 is equal to the inverse tan of 𝐅 𝑦 over 𝐅 𝑥.

Let us now look at an example which includes multiple forces acting at a point.

A body has a force of 10 newtons acting on it horizontally, 25 newtons acting on it vertically upward, and five newtons acting on it at an angle of 45 degrees to the horizontal, as shown in the figure. What is the magnitude of the single resultant force acting on the body, and at what angle to the horizontal does it act? Give your answers correct to one decimal place.

To begin, the five-newton force can be decomposed into its horizontal and vertical components. We can then add these components to the other horizontal and vertical forces to find the resultant force. The horizontal component is equal to the magnitude, five, multiplied by the cosine of the adjacent angle, 45 degrees. The vertical component is equal to five multiplied by the sine of the adjacent angle, 45 degrees. Adding together all horizontal components, we get the horizontal component of the resultant force, 𝐅 𝑥, equals 10 plus five times cos 45 degrees.

Likewise, the vertical component of the resultant force, 𝐅 𝑦, is equal to 25 plus five times sin 45 degrees. Both cos of 45 degrees and sin of 45 degrees are equal to the square root of two over two. So these become 10 plus five times the square root of two over two and 25 plus five times the square root of two over two, respectively.

Now recall that the magnitude of any force in two dimensions is given by the positive square root of the sum of its horizontal and vertical components squared. The inclination angle 𝜃 is given by the inverse tangent of the vertical component, 𝐅 𝑦, over the horizontal component, 𝐅 𝑥. Therefore, the magnitude of the resultant force, 𝐅 𝑅, is given by the square root of 10 plus five times the square root of two over two squared plus 25 plus five times the square root of two over two squared. Performing this calculation, we obtain the magnitude of the resultant force to one decimal place is 31.6 newtons.

The angle of inclination 𝜃 is equal to the inverse tan of 25 plus five times the square root of two over two divided by 10 plus five times the square root of two over two. Performing this calculation gives us the angle the resultant force makes with the horizontal is 64.6 degrees to one decimal place.

Note, there is no limit to the number of forces that can act at a point. And forces that act at a point can be resolved into perpendicular components.

Let’s now look at an example of six forces acting at a point.

The diagram shows a regular hexagon, 𝐴𝐵𝐶𝐷𝐸𝐹, whose diagonals intersect at point 𝑀. The six forces shown acting at 𝑀 are measured in newtons. Find 𝑅, the magnitude of their resultant, and 𝜃, the angle between their resultant and the positive 𝑥-axis. Round your value of 𝜃 to the nearest minute if necessary.

The regular hexagon consists of six congruent equilateral triangles, which have an internal angle of 60 degrees. Consider the force acting along the line 𝑀𝐹. The horizontal component of the force will be given by the magnitude, 60 newtons, multiplied by the cosine of the angle the force makes with the positive 𝑥-axis, 60 degrees. Similarly, the vertical component of the force is equal to the magnitude, 60 newtons, multiplied by the sine of the angle the force makes with the positive 𝑥-axis, 60 degrees.

For the rest of the forces, the same formula will apply, but the angle will change. The angles the forces in the directions of 𝐸, 𝐷, 𝐶, 𝐵, and 𝐴 make with the positive 𝑥-axis are each 60 degrees greater than the last: 120 degrees, 180 degrees, 240 degrees, 300 degrees, and 360 degrees, respectively.

We can add these horizontal components together to give the horizontal component of the resultant force as shown. We can do the same for the vertical component of the force, this time taking the sine of the angles with the positive 𝑥-axis, as shown. These trigonometric components evaluate to fairly simple values, which we can substitute in.

We now perform some simplification on the horizontal component of the force 𝐅 𝑥. Performing this calculation, we find 𝐅 𝑥 is equal to five. We can do the same with the vertical components, for which the trigonometric components are also quite simple. Again, we perform some simplification steps. Here we find we can take out a common factor of the square root of three to simplify our answer. We then find that 𝐅 𝑦 is equal to nine times the square root of three. We will leave this in surd form to calculate the resultant force.

Now that we have the horizontal and vertical components of the resultant force, we can find its magnitude by taking the positive square root of the sum of these components squared. And we can find the inclination angle of the force by taking the inverse tangent of the vertical component over the horizontal component. For the magnitude of the force, this gives us the square root of five squared plus nine times the square root of three squared. This simplifies to two times the square root of 63 newtons.

