05.1 Resultant of Force Concept Overview

RESULTANT OF A FORCE | CONCEPT OVERVIEW The TOPICS OF RESULTANT OF A FORCE can be referenced on page 67 of the NCEES Supplied Reference Handbook, 9.4 version for Computer Based Testing.

CONCEPT INTRO: A FORCE can be broken down into individual components that represent the force in a particular direction or along a particular axis. The VECTOR FORM of a THREE-DIMENSIONAL FORCE can be written as: 𝐹 = 𝐹# 𝑖 + 𝐹& 𝑗 + 𝐹( 𝑘 Any vector, or FORCE for that matter, can be broken down into two or more individual vectors, which are the COMPONENT VECTORS of the original vector. The process of defining a VECTOR COMPONENT is known as the RESOLUTION OF A VECTOR. When asked to RESOLVE A VECTOR into its components, this is the answer they are looking for. In working with vectors in STATICS, you always resolve ACTIONS, such as FORCES and DISPLACEMENTS, into exactly TWO PIECES for TWO-DIMENSIONAL problems and into exactly THREE PIECES for THREE-DIMENSIONAL problems. You can also depict TWO-DIMENSIONAL ACTIONS with THREE-DIMENSIONAL Made with by Prepineer | Prepineer.com

COMPONENTS by assigning a THIRD COMPONENT with a ZERO MAGNITUDE. When resolving a VECTOR in to its individual VECTOR COMPONENTS, these COMPONENTS will typically fall in to one of TWO CATEGORIES: CARTESIAN COMPONENTS and NON-CARTESIAN COMPONENTS.

CARTESIAN COMPONENTS: As the name CARTESIAN COMPONENTS implies, all of the resolved vector components are aligned with the CARTESIAN 𝑥−, 𝑦−, and 𝑧 − axes of the coordinate system. You may also see these CARTERSIAN COMPONENTS referred to as RECTANGULAR COMPONENTS, which will probably be the most common components you will calculate, and are fortunately usually the easiest to compute. In THREE-DIMENSIONAL problems, the RECTANGULAR COMPONENTS of a vector can be expressed in terms of 𝑖, 𝑗, and 𝑘 UNIT VECTORS in the 𝑥, 𝑦, and 𝑧 directions, respectively.



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Illustrating how this all plays out in a generalized CARTESIAN COORDINATE SYSTEM, we have:

NON-CARTESIAN COMPONENTS: NON-CARTESIAN COMPONENTS occur when the RESOLUTION of a VECTOR includes one or more COMPONENT that DOES NOT fall directly on a CARTESIAN AXES. If all COMPONENTS were aligned with the axes, they’d be CARTESIAN COMPONENTS.



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RESULTANT OF A FORCE: The FORMULAS for the RESULTANT of a FORCE can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Forces acting at some angle relevant to an established COORDINATE SYSTEM can be resolved into mutually perpendicular forces called COMPONENTS. The COMPONENTS lay out like this: • The component of a force parallel to the X-AXIS is called the X-COMPONENT • The component of a force parallel to the Y-AXIS is the called the Y-COMPONENT • The component of a force parallel to the Z-AXIS is the called the Z-COMPONENT Equivalently, these COMPONENTS can also be represented using UNIT VECTORS, such that: • The component of a force parallel to the X-AXIS represents the i-COMPONENT • The component of a force parallel to the Y-AXIS represents the j-COMPONENT • The component of a force parallel to the Z-AXIS represents the k-COMPONENT



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The most common RESOLUTION that we will be working with is one in a TWODIMENSIONAL space. In this case, the FORCE is RESOLVED along the x and y-axis of the typical CARTESIAN COORDINATE SYSTEM, illustrated as:

We will use, and need to be comfortable with, the BASIC TRIGONOMETRIC FUNCTIONS allowing us to RESOLVE a single FORCE in to its individual COMPONENTS. Generally speaking, this is done by MULTIPLYING the MAGNITUDE, or resultant of the force, by the appropriate TRIGONOMETRIC FUNCTION and the associated ANGLE specified.



