Lesson Explainer: Resultant of Coplanar Forces Mathematics

In this explainer, 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 below.

We call the net force exerted by the combination of all the forces the resultant force. In this instance, this is vector ⃑𝑅 shown below.

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 ⃑𝐹īŠ¨ to the finishing point (or tip) of ⃑𝐹īŠ§, we move the tail of ⃑𝐹īŠŠ to the tip of ⃑𝐹īŠ¨, and so on for however many forces we have. The result of this is shown below.

We note that this gives us a polygon of forces where the resultant force is the side formed by connecting the starting point to the tip of the final force, ⃑𝐹īŠŠ.

Another way to find the resultant is to add the perpendicular components of each force together. Suppose we have a force ⃑𝐹īŠ§ that can be decomposed into the horizontal and vertical components ⃑𝐹īŠ§ī‚ and ⃑𝐹īŠ§ī“, and similarly a force ⃑𝐹īŠ¨ that can be decomposed into ⃑𝐹īŠ¨ī‚ and ⃑𝐹īŠ¨ī“. Then, we can find the resultant of these two forces by adding the components, as shown.

The perpendicular components of a force 𝐹 are labeled 𝐹īŠ§ and 𝐹īŠ¨, as shown in the following figure.

The magnitude of 𝐹īŠ§ is given by 𝐹=𝐹(𝜃)=𝐹(𝜙),īŠ§cossin where 𝜃 is the angle between 𝐹 and 𝐹īŠ§ and 𝜙 is the angle between 𝐹 and 𝐹īŠ¨.

𝐹īŠ¨, the component perpendicular to 𝐹īŠ§, is given by 𝐹=𝐹(𝜃)=𝐹(𝜙).īŠ¨sincos

The magnitude of the resultant of the perpendicular components of a force is given by 𝐹=ī„đš+𝐹.īŠ¨īŠ§īŠ¨īŠ¨

Let us look at an example of multiple forces acting at a point.

Example 1: Finding the Magnitude and Direction of the Resultant Force of Three Forces Acting on a Body

A body has a force of 10 newtons acting on it horizontally, 25 newtons acting on it vertically upward, and 5 newtons acting on it at an angle of 45∘ 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.

Answer

The 5-newton force can be resolved into perpendicular components parallel to the đ‘Ĩ- and đ‘Ļ-axes, as shown in the following figure.

The net force acting in the đ‘Ĩ-direction is given by 𝐹=10+5(45)=10+5ī€ŋ√22ī‹.ī—cos

The net force acting in the đ‘Ļ-direction is given by 𝐹=25+5(45)=25+5ī€ŋ√22ī‹.ī˜sin

⃑𝐹ī— and ⃑𝐹ī˜ correspond to the two shorter legs of a right triangle, as shown in the following figure.

The length of the hypotenuse of this triangle equals the magnitude of the resultant force, which is, therefore, given by 𝐹=ī€ŋ10+5ī€ŋ√22ī‹ī‹+ī€ŋ25+5ī€ŋ√22ī‹ī‹đš=ī„Ąī„Ŗī„Ŗī„ ī€ŋ10+5ī€ŋ√22ī‹ī‹+ī€ŋ25+5ī€ŋ√22ī‹ī‹.īŠ¨īŒąīŠ¨īŠ¨īŒąīŠ¨īŠ¨

Only the positive root of 𝐹īŠ¨īŒą is used, as the magnitude of a force is necessarily positive.

To one decimal place, this is 31.6 newtons.

The angle 𝜃 from the horizontal at which the resultant force acts can be determined using the equation tantan𝜃=𝐹𝐹𝜃=ī€Ŋ25+5ī€Ŋī‰ī‰ī€Ŋ10+5ī€Ŋī‰ī‰.ī˜ī—√īŠ¨īŠ¨âˆšīŠ¨īŠ¨

We can use this formula to obtain 𝜃: 𝜃=⎛⎜⎜⎝ī€Ŋ25+5ī€Ŋī‰ī‰ī€Ŋ10+5ī€Ŋī‰ī‰âŽžâŽŸâŽŸâŽ .arctan√īŠ¨īŠ¨âˆšīŠ¨īŠ¨

To one decimal place, this is 64.6∘.

There is no limit to the number of forces that can act at a point. Each force that acts at a point can be resolved into perpendicular components. Let us look at an example of six forces acting at a point.

Example 2: Finding the Resultant of Six Forces Acting on a Regular Hexagon at a Point

The diagram shows a regular hexagon, 𝐴đĩđļ𝐷𝐸𝐹, whose diagonals intersect at point 𝑀. The 6 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 đ‘Ĩ-axis. Round your value of 𝜃 to the nearest minute if necessary.

