06.1 Resultant of Force Problem Set
RESULTANT OF A FORCE | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: The magnitude and direction of the resultant force, stemming from the force system illustrated, is most close to:
A. 19.83 ππ; 14.09Β° B. 19.83 ππ; 24.09Β° C. 29.83 ππ; 14.09Β° D. 29.83 ππ; 24.09Β°
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PROBLEM 2: The magnitude and direction of the resultant force, stemming from the force system illustrated, is most close to:
A. 45.29 ππ; 22.22Β° B. 45.29 ππ; 67.76Β° C. 53.10 ππ; 33.09Β° D. 53.10 ππ; 43.09Β°
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PROBLEM 3: The free-body diagram below illustrates a force system acting from point C which is in complete equilibrium. Given the data presented, the sum of the forces along the π₯ β ππ₯ππ is represented by the expression:
A. πΉ7 sin 50Β° β 20 = 0 B. πΉ7 cos 50Β° β 20 = 0 C. πΉ7 sin 50Β° β πΉ> = 0 D. πΉ7 cos 50Β° + 20 = 0
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PROBLEM 4: The free-body diagram below illustrates a force system acting from point C which is in complete equilibrium. Given the data presented, the sum of the forces along the π¦ β ππ₯ππ is represented by the expression:
A. πΉ7 sin 50Β° β 20 = 0 B. πΉ7 cos 50Β° β 20 = 0 C. πΉ7 sin 50Β° β πΉ> = 0 D. πΉ7 cos 50Β° + 20 = 0
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RESULTANT OF A FORCE | SOLUTIONS SOLUTION 1: 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 from a single point of application in a FORCE SYSTEM. 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.
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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. Everything is already laid out in the format that we want, so just highlighting our CARTESIAN COORDINATE AXES in blue, we have:
The next step is to cycle through each individual force, one at a time, and break the force into its HORIZONTAL and VERTICAL COMPONENTS.
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We will start with the 100 ππ force highlighted here in red:
The LINE OF ACTION of the force is provided, as well as the SENSE OF THE ANGLE relative to the origin, given as 30Β°. We can use this ANGLE to determine the COMPONENTS, illustrated now in GREEN and PURPLE:
We will use TRIGONOMETRIC IDENTITIES to solve for each COMPONENT.
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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. Knowing where our ANGLE of 30Β° resides in relation to the HORIZONTAL X-AXIS, we can identify the X-AXIS as our ADJACENT side, Y-AXIS as our OPPOSITE side, and the FORCE vector as our HYPOTENUSE. Working within a TWO-DIMENSIONAL CARTESIAN COORDINATE SYSTEM, this means that that each COMPONENT can be defined as: β’ The HORIZONTAL COMPONENT of the force: πΉA = πΉB cos π β’ The VERTICAL COMPONENT of the force: πΉD = πΉB sin π Letβs start with calculating the HORIZONTAL COMPONENT. Pulling the data we have and plugging it in to the general formula established, we get: πΉAEFF = (100 ππ) cos(30Β°) Or: πΉAEFF = 86.6 ππ
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Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA that we have established, giving us: πΉDEFF = (100 ππ) sin(30Β°) Or: πΉDEFF = 50 ππ We now have both the horizontal and vertical components for the 100 lb force and will rinse and repeat this same process for the two remaining FORCES acting on the system. Letβs move on to the 50 lb FORCE, muting out everything else.
This FORCE acts from the same POINT OF APPLICATION, but we are given no ANGLE. However, we are given two VALUES that represent the RISE and RUN of this FORCEβ¦or in other words, the SLOPE.
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We need to establish the ANGLE at the POINT OF APPLICATION, between the 50 lb FORCE and the X-AXIS. With the RISE and RUN of this FORCE defined, we can use the TRIGNOMETRIC IDENTITY of the TANGENT FUNCTION to quantify this value. Doing so we get: π π = π‘ππK> ( ) π Or:
π = π‘ππK>
3 = 36.9Β° 4
Therefore, the ANGLE between the 50 lb FORCE and the X-AXIS is 36.9Β°. Alternatively, we could have recognized that this is a special case 3-4-5 TRIANGLE and used that relationship to determine the COMPONENTS moving forward. With this information, we determine that the HORIZONTAL COMPONENT can be calculated as: πΉAMF = πΉ cos π = (β50 ππ)cos (36.9Β°) Or: πΉAMF = β40 ππ
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Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA, giving us: πΉDMF = πΉ sin π = (50 ππ)sin (36.9Β°) Or: πΉDMF = 30 ππ 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 NEGATIVE value for the MAGNITUDE of the ORIGINAL FORCE when deriving our HORIZONTAL COMPONENT and a POSITIVE 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 LEFT is the NEGATIVE XDIRECTION and running VERTICALLY UP is the POSITIVE Y-DIRECTION. Letβs move on to the 80 lb FORCE. This FORCE acts in the THIRD QUADRANT, with an angle relative to the Y-AXIS, given as 20Β°.