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In TWO DIMENSIONS, two components will result from a single force vector, given that this force vector is at some ANGLE relative to the COORDINATE SYSTEM in which it resides. In THREE DIMENSIONS, three components will result from a single force vector, again, given that this force vector is at some ANGLE relative to each axis within the COORDINATE SYSTEM it resides. However, for both TWO and THREE DIMENSIONAL cases, these individual components must always result in the SAME ORIGINAL COMBINED BEHAVIOR when added back together, otherwise, it would be changing the dynamics of the stated scenario. A component of a resultant vector must also be the same type of vector as the resultant itself. For example, if we are defining the components of a FORCE VECTOR, then the COMPONENTS are also FORCE VECTORS. The ANGLE between the POINT OF APPLICATION of the ORIGINAL FORCE and the AXES will drive defining each INDIVIDUAL COMPONENT of the FORCE. The GENERAL FORMULAS revolving around the COMPONENTS of a VECTOR are not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For this reason, we must memorize this process and the various formulas, understanding fully its application independent of the NCEES Supplied Reference Handbook.



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Working within a TWO-DIMENSIONAL CARTESIAN COORDINATE SYSTEM, each COMPONENT can be defined as: • The HORIZONTAL COMPONENT of the force: 𝐹# = 𝐹6 cos 𝜃 • The VERTICAL COMPONENT of the force: 𝐹& = 𝐹6 sin 𝜃 As we see, the COMPONENTS can be RESOLVED both HORIZONTALLY and VERTICALLY using the SINE FUNCTION to represent the VERTICAL COMPONENT and the COSINE FUNCTION to represent the HORIZONTAL COMPONENT. This can generally be illustrated as:

It is important to note here that these formulas assume an ANGLE provided at the POINT OF APPLICATION, between the FORCE and the X-AXIS…as illustrated. These

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COMPONENTS would be derived differently had the ANGLE been specified elsewhere away from this AXIS. We will further discuss this moving forward. The GENERAL FORMULA that allows us to take the INDIVIDUAL COMPONENTS after they have been resolved, and combine them back together in to a single RESULTANT FORCE can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The RESULTANT, or MAGNITUDE, of a FORCE, (𝐹), of any number of FORCES, n, broken down in to COMPONENTS 𝐹#,; and 𝐹&,; can be found using:

?




?


?

𝐹&,; ;=>

The RESULTANT DIRECTION with respect to the X-AXIS is calculated as the ARCTANGENT, or INVERSE TANGENT, of the RATIO of the SUMMATION of all the VERTICAL COMPONENTS to the SUMMATION of all the HORIZONTAL COMPONENTS, or otherwise written as:

𝜃 = arctan

< ;=> 𝐹&,; < ;=> 𝐹#,;

It is important to remember that each COMPONENT reflects a direct relationship with the ORIGIAL FORCE using TRIGONOMETRY of a provided ANGLE. Made with by Prepineer | Prepineer.com

Depending on the geometry given in a problem, it is possible to use either the SINE or COSINE function to determine the appropriate COMPONENT, either VERTICAL or HORIZONTAL. The following illustrations provide an example when trigonometry was used to solve for the components of a force given ANGLES located in different areas. In the first illustration, we are given an angle that is a common scenario where the SINE function is used to calculate the VERTICAL COMPONENT, and the COSINE function is used to calculate the HORIZONTAL COMPONENT:

In the next illustration, we are given a second angle where the COSINE function is used to calculate the VERTICAL COMPONENT, and the SINE function is used to calculate the HORIZONTAL COMPONENT:



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The COMPONENT VECTORS will RESOLVE exactly the same, regardless of the direction you take…however, the lesson here is this, use your understanding of TRIGNOMETRIC FUNCTIONS to ensure you are relating the correct information to obtain the correct result.

FREE-BODY DIAGRAMS: A FREE BODY DIAGRAM is used to represent the physical conditions of any RIGID OBJECT you want to analyze, including dimensional data, and the forces acting on the system. FREE BODY DIAGRAMS can be complex pictures of multiple objects and systems, or diagrams of a small subcomponent of a larger piece within a system.



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We use Free Body Diagrams, commonly abbreviated as FBD, to assess the interaction of forces on bodies of mass. A FBD is essentially a sketch of a body separated from its surroundings and is in COMPLETE EQUILIBRIUM. All forces acting on a body at any specific moment in time can be represented on a FBD.