Answer

The hexagon consists of 6 equilateral triangles, hence the angle between two forces that correspond to sides of any one of these triangles is 60∘.

Taking angles from the đ‘Ĩ-axis clockwise, the net force acting in the đ‘Ĩ-direction is given by 𝐹=63+27(60)+44(120)−65+29(240)+60(300)𝐹=63+272−22−65−292+30=5.ī—ī—coscoscoscosN

Recalling that coscos(90)=(270)=0, the net force acting in the đ‘Ļ-direction is given by 𝐹=60(30)+27(150)+44(210)+29(330)𝐹=60ī€ŋ√32ī‹âˆ’27ī€ŋ√32ī‹âˆ’44ī€ŋ√32ī‹+29ī€ŋ√32ī‹đš=(60+29−27−44)ī€ŋ√32ī‹=18ī€ŋ√32ī‹=9√3.ī˜ī˜ī˜coscoscoscos

⃑𝐹ī— and ⃑𝐹ī˜ correspond to the two shorter legs of a right triangle. The length of the hypotenuse of this triangle equals the magnitude of the resultant force, which is, therefore, given by 𝐹=(5)+ī€ģ9√3ī‡đš=25+243=268𝐹=√268=2ī„ž2682=2√67.īŠ¨īŒąīŠ¨īŠ¨īŠ¨īŒąīŒąīŠ¨N

Only the positive root of 𝐹īŠ¨īŒą is used, as the magnitude of a force is necessarily positive.

The angle 𝜃 from the horizontal at which the resultant force acts can be determined using the equation 𝜃=ī€ŋ9√35ī‹.arctan

To the nearest minute, this is 7213′∘.

A force can be expressed in terms of its perpendicular components, where each component is expressed as the length of a unit vector in a direction perpendicular to that of the other component. Let us look at an example in which the resultant of forces expressed in such a way is determined.

Example 3: Finding the Magnitude and Direction of the Resultant of Three Forces Acting at a Point

The resultant of forces ⃑𝐹=ī€ē−4⃑𝑖+2⃑𝑗ī†īŠ§N, ⃑𝐹=ī€ē5⃑𝑖−7⃑𝑗ī†īŠ¨N, and ⃑𝐹=ī€ē2⃑𝑖+9⃑𝑗ī†īŠŠN makes an angle 𝜃 with the positive đ‘Ĩ-axis. Determine 𝑅, the magnitude of the resultant, and the value of tan𝜃.

Answer

The resultant of ⃑𝐹īŠ§, ⃑𝐹īŠ¨, and ⃑𝐹īŠŠ in the đ‘Ĩ-direction is the sum of the ⃑𝑖 components of these forces, which is given by 𝐹=−4+5+2=3.ī—

The resultant of ⃑𝐹īŠ§, ⃑𝐹īŠ¨, and ⃑𝐹īŠŠ in the đ‘Ļ-direction is the sum of the ⃑𝑗 components of these forces, which is given by 𝐹=2−7+9=4.ī˜

The summed components of ⃑𝐹ī— and ⃑𝐹ī˜ are shown as force vectors in the following figure. The line from the tail of ⃑𝐹īŠ§ī— to the head of ⃑𝐹īŠŠī˜ is the resultant, 𝑅, of the components.

The magnitude of 𝑅 can be determined using the Pythagorean theorem: 𝑅=3+4=9+16=25𝑅=√25=5.īŠ¨īŠ¨īŠ¨N

The tangent of angle 𝜃 from the horizontal at which the resultant force acts can be determined using the formula tan𝜃=𝐹𝐹=43.ī˜ī—

Now, let us look at an example in which the angles at which forces act are not directly given.

Example 4: Finding the Resultant of Four Forces Acting at a Point on a Triangle

𝐴đĩđļ is a triangle with a right angle at đĩ, where 𝐴đĩ=32cm, đĩđļ=24cm, 𝐷∈𝐴đļ, and đĩ𝐷=𝐷đļ. Four forces of magnitudes 2, 3, 19, and 14 newtons are acting at point đĩ in the directions īƒĢ𝐴đĩ, īƒĒđĩđļ, īƒĢđļ𝐴, and īƒĢđĩ𝐷 respectively. Find the magnitude of the resultant of these forces.

Answer

It is helpful to first draw 𝐴đĩđļ to determine what this can reveal about the directions that the 19 N and 14 N forces act in. The following figure shows 𝐴đĩđļ.

The tangent of 𝜃 is given by tan𝜃=2432=34.