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Muting out all the other FORCES and focusing in on this 80 lb FORCE, we have:
This is a situation where we need to adjust the typical way that we RESOLVE A FORCE. Most times we will be given an ANGLE that is relative to the X-AXIS, as we did for the 100 lb FORCE. However, in this case, we have an ANGLE relative to the Y-AXIS. This doesnβt complicate things other than a quick mental adjustment in the way we are relating the ANGLE and the COMPONENTS with the TRIGNOMETRIC IDENTITIES that we rely on. In the case that we are working within a TWO-DIMENSIONAL CARTESIAN COORDINATE SYSTEM, and the ANGLE we are given is RELATIVE to the Y-AXIS, then each COMPONENT can be defined as: β’ The HORIZONTAL COMPONENT of the force: πΉA = πΉB sin π
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β’ The VERTICAL COMPONENT of the force: πΉD = πΉB cos π With all the data we need already defined, letβs start with calculating the HORIZONTAL COMPONENT. Plugging our data in to the general formula, we get: πΉANF = πΉ sin π = (β80 ππ)(sin 20Β°) Or: πΉANF = β27.36 ππ Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA, giving us: πΉDNF = πΉ cos π = (β80 ππ)(cos 20Β°) Or: πΉDNF = β75.17 ππ We now have all of our INDIVIDUAL FORCES broken down in to their X and Y COMPONENT FORCES.
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We are now able to analyze the CUMULATIVE EFFECTS of the FORCES on this SYSTEM because we have components that can be related to one another, whereas before, we had FORCES that were at random, unrelated, angles. 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 Letβs 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, (πΉ), of any number of FORCES, n, broken down in to COMPONENTS πΉA,P and πΉD,P can be found using:
7
S
πΉ=
πΉA,P PT>
7
S
+
πΉD,P PT>
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7
S
=
πΉA,P PT>
7
S
+
> 7
πΉD,P PT>
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The first thing we need to do is gather all of the HORIZONTAL COMPONENTS and add them together, giving us:
πΉA = πΉAEFF + πΉAMF + πΉANF = 86.6 β 40 β 27.36 Or:
πΉA = 19.24 ππ
Gathering all of the VERTICAL COMPONENTS and adding them together, we get:
πΉD = πΉDEFF + πΉDMF + πΉDNF = 50 + 30 β 75.17 Or:
πΉD = 4.83 ππ
Taking this data, we can plug it in to our GENERAL FORMULA to define the RESULTANT FORCE, such that:
πΉ=
19.24
7
+ 4.83
7
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Which gives us a RESULTANT FORCE, which is the CUMULATIVE EFFECT of all the defined INDIVIDUAL FORCES, as: πΉ = 19.83 ππ With our RESULTANT FORCE now defined, letβs determine the ANGLE in which it acts RELATIVE to the X-AXIS. 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:
π = arctan
S PT> πΉD,P S PT> πΉA,P
This may be a bit confusing as it is presented, so letβs rewrite it equivalently as:
π = tanK>
πΉD πΉA
We have all the data we need already defined at this point, letβs plug it in, giving us:
π = tanK>
4.83 = 14.09Β° 19.24
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Which tells us that the RESULTANT DIRECTION, or ANGLE, of our RESULTANT FORCE, relative to the X-AXIS, is: π = 14.09Β° This allows us to illustrate our RESULTANT FORCE, which is EQUIVALENT to all the FORCES acting on this system, as:
The correct answer choice is A. ππ. ππ π₯π; ππ. ππΒ°
SOLUTION 2: 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.
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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 from a single point of application in a FORCE SYSTEM. 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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Everything is already laid out in the format that we want, so just highlighting our CARTESIAN COORDINATE AXES in blue, we have:
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 120 ππ 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 40Β°:
We will use the specified ANGLE along with TRIGONOMETRIC IDENTITIES to solve for each COMPONENT. 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. Knowing where our ANGLE of 40Β° resides in relation to the HORIZONTAL X-AXIS, 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, this means that that each COMPONENT can be defined as: β’ The HORIZONTAL COMPONENT of the force: πΉA = πΉB cos π β’ The VERTICAL COMPONENT of the force: πΉD = πΉB sin π Letβs start with calculating the HORIZONTAL COMPONENT. Pulling the data we have and plugging it in to the general formula established, we get: πΉAE`F = (120 ππ) cos(40Β°) Or: πΉAE`F = 91.93 ππ Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA that we have established, giving us: πΉDE`F = (120 ππ) sin(40Β°) Or: πΉDE`F = 77.13 ππ
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We now have both the horizontal and vertical components for the 120 lb force and will rinse and repeat this same process for the two remaining FORCES acting on the system. Letβs move on to the 60 lb FORCE, muting out everything else.
This FORCE acts from the same POINT OF APPLICATION, but we are given no ANGLE. However, through observation, we can conclude that this FORCE is acting ONLY in the VERTICAL DIRECTION; there will be no HORIZONTAL COMPONENT. The FORCE is what it is, there is no need to do any further work in breaking it down in to its INDIVIDUAL COMPONENTS. For documentation purposes and calculations down the road, we can define the HORIZONTAL COMPONENT as: πΉAaF = 0 ππ
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We can also define the VERTICAL COMPONENT as: πΉDaF = 60 ππ Letβs move on to the 130 lb FORCE.