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RESULTANT OF A FORCE | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. A hook has three cables attached to it, as illustrated, where 𝛼 = 35°. The resultant magnitude and resultant direction of the three forces is most close to:

A. 137 𝑁; 86.6° B. 217 𝑁; 176.6° C. 309 𝑁; 86.6° D. 217 𝑁; 176.6°



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SOLUTION: The TOPIC of the RESULTANT of a FORCE can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. This problem may come across as very complicated, however, the process is a simple rinse and repeat, just as we would do with a single force. Rest assured, that you will not see a problem as complex as this one come exam day, but for edification and repetition purposes, we feel that presenting and having you work through it will be incredibly powerful moving forward. In this problem, we have three separate forces acting on a single rigid hook in complete equilibrium. The process to determine the RESULTANT FORCE, along with the RESULTANT DIRECTION, requires that we break each individual force up into components, sum them together, and then plug those components in to the GENERAL FORMULAS that we are given in the NCEES REFERNCE HANDBOOK to calculate the resultant. The first step in this problem is to define our Coordinate System. When working with two-dimensional problems, it is typical to use a CARTESIAN COORDINATE SYSTEM such that the x-axis represents the horizontal component, and the y-axis represents the vertical component.

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Laying our hook out on this CARTESIAN COORDINATE SYSTEM, we get:

The next step is to cycle through each individual force, one at a time, and break the force into its HORIZONTAL and VERTICAL COMPONENTS. We will start with the 100 𝑁 force highlighted here in red:



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The line of action of the force is provided, as well as the sense of the angle relative to the origin, given as 𝛼 = 35°. We can use this angle to determine the components, illustrated in GREEN and PURPLE, such that:

We will use TRIGONOMETRIC IDENTITIES to solve for each of the components. The GENERAL FORMULAS revolving around the COMPONENTS of a VECTOR are not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. For this reason, we must memorize this process and the various formulas, understanding fully its application independent of the NCEES Supplied Reference Handbook. With 𝛼 = 35°, we can identify the 𝑥 − 𝑎𝑥𝑖𝑠 as our adjacent side, 𝑦 − 𝑎𝑥𝑖𝑠 as our opposite side, and the force vector as our hypotenuse.



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Working within a TWO-DIMENSIONAL CARTESIAN COORDINATE SYSTEM, each COMPONENT can be defined as: • The HORIZONTAL COMPONENT of the force: 𝐹# = 𝐹6 cos 𝜃 • The VERTICAL COMPONENT of the force: 𝐹& = 𝐹6 sin 𝜃 Let’s start with calculating the HORIZONTAL COMPONENT. Pulling the data we have and plugging it in to the general formula established, we get: 𝐹#RSS = 𝐹 cos 𝜃 = (100 𝑁) cos(35°) Or: 𝐹#RSS = 81.92 𝑁 Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA that we have established, giving us: 𝐹&RSS = 𝐹 sin 𝜃 = (−100) sin( 35°) Or: 𝐹&RSS = −57.36 𝑁



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There are a few things to NOTE at this point: 1. Make sure that your calculator is in DEGREE MODE and not in RADIAN MODE. 2. We used a POSITIVE value for the MAGNITUDE of the ORIGINAL FORCE when deriving our HORIZONTAL COMPONENT and a NEGATIVE value for the MAGNITUDE when deriving our VERTICAL COMPONENT. This runs in line with the standard convention for a CARTESIAN COORDINATE SYSTEM, where running HORIZONTALLY to the RIGHT is the POSITIVE XDIRECTION and running VERTICALLY DOWN is the NEGATIVE YDIRECTION. We now have both the horizontal and vertical components for the 100 N force and will rinse and repeat this same process for the two remaining FORCES acting on the hook. Let’s move on to the 150 N FORCE. This FORCE acts on the hook at an angle that is not directly given in the problem statement. However, we know the ANGLE between this 150 N FORCE and the 100 N FORCE, as well as the ANGLE from the 100 N FORCE back to the X-AXIS.