For the inclination angle 𝜃, we have the inverse tan of nine times the square root of three over five. This is equal to approximately 72.21 degrees, which to the nearest minute is equal to 72 degrees and 13 minutes.

Forces can also be expressed in terms of their perpendicular components. In a 2D Cartesian coordinate system, this is done with scalar multiples of the unit vectors 𝐢 and 𝐣, which act in the 𝑥- and 𝑦-directions, respectively.

Let’s now take a look at an example of determining the resultant of multiple forces expressed in this way.

The resultant of forces 𝐅 one equals minus four 𝐢 plus two 𝐣 newtons, 𝐅 two equals five 𝐢 minus seven 𝐣 newtons, and 𝐅 three equals two 𝐢 plus nine 𝐣 newtons makes an angle of 𝜃 with the positive 𝑥-axis. Determine 𝑅, the magnitude of the resultant, and find the value of tan 𝜃.

Let us start with a force diagram of 𝐅 one, 𝐅 two, and 𝐅 three acting from a point. We can determine the horizontal and vertical components of the resultant by summing the horizontal and vertical components of the individual forces. The horizontal components of 𝐅 one, 𝐅 two, and 𝐅 three are given by the question as negative four, five, and two newtons, respectively. Likewise, the vertical components are given by two, negative seven, and nine newtons, respectively.

Therefore, the horizontal component of the resultant, 𝑅 𝑥, is given by negative four plus five plus two, which is equal to three newtons. Similarly, the vertical component of the resultant, 𝑅 𝑦, is given by two minus seven plus nine, which is equal to four newtons. This gives us that the resultant force 𝑅 is equal to three 𝐢 plus four 𝐣 newtons.

We can see a visual representation of this to scale on the diagram and note that seems correct. The magnitude of the resultant 𝑅 is given by the square root of the sum of the horizontal and vertical components, 𝑅 𝑥 and 𝑅 𝑦, squared. In this case, we have the square root of three squared plus four squared. This simplifies to the square root of 25, which is equal to five newtons.

The tangent of the angle that the resultant makes with the positive 𝑥-axis is given by the vertical component, 𝑅 𝑦, over the horizontal component, 𝑅 𝑥. In this case, therefore, tan 𝜃 is equal to four over three, which is four-thirds.

Since we have the precise components of these forces and we can draw them to scale, we can also find the resultant using the tip-to-tail method. We start by drawing 𝐅 one. We then draw 𝐅 two starting from the tip of 𝐅 one. Finally, we draw 𝐅 three starting from the tip of 𝐅 two.

The resultant force is given by the vector from the original starting point to the tip of the final force, 𝐅 three. On a scale diagram, we could measure the 𝑥- and 𝑦-components of 𝑅 to get the resultant force, which is equal to three 𝐢 plus four 𝐣 newtons. We note that this agrees with our previous result. We would calculate the magnitude of this vector using exactly the same method as we did previously. And we would find this is equal to five newtons. Likewise, we’d reach the same answer for tan 𝜃, which we’d calculate by taking the vertical component over the horizontal component. And this is four over three.

Let’s finish this video by recapping some key points. Multiple forces can be summed by summing the perpendicular components of the forces and determining the resultant of these components. If 𝐅 𝑥 is the horizontal component of a force 𝐅, the magnitude of 𝐅 𝑥 is given by the magnitude of 𝐅 multiplied by the cosine of the angle between 𝐅 and 𝐅 𝑥. We call this angle 𝜃.

Similarly, if 𝐅 𝑦 is the vertical component of the force 𝐅, which is perpendicular to 𝐅 𝑥, this is given by 𝐅 times sin 𝜃. The magnitude of the resultant of the perpendicular components of a force is given by 𝐅 equals the square root of 𝐅 𝑥 squared plus 𝐅 𝑦 squared. If a force 𝐅 𝐴 is given by 𝑥 𝑎 times 𝐢 plus 𝑦 𝑎 times 𝐣 plus 𝑧 𝑎 times 𝐤 and a force 𝐅 𝐵 is given by 𝑥 𝑏 times 𝐢 plus 𝑦 𝑏 times 𝐣 plus 𝑧 𝑏 times 𝐤, the resultant of these forces is given by 𝐅 𝐴 plus 𝐅 𝐵. This is equal to 𝑥 𝑎 plus 𝑥 𝑏 times 𝐢 plus 𝑦 𝑎 plus 𝑦 𝑏 times 𝐣 plus 𝑧 𝑎 plus 𝑧 𝑏 times 𝐤.

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