This FORCE acts from the same POINT OF APPLICATION, but we are given no ANGLE. However, we are given two VALUES that represent the RISE and RUN of this FORCEβ¦or in other words, the SLOPE. We need to establish the ANGLE at the POINT OF APPLICATION, between the 130 lb FORCE and the X-AXIS. With the RISE and RUN of this FORCE defined, we can use the TRIGNOMETRIC IDENTITY of the TANGENT FUNCTION to quantify this value.
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Doing so we get: π π = π‘ππK> ( ) π Or:
π = π‘ππK>
12 = 67.4Β° 5
Therefore, the ANGLE between the 130 lb FORCE and the X-AXIS is 67.4Β°. With this information, we determine that the HORIZONTAL COMPONENT can be calculated as: πΉAMF = πΉ cos π = (β130 ππ)cos (67.4Β°) Or: πΉAMF = β50 ππ Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA, giving us: πΉDMF = πΉ sin π = (β130 ππ)sin (67.4Β°)
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Or: πΉDMF = β120 ππ 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 NEGATIVE value for the MAGNITUDE of the ORIGINAL FORCE when deriving our HORIZONTAL and VERTICAL COMPONENT. This runs in line with the standard convention for a CARTESIAN COORDINATE SYSTEM, where running HORIZONTALLY to the LEFT is the NEGATIVE XDIRECTION and running VERTICALLY DOWN is the NEGATIVE YDIRECTION. 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 SYSTEM because we have components that can be related to one another, whereas before, we had FORCES that were at random, unrelated, angles. From here we need to complete two more steps, which are: 1. Derive the RESULTANT FORCE
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2. Determine the ANGLE in which this RESULTANT FORCE acts relative to the XAXIS Letβs 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, (πΉ), of any number of FORCES, n, broken down in to COMPONENTS πΉA,P and πΉD,P can be found using:
7
S
πΉ=
πΉA,P PT>
7
S
+
πΉD,P PT>
7
S
=
πΉA,P PT>
7
S
+
> 7
πΉD,P PT>
The first thing we need to do is gather all of the HORIZONTAL COMPONENTS and add them together, giving us:
πΉA = πΉAE`F + πΉAaF + πΉAEbF = 91.93 + 0 β 50
Or:
πΉA = 41.93 ππ
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Gathering all of the VERTICAL COMPONENTS and adding them together, we get:
πΉD = πΉDE`F + πΉDaF + πΉDEbF = 77.13 + 60 β 120
Or:
πΉD = 17.13 ππ
Taking this data, we can plug it in to our GENERAL FORMULA to define the RESULTANT FORCE, such that:
πΉ=
41.93
7
+ 17.13
7
Which gives us a RESULTANT FORCE, which is the CUMULATIVE EFFECT of all the defined INDIVIDUAL FORCES, as: πΉ = 45.29 ππ With our RESULTANT FORCE now defined, letβs 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:
π = arctan
S PT> πΉD,P S PT> πΉA,P
This may be a bit confusing as it is presented, so letβs rewrite it equivalently as:
π = tanK>
πΉD πΉA
We have all the data we need already defined at this point, letβs plug it in, giving us:
π = tanK>
17.13 41.93
Which tells us that the RESULTANT DIRECTION, or ANGLE, of our RESULTANT FORCE, relative to the X-AXIS, is: π = 22.22Β°
The correct answer choice is A. ππ. ππ π₯π; ππ. ππΒ°
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SOLUTION 3: 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. In this problem, we have three separate forces acting from a single point of application in a FORCE SYSTEM. The process to determine the SUM of the FORCES in the HORIZONTAL XDIRECTION, requires that we break each individual force up into components and sum them together. 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 out our FORCE SYSTEM on a typical CARTESIAN COORDINATE AXES, 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 COMPONENTSβ¦though at the end, we will only be concerned with the HORIZONTAL COMPONENTS. We will start with the πΉ> . This FORCE acts from the POINT OF APPLICATION, but we are given no ANGLE. However, through observation, we can conclude that this FORCE is acting ONLY in the VERTICAL DIRECTION; there will be no HORIZONTAL COMPONENT.
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The FORCE is what it is, there is no need to do any further work in breaking it down in to its INDIVIDUAL COMPONENTS. For documentation purposes and calculations down the road, we can define the HORIZONTAL COMPONENT as: πΉAE = 0 ππ We can also define the VERTICAL COMPONENT as: πΉDE = βπΉD Letβs move on to the πΉ7 FORCE. The LINE OF ACTION of the force is provided, as well as the SENSE OF THE ANGLE relative to the origin, given as 50Β°. We will use this specified ANGLE along with TRIGONOMETRIC IDENTITIES to solve for each COMPONENT. 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. Knowing where our ANGLE of 50Β° resides in relation to the HORIZONTAL X-AXIS, 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, this means that that each COMPONENT can be defined as: β’ The HORIZONTAL COMPONENT of the force: πΉA = πΉB cos π β’ The VERTICAL COMPONENT of the force: πΉD = πΉB sin π Letβs start with calculating the HORIZONTAL COMPONENT. Pulling the data we have and plugging it in to the general formula established, we get: πΉA` = πΉ7 cos(50Β°) Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA that we have established, giving us: πΉD` = πΉ7 sin(50Β°) We now have both the horizontal and vertical components for the πΉ7 FORCE and will move on to our final 20 lb FORCE. This FORCE acts from the POINT OF APPLICATION but we are not given an ANGLE. However, through observation, we can conclude that this FORCE is acting ONLY in the HORIZONTAL DIRECTION; there will be no VERTICAL COMPONENT.