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With this information, we can add the given angles of 30° and 𝛼 = 35° to calculate the angle at which the 150 N FORCE acts relative to the 𝑥 − 𝑎𝑥𝑖𝑠, such that:

Or in other terms: 𝜃 = 30° + 35° = 65° Starting again with calculating the HORIZONTAL COMPONENT, we can pull the data we have and plug it in to the general formula established, giving us: 𝐹#RWS = 𝐹 cos 𝜃 = (+150 𝑁) cos(65°) Or: 𝐹#RWS = 63.39 𝑁



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Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA, giving us: 𝐹&RWS = 𝐹 sin 𝜃 = (−150) sin(65°) Or: 𝐹&RWS = −135.95 𝑁 Again, make sure that you are following the proper SIGN CONVENTION for a standard CARTESIAN COORDINATE SYSTEM and that your calculator is in DEGREE MODE and not RADIAN MODE. Let’s move on to the 200 N FORCE. This FORCE acts with an opposite sense, but similar angle relative to the origin, given as 𝛼 = 35°.



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With all the data we need already defined, let’s start with calculating the HORIZONTAL COMPONENT. Plugging our data in to the general formula, we get: 𝐹#XSS = 𝐹 cos 𝜃 = (−200 𝑁) cos(35°) Or: 𝐹#XSS = −163.83 𝑁 Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA, giving us: 𝐹&XSS = 𝐹 sin 𝜃 = (−200) sin(35°) = −114.72 𝑁 Or: 𝐹&XSS = −114.72 𝑁 We now have all of our INDIVIDUAL FORCES broken down in to their X and Y COMPONENT FORCES. We are now able to analyze the CUMULATIVE EFFECTS of the FORCES on this hook because we have components that can be related to one another, whereas before, we had FORCES that were at random, unrelated, angles.

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From here we need to complete two more steps, which are: 1. Derive the RESULTANT FORCE 2. Determine the ANGLE in which this RESULTANT FORCE acts relative to the XAXIS Let’s derive the RESULTANT FORCE. The GENERAL FORMULA that allows us to take the INDIVIDUAL COMPONENTS and define a single RESULTANT FORCE can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The RESULTANT, or MAGNITUDE, of a FORCE, (𝐹), of any number of FORCES, n, broken down in to COMPONENTS 𝐹#,; and 𝐹&,; can be found using:

?




?




?




?


?

𝐹&,; ;=>

The first thing we need to do is gather all of the HORIZONTAL COMPONENTS and add them together, giving us:

𝐹# = 𝐹#RSS + 𝐹#RWS + 𝐹#XSS = 81.92 + 63.39 − 163.83



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Or:

𝐹# = − 18.52 𝑁

Gathering all of the VERTICAL COMPONENTS and adding them together, we get:

𝐹& = 𝐹&RSS + 𝐹&RWS + 𝐹&XSS = −57.36 − 135.95 − 114.72 Or:

𝐹& = −308.03 𝑁

Taking this data, we can plug it in to our GENERAL FORMULA to define the RESULTANT FORCE, such that:

𝐹=

−18.52 𝑁

?

+ −308.03 𝑁

?

Which gives us a RESULTANT FORCE, which is the CUMULATIVE EFFECT of all the defined INDIVIDUAL FORCES, as: 𝐹 = 309 𝑁 With our RESULTANT FORCE now defined, let’s determine the ANGLE in which it acts RELATIVE to the X-AXIS.



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The RESULTANT DIRECTION with respect to the X-AXIS is calculated as the ARCTANGENT, or INVERSE TANGENT, of the RATIO of the VERTICAL COMPONENT to the HORIZONTAL COMPONENT, or otherwise written as:

𝜃 = arctan

< ;=> 𝐹&,; < ;=> 𝐹#,;

This may be a bit confusing as it is presented, so let’s rewrite it equivalently as:

𝜃 = tanZ>

𝐹& 𝐹#

We have all the data we need already defined at this point, let’s plug it in, giving us:

𝜃 = tanZ>

−308.03 𝑁 −18.52 𝑁

Which tells us that the RESULTANT DIRECTION, or ANGLE, of our RESULTANT FORCE, relative to the X-AXIS, is: 𝜃 = 86.6°



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This allows us to illustrate our RESULTANT FORCE, which is EQUIVALENT to all the FORCES acting on this hook, as:

The correct answer choice is C. 𝟑𝟎𝟗 𝐍; 𝟖𝟔. 𝟔°



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