Taking the sides of 𝐴đĩđļ opposite and adjacent to 𝜃 as having lengths of 3𝑙 and 4𝑙, respectively, the hypotenuse of 𝐴đĩđļ has a length given by ℎ=ī„(3𝑙)+(4𝑙)=ī„(25𝑙)=5𝑙.īŠ¨īŠ¨īŠ¨

From this, we see that sin𝜃=35 and cos𝜃=45.

The directions in which the forces act at đĩ are shown in the following figure.

From this, we see that the net force acting in the đ‘Ĩ-direction, taking īƒĢ𝐴đĩ as positive, is given by 𝐹=2−14𝜃−19𝜃𝐹=2−33𝜃𝐹=2−33ī€ŧ45īˆ=−24.4.ī—ī—ī—coscoscosN

The net force acting in the đ‘Ļ-direction, taking īƒĒđĩđļ as positive, is given by 𝐹=3+14𝜃−19𝜃𝐹=3−5𝜃𝐹=3−5ī€ŧ35īˆ=3−3=0.ī˜ī˜ī˜sinsinsinN

The magnitude of the resultant force equals the magnitude of the net force in the đ‘Ĩ-direction, which is 24.4 N.

Let us now look at an example in which the resultant of a set of forces is known but some of the forces contributing to the resultant are of unknown magnitude.

Example 5: Finding the Magnitudes of Two Unknown Forces out of a Group of Forces given the Free-Body Diagram

Forces of magnitudes 𝐹, 16, 𝐾, 18, and 9√3 newtons act at a point in the directions shown on the diagram. Their resultant, 𝑅, has a magnitude of 20 N. Find the values of 𝐹 and 𝐾.

Answer

The angle between the horizontal and the 16 N force is given by 90−30=60.∘

The net force acting in the đ‘Ĩ-direction is given by 𝐹=𝐹+16(60)−18=𝐹+8−18=𝐹−10.ī—cos

The net force acting in the đ‘Ļ-direction is given by 𝐹=𝐾+16(60)−9√3=𝐾+8√3−9√3=𝐾−√3.ī˜sin

The resultant, 𝑅, has a magnitude of 20 N. 𝑅 has a horizontal component equal to 𝐹ī— and a vertical component equal to 𝐹ī˜. From this, we see that 𝐹−10=20(30)𝐹−10=10√3𝐹=10+10√3.cos

We also see that 𝐾−√3=20(30)𝐾−√3=10𝐾=10+√3.sin

We can confirm that these values of 𝐹 and 𝐾 are correct by substituting these values into the expressions for 𝐹ī— and 𝐹ī˜: 𝐹=10+10√3−10=10√3,𝐹=10+√3−√3=10.ī—ī˜

The magnitude of the resultant is given by 𝑅=ī„đš+𝐹=ī„žī€ģ10√3ī‡+10=√100(3)+100=√400=20,īŠ¨ī—īŠ¨ī˜īŠ¨īŠ¨N which is the value of 𝑅 stated in the question.

Let us summarize what we have learned in these examples.

Key Points

  • Multiple forces can be summed by summing the perpendicular components of the forces and determining the resultant of the components.
  • If 𝐹īŠ§ is one of the perpendicular components of a force ⃑𝐹, the magnitude of 𝐹īŠ§ is given by 𝐹=𝐹(𝜃),īŠ§cos where 𝜃 is the angle between ⃑𝐹 and 𝐹īŠ§. 𝐹īŠ¨, the component perpendicular to 𝐹īŠ§, is given by 𝐹=𝐹(𝜃).īŠ¨sin
  • The magnitude of the resultant of the perpendicular components of a force is given by 𝐹=ī„đš+𝐹.īŠ¨īŠ§īŠ¨īŠ¨
  • If a force ⃑𝐹īŒ  is given by ⃑𝐹=đ‘Ĩ⃑𝑖+đ‘Ļ⃑𝑗+𝑧⃑𝑘īŒ īŒēīŒēīŒē and a force ⃑𝐹īŒĄ is given by ⃑𝐹=đ‘Ĩ⃑𝑖+đ‘Ļ⃑𝑗+𝑧⃑𝑘,īŒĄīŒģīŒģīŒģ the resultant of these forces is given by ⃑𝐹+⃑𝐹=(đ‘Ĩ+đ‘Ĩ)⃑𝑖+(đ‘Ļ+đ‘Ļ)⃑𝑗+(𝑧+𝑧)⃑𝑘.īŒ īŒĄīŒēīŒģīŒēīŒģīŒēīŒģ

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