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The FORCE is what it is, there is no need to do any further work in breaking it down in to its INDIVIDUAL COMPONENTS. For documentation purposes and calculations down the road, we can define the HORIZONTAL COMPONENT as: πΉA`F = β20 ππ We can also define the VERTICAL COMPONENT as: πΉD`F = 0 SIGN CONVENTION is important, donβt overlook! We used a NEGATIVE value for the MAGNITUDE of the HORIZONTAL COMPONENT, which runs in line with the standard convention for a CARTESIAN COORDINATE SYSTEM, where running HORIZONTALLY to the LEFT is the NEGATIVE X-DIRECTION We now have all of our INDIVIDUAL FORCES broken down in to their X and Y COMPONENT FORCESβ¦however, letβs dial in only on the HORIZONTAL COMPONENTS. This problem asks us to determine the SUM of the forces along the π₯ β ππ₯ππ , or the SUM of all the HORIZONTAL COMPONENTS.
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Gathering all of the HORIZONTAL COMPONENTS and adding them together, we get:
πΉA = πΉ7 cos 50Β° β 20
And because our FORCE SYSTEM is in complete EQUILIBRIUM, we can write the expression as: πΉ7 cos 50Β° β 20 = 0 The correct answer choice is B. ππ πππ ππΒ° β ππ = π
SOLUTION 4: 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. In this problem, we have three separate forces acting from a single point of application in a FORCE SYSTEM. The process to determine the SUM of the FORCES in the VERTICAL Y-DIRECTION, requires that we break each individual force up into components and sum them together.
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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. Laying out our FORCE SYSTEM on a typical CARTESIAN COORDINATE AXES, 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 COMPONENTSβ¦though at the end, we will only be concerned with the VERTICAL COMPONENTS. We will start with the πΉ> . This FORCE acts from the POINT OF APPLICATION, but we are given no ANGLE.
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However, through observation, we can conclude that this FORCE is acting ONLY in the VERTICAL DIRECTION; there will be no HORIZONTAL COMPONENT. The FORCE is what it is, there is no need to do any further work in breaking it down in to its INDIVIDUAL COMPONENTS. For documentation purposes and calculations down the road, we can define the HORIZONTAL COMPONENT as: πΉAE = 0 ππ We can also define the VERTICAL COMPONENT as: πΉDE = βπΉD Letβs move on to the πΉ7 FORCE. The LINE OF ACTION of the force is provided, as well as the SENSE OF THE ANGLE relative to the origin, given as 50Β°. We will use this specified ANGLE along with TRIGONOMETRIC IDENTITIES to solve for each COMPONENT. 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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Knowing where our ANGLE of 50Β° resides in relation to the HORIZONTAL X-AXIS, we can identify the X-AXIS as our ADJACENT side, Y-AXIS as our OPPOSITE side, and the FORCE vector as our HYPOTENUSE. Working within a TWO-DIMENSIONAL CARTESIAN COORDINATE SYSTEM, this means that that each COMPONENT can be defined as: β’ The HORIZONTAL COMPONENT of the force: πΉA = πΉB cos π β’ The VERTICAL COMPONENT of the force: πΉD = πΉB sin π Letβs start with calculating the HORIZONTAL COMPONENT. Pulling the data we have and plugging it in to the general formula established, we get: πΉA` = πΉ7 cos(50Β°) Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA that we have established, giving us: πΉD` = πΉ7 sin(50Β°) We now have both the horizontal and vertical components for the πΉ7 FORCE and will move on to our final 20 lb FORCE. This FORCE acts from the POINT OF APPLICATION but we are not given an ANGLE.
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However, through observation, we can conclude that this FORCE is acting ONLY in the HORIZONTAL DIRECTION; there will be no VERTICAL COMPONENT. The FORCE is what it is, there is no need to do any further work in breaking it down in to its INDIVIDUAL COMPONENTS. For documentation purposes and calculations down the road, we can define the HORIZONTAL COMPONENT as: πΉA`F = β20 ππ We can also define the VERTICAL COMPONENT as: πΉD`F = 0 SIGN CONVENTION is important, donβt overlook! We used a NEGATIVE value for the MAGNITUDE of the HORIZONTAL COMPONENT, which runs in line with the standard convention for a CARTESIAN COORDINATE SYSTEM, where running HORIZONTALLY to the LEFT is the NEGATIVE X-DIRECTION We now have all of our INDIVIDUAL FORCES broken down in to their X and Y COMPONENT FORCESβ¦however, letβs dial in only on the VERTICAL COMPONENTS. This problem asks us to determine the SUM of the forces along the π¦ β ππ₯ππ , or the SUM of all the VERTICAL COMPONENTS.
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Gathering all of the VERTICAL COMPONENTS and adding them together, we get:
πΉD = πΉ7 sin 50Β° β πΉ>
And because our FORCE SYSTEM is in complete EQUILIBRIUM, we can write the expression as: πΉ7 sin 50Β° β πΉ> = 0 The correct answer choice is C. ππ πππ ππΒ° β ππ = π
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PROBLEM 1: The magnitude and direction of the resultant force, stemming from the force system illustrated, is most close to:
A. 19.83 ππ; 14.09Β° B. 19.83 ππ; 24.09Β° C. 29.83 ππ; 14.09Β° D. 29.83 ππ; 24.09Β°
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PROBLEM 2: The magnitude and direction of the resultant force, stemming from the force system illustrated, is most close to:
A. 45.29 ππ; 22.22Β° B. 45.29 ππ; 67.76Β° C. 53.10 ππ; 33.09Β° D. 53.10 ππ; 43.09Β°
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PROBLEM 3: The free-body diagram below illustrates a force system acting from point C which is in complete equilibrium. Given the data presented, the sum of the forces along the π₯ β ππ₯ππ is represented by the expression:
A. πΉ7 sin 50Β° β 20 = 0 B. πΉ7 cos 50Β° β 20 = 0 C. πΉ7 sin 50Β° β πΉ> = 0 D. πΉ7 cos 50Β° + 20 = 0
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PROBLEM 4: The free-body diagram below illustrates a force system acting from point C which is in complete equilibrium. Given the data presented, the sum of the forces along the π¦ β ππ₯ππ is represented by the expression:
A. πΉ7 sin 50Β° β 20 = 0 B. πΉ7 cos 50Β° β 20 = 0 C. πΉ7 sin 50Β° β πΉ> = 0 D. πΉ7 cos 50Β° + 20 = 0
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RESULTANT OF A FORCE | SOLUTIONS SOLUTION 1: 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 from a single point of application in a FORCE SYSTEM. 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.
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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. Everything is already laid out in the format that we want, so just highlighting our CARTESIAN COORDINATE AXES in blue, we have:
The next step is to cycle through each individual force, one at a time, and break the force into its HORIZONTAL and VERTICAL COMPONENTS.
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We will start with the 100 ππ force highlighted here in red:
The LINE OF ACTION of the force is provided, as well as the SENSE OF THE ANGLE relative to the origin, given as 30Β°. We can use this ANGLE to determine the COMPONENTS, illustrated now in GREEN and PURPLE:
We will use TRIGONOMETRIC IDENTITIES to solve for each COMPONENT.
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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. Knowing where our ANGLE of 30Β° resides in relation to the HORIZONTAL X-AXIS, we can identify the X-AXIS as our ADJACENT side, Y-AXIS as our OPPOSITE side, and the FORCE vector as our HYPOTENUSE. Working within a TWO-DIMENSIONAL CARTESIAN COORDINATE SYSTEM, this means that that each COMPONENT can be defined as: β’ The HORIZONTAL COMPONENT of the force: πΉA = πΉB cos π β’ The VERTICAL COMPONENT of the force: πΉD = πΉB sin π Letβs start with calculating the HORIZONTAL COMPONENT. Pulling the data we have and plugging it in to the general formula established, we get: πΉAEFF = (100 ππ) cos(30Β°) Or: πΉAEFF = 86.6 ππ
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Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA that we have established, giving us: πΉDEFF = (100 ππ) sin(30Β°) Or: πΉDEFF = 50 ππ We now have both the horizontal and vertical components for the 100 lb force and will rinse and repeat this same process for the two remaining FORCES acting on the system. Letβs move on to the 50 lb FORCE, muting out everything else.
This FORCE acts from the same POINT OF APPLICATION, but we are given no ANGLE. However, we are given two VALUES that represent the RISE and RUN of this FORCEβ¦or in other words, the SLOPE.
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We need to establish the ANGLE at the POINT OF APPLICATION, between the 50 lb FORCE and the X-AXIS. With the RISE and RUN of this FORCE defined, we can use the TRIGNOMETRIC IDENTITY of the TANGENT FUNCTION to quantify this value. Doing so we get: π π = π‘ππK> ( ) π Or:
π = π‘ππK>
3 = 36.9Β° 4
Therefore, the ANGLE between the 50 lb FORCE and the X-AXIS is 36.9Β°. Alternatively, we could have recognized that this is a special case 3-4-5 TRIANGLE and used that relationship to determine the COMPONENTS moving forward. With this information, we determine that the HORIZONTAL COMPONENT can be calculated as: πΉAMF = πΉ cos π = (β50 ππ)cos (36.9Β°) Or: πΉAMF = β40 ππ
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Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA, giving us: πΉDMF = πΉ sin π = (50 ππ)sin (36.9Β°) Or: πΉDMF = 30 ππ 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 NEGATIVE value for the MAGNITUDE of the ORIGINAL FORCE when deriving our HORIZONTAL COMPONENT and a POSITIVE 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 LEFT is the NEGATIVE XDIRECTION and running VERTICALLY UP is the POSITIVE Y-DIRECTION. Letβs move on to the 80 lb FORCE. This FORCE acts in the THIRD QUADRANT, with an angle relative to the Y-AXIS, given as 20Β°.
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Muting out all the other FORCES and focusing in on this 80 lb FORCE, we have:
This is a situation where we need to adjust the typical way that we RESOLVE A FORCE. Most times we will be given an ANGLE that is relative to the X-AXIS, as we did for the 100 lb FORCE. However, in this case, we have an ANGLE relative to the Y-AXIS. This doesnβt complicate things other than a quick mental adjustment in the way we are relating the ANGLE and the COMPONENTS with the TRIGNOMETRIC IDENTITIES that we rely on. In the case that we are working within a TWO-DIMENSIONAL CARTESIAN COORDINATE SYSTEM, and the ANGLE we are given is RELATIVE to the Y-AXIS, then each COMPONENT can be defined as: β’ The HORIZONTAL COMPONENT of the force: πΉA = πΉB sin π
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β’ The VERTICAL COMPONENT of the force: πΉD = πΉB cos π With all the data we need already defined, letβs start with calculating the HORIZONTAL COMPONENT. Plugging our data in to the general formula, we get: πΉANF = πΉ sin π = (β80 ππ)(sin 20Β°) Or: πΉANF = β27.36 ππ Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA, giving us: πΉDNF = πΉ cos π = (β80 ππ)(cos 20Β°) Or: πΉDNF = β75.17 ππ We now have all of our INDIVIDUAL FORCES broken down in to their X and Y COMPONENT FORCES.
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We are now able to analyze the CUMULATIVE EFFECTS of the FORCES on this SYSTEM because we have components that can be related to one another, whereas before, we had FORCES that were at random, unrelated, angles. 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 Letβs 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, (πΉ), of any number of FORCES, n, broken down in to COMPONENTS πΉA,P and πΉD,P can be found using:
7
S
πΉ=
πΉA,P PT>
7
S
+
πΉD,P PT>
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7
S
=
πΉA,P PT>
7
S
+
> 7
πΉD,P PT>
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The first thing we need to do is gather all of the HORIZONTAL COMPONENTS and add them together, giving us:
πΉA = πΉAEFF + πΉAMF + πΉANF = 86.6 β 40 β 27.36 Or:
πΉA = 19.24 ππ
Gathering all of the VERTICAL COMPONENTS and adding them together, we get:
πΉD = πΉDEFF + πΉDMF + πΉDNF = 50 + 30 β 75.17 Or:
πΉD = 4.83 ππ
Taking this data, we can plug it in to our GENERAL FORMULA to define the RESULTANT FORCE, such that:
πΉ=
19.24
7
+ 4.83
7
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Which gives us a RESULTANT FORCE, which is the CUMULATIVE EFFECT of all the defined INDIVIDUAL FORCES, as: πΉ = 19.83 ππ With our RESULTANT FORCE now defined, letβs determine the ANGLE in which it acts RELATIVE to the X-AXIS. 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:
π = arctan
S PT> πΉD,P S PT> πΉA,P
This may be a bit confusing as it is presented, so letβs rewrite it equivalently as:
π = tanK>
πΉD πΉA
We have all the data we need already defined at this point, letβs plug it in, giving us:
π = tanK>
4.83 = 14.09Β° 19.24
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Which tells us that the RESULTANT DIRECTION, or ANGLE, of our RESULTANT FORCE, relative to the X-AXIS, is: π = 14.09Β° This allows us to illustrate our RESULTANT FORCE, which is EQUIVALENT to all the FORCES acting on this system, as:
The correct answer choice is A. ππ. ππ π₯π; ππ. ππΒ°
SOLUTION 2: 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.
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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 from a single point of application in a FORCE SYSTEM. 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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Everything is already laid out in the format that we want, so just highlighting our CARTESIAN COORDINATE AXES in blue, we have:
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 120 ππ 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 40Β°:
We will use the specified ANGLE along with TRIGONOMETRIC IDENTITIES to solve for each COMPONENT. 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. Knowing where our ANGLE of 40Β° resides in relation to the HORIZONTAL X-AXIS, 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, this means that that each COMPONENT can be defined as: β’ The HORIZONTAL COMPONENT of the force: πΉA = πΉB cos π β’ The VERTICAL COMPONENT of the force: πΉD = πΉB sin π Letβs start with calculating the HORIZONTAL COMPONENT. Pulling the data we have and plugging it in to the general formula established, we get: πΉAE`F = (120 ππ) cos(40Β°) Or: πΉAE`F = 91.93 ππ Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA that we have established, giving us: πΉDE`F = (120 ππ) sin(40Β°) Or: πΉDE`F = 77.13 ππ
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We now have both the horizontal and vertical components for the 120 lb force and will rinse and repeat this same process for the two remaining FORCES acting on the system. Letβs move on to the 60 lb FORCE, muting out everything else.
This FORCE acts from the same POINT OF APPLICATION, but we are given no ANGLE. However, through observation, we can conclude that this FORCE is acting ONLY in the VERTICAL DIRECTION; there will be no HORIZONTAL COMPONENT. The FORCE is what it is, there is no need to do any further work in breaking it down in to its INDIVIDUAL COMPONENTS. For documentation purposes and calculations down the road, we can define the HORIZONTAL COMPONENT as: πΉAaF = 0 ππ
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We can also define the VERTICAL COMPONENT as: πΉDaF = 60 ππ Letβs move on to the 130 lb FORCE.
This FORCE acts from the same POINT OF APPLICATION, but we are given no ANGLE. However, we are given two VALUES that represent the RISE and RUN of this FORCEβ¦or in other words, the SLOPE. We need to establish the ANGLE at the POINT OF APPLICATION, between the 130 lb FORCE and the X-AXIS. With the RISE and RUN of this FORCE defined, we can use the TRIGNOMETRIC IDENTITY of the TANGENT FUNCTION to quantify this value.
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Doing so we get: π π = π‘ππK> ( ) π Or:
π = π‘ππK>
12 = 67.4Β° 5
Therefore, the ANGLE between the 130 lb FORCE and the X-AXIS is 67.4Β°. With this information, we determine that the HORIZONTAL COMPONENT can be calculated as: πΉAMF = πΉ cos π = (β130 ππ)cos (67.4Β°) Or: πΉAMF = β50 ππ Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA, giving us: πΉDMF = πΉ sin π = (β130 ππ)sin (67.4Β°)
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Or: πΉDMF = β120 ππ 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 NEGATIVE value for the MAGNITUDE of the ORIGINAL FORCE when deriving our HORIZONTAL and VERTICAL COMPONENT. This runs in line with the standard convention for a CARTESIAN COORDINATE SYSTEM, where running HORIZONTALLY to the LEFT is the NEGATIVE XDIRECTION and running VERTICALLY DOWN is the NEGATIVE YDIRECTION. 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 SYSTEM because we have components that can be related to one another, whereas before, we had FORCES that were at random, unrelated, angles. From here we need to complete two more steps, which are: 1. Derive the RESULTANT FORCE
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2. Determine the ANGLE in which this RESULTANT FORCE acts relative to the XAXIS Letβs 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, (πΉ), of any number of FORCES, n, broken down in to COMPONENTS πΉA,P and πΉD,P can be found using:
7
S
πΉ=
πΉA,P PT>
7
S
+
πΉD,P PT>
7
S
=
πΉA,P PT>
7
S
+
> 7
πΉD,P PT>
The first thing we need to do is gather all of the HORIZONTAL COMPONENTS and add them together, giving us:
πΉA = πΉAE`F + πΉAaF + πΉAEbF = 91.93 + 0 β 50
Or:
πΉA = 41.93 ππ
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Gathering all of the VERTICAL COMPONENTS and adding them together, we get:
πΉD = πΉDE`F + πΉDaF + πΉDEbF = 77.13 + 60 β 120
Or:
πΉD = 17.13 ππ
Taking this data, we can plug it in to our GENERAL FORMULA to define the RESULTANT FORCE, such that:
πΉ=
41.93
7
+ 17.13
7
Which gives us a RESULTANT FORCE, which is the CUMULATIVE EFFECT of all the defined INDIVIDUAL FORCES, as: πΉ = 45.29 ππ With our RESULTANT FORCE now defined, letβs 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:
π = arctan
S PT> πΉD,P S PT> πΉA,P
This may be a bit confusing as it is presented, so letβs rewrite it equivalently as:
π = tanK>
πΉD πΉA
We have all the data we need already defined at this point, letβs plug it in, giving us:
π = tanK>
17.13 41.93
Which tells us that the RESULTANT DIRECTION, or ANGLE, of our RESULTANT FORCE, relative to the X-AXIS, is: π = 22.22Β°
The correct answer choice is A. ππ. ππ π₯π; ππ. ππΒ°
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SOLUTION 3: 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. In this problem, we have three separate forces acting from a single point of application in a FORCE SYSTEM. The process to determine the SUM of the FORCES in the HORIZONTAL XDIRECTION, requires that we break each individual force up into components and sum them together. 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 out our FORCE SYSTEM on a typical CARTESIAN COORDINATE AXES, 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 COMPONENTSβ¦though at the end, we will only be concerned with the HORIZONTAL COMPONENTS. We will start with the πΉ> . This FORCE acts from the POINT OF APPLICATION, but we are given no ANGLE. However, through observation, we can conclude that this FORCE is acting ONLY in the VERTICAL DIRECTION; there will be no HORIZONTAL COMPONENT.
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The FORCE is what it is, there is no need to do any further work in breaking it down in to its INDIVIDUAL COMPONENTS. For documentation purposes and calculations down the road, we can define the HORIZONTAL COMPONENT as: πΉAE = 0 ππ We can also define the VERTICAL COMPONENT as: πΉDE = βπΉD Letβs move on to the πΉ7 FORCE. The LINE OF ACTION of the force is provided, as well as the SENSE OF THE ANGLE relative to the origin, given as 50Β°. We will use this specified ANGLE along with TRIGONOMETRIC IDENTITIES to solve for each COMPONENT. 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. Knowing where our ANGLE of 50Β° resides in relation to the HORIZONTAL X-AXIS, 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, this means that that each COMPONENT can be defined as: β’ The HORIZONTAL COMPONENT of the force: πΉA = πΉB cos π β’ The VERTICAL COMPONENT of the force: πΉD = πΉB sin π Letβs start with calculating the HORIZONTAL COMPONENT. Pulling the data we have and plugging it in to the general formula established, we get: πΉA` = πΉ7 cos(50Β°) Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA that we have established, giving us: πΉD` = πΉ7 sin(50Β°) We now have both the horizontal and vertical components for the πΉ7 FORCE and will move on to our final 20 lb FORCE. This FORCE acts from the POINT OF APPLICATION but we are not given an ANGLE. However, through observation, we can conclude that this FORCE is acting ONLY in the HORIZONTAL DIRECTION; there will be no VERTICAL COMPONENT.
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The FORCE is what it is, there is no need to do any further work in breaking it down in to its INDIVIDUAL COMPONENTS. For documentation purposes and calculations down the road, we can define the HORIZONTAL COMPONENT as: πΉA`F = β20 ππ We can also define the VERTICAL COMPONENT as: πΉD`F = 0 SIGN CONVENTION is important, donβt overlook! We used a NEGATIVE value for the MAGNITUDE of the HORIZONTAL COMPONENT, which runs in line with the standard convention for a CARTESIAN COORDINATE SYSTEM, where running HORIZONTALLY to the LEFT is the NEGATIVE X-DIRECTION We now have all of our INDIVIDUAL FORCES broken down in to their X and Y COMPONENT FORCESβ¦however, letβs dial in only on the HORIZONTAL COMPONENTS. This problem asks us to determine the SUM of the forces along the π₯ β ππ₯ππ , or the SUM of all the HORIZONTAL COMPONENTS.
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Gathering all of the HORIZONTAL COMPONENTS and adding them together, we get:
πΉA = πΉ7 cos 50Β° β 20
And because our FORCE SYSTEM is in complete EQUILIBRIUM, we can write the expression as: πΉ7 cos 50Β° β 20 = 0 The correct answer choice is B. ππ πππ ππΒ° β ππ = π
SOLUTION 4: 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. In this problem, we have three separate forces acting from a single point of application in a FORCE SYSTEM. The process to determine the SUM of the FORCES in the VERTICAL Y-DIRECTION, requires that we break each individual force up into components and sum them together.
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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. Laying out our FORCE SYSTEM on a typical CARTESIAN COORDINATE AXES, 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 COMPONENTSβ¦though at the end, we will only be concerned with the VERTICAL COMPONENTS. We will start with the πΉ> . This FORCE acts from the POINT OF APPLICATION, but we are given no ANGLE.
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However, through observation, we can conclude that this FORCE is acting ONLY in the VERTICAL DIRECTION; there will be no HORIZONTAL COMPONENT. The FORCE is what it is, there is no need to do any further work in breaking it down in to its INDIVIDUAL COMPONENTS. For documentation purposes and calculations down the road, we can define the HORIZONTAL COMPONENT as: πΉAE = 0 ππ We can also define the VERTICAL COMPONENT as: πΉDE = βπΉD Letβs move on to the πΉ7 FORCE. The LINE OF ACTION of the force is provided, as well as the SENSE OF THE ANGLE relative to the origin, given as 50Β°. We will use this specified ANGLE along with TRIGONOMETRIC IDENTITIES to solve for each COMPONENT. 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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Knowing where our ANGLE of 50Β° resides in relation to the HORIZONTAL X-AXIS, we can identify the X-AXIS as our ADJACENT side, Y-AXIS as our OPPOSITE side, and the FORCE vector as our HYPOTENUSE. Working within a TWO-DIMENSIONAL CARTESIAN COORDINATE SYSTEM, this means that that each COMPONENT can be defined as: β’ The HORIZONTAL COMPONENT of the force: πΉA = πΉB cos π β’ The VERTICAL COMPONENT of the force: πΉD = πΉB sin π Letβs start with calculating the HORIZONTAL COMPONENT. Pulling the data we have and plugging it in to the general formula established, we get: πΉA` = πΉ7 cos(50Β°) Moving on to the VERTICAL COMPONENT, we take our data and plug it in to the GENERAL FORMULA that we have established, giving us: πΉD` = πΉ7 sin(50Β°) We now have both the horizontal and vertical components for the πΉ7 FORCE and will move on to our final 20 lb FORCE. This FORCE acts from the POINT OF APPLICATION but we are not given an ANGLE.
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However, through observation, we can conclude that this FORCE is acting ONLY in the HORIZONTAL DIRECTION; there will be no VERTICAL COMPONENT. The FORCE is what it is, there is no need to do any further work in breaking it down in to its INDIVIDUAL COMPONENTS. For documentation purposes and calculations down the road, we can define the HORIZONTAL COMPONENT as: πΉA`F = β20 ππ We can also define the VERTICAL COMPONENT as: πΉD`F = 0 SIGN CONVENTION is important, donβt overlook! We used a NEGATIVE value for the MAGNITUDE of the HORIZONTAL COMPONENT, which runs in line with the standard convention for a CARTESIAN COORDINATE SYSTEM, where running HORIZONTALLY to the LEFT is the NEGATIVE X-DIRECTION We now have all of our INDIVIDUAL FORCES broken down in to their X and Y COMPONENT FORCESβ¦however, letβs dial in only on the VERTICAL COMPONENTS. This problem asks us to determine the SUM of the forces along the π¦ β ππ₯ππ , or the SUM of all the VERTICAL COMPONENTS.
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Gathering all of the VERTICAL COMPONENTS and adding them together, we get:
πΉD = πΉ7 sin 50Β° β πΉ>
And because our FORCE SYSTEM is in complete EQUILIBRIUM, we can write the expression as: πΉ7 sin 50Β° β πΉ> = 0 The correct answer choice is C. ππ πππ ππΒ° β ππ = π
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