08.1 Resolution of Force Problem Set
RESOLUTION OF A FORCE | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: A hurricane wind blowing at 300 mph is acting on a building in Sunrise, FL with a defined force vector πΉ = 3.5π β 1.5π + 2.0π. The angle by which the force makes with the positive y-axis is most close to: A. 20.4Β° B. 66.4Β° C. 69.6Β° D. 110Β°
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PROBLEM 2: A transmission tower is supported by a guy wire that anchored to the ground at point A and attached to the top of the tower at point B. If the cable has a tension of 2,500, the resultant of the tensile force is most close to:
A. 235,850 B. 243,280 C. 246,800 D. 268,000
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PROBLEM 3: A force is represented by the force vector πΉ = β1060π + 2120π + 795π, the angle by which the force makes with the positive x-axis is most close to: A. 20.4Β° B. 32.0Β° C. 71.5Β° D. 115.1Β°
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PROBLEM 4: A force is represented by the force vector πΉ = β1060π + 2120π + 795π, the angle by which the force makes with the positive y-axis is most close to: A. 20.4Β° B. 32.0Β° C. 71.5 D. 115.1Β°
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PROBLEM 5: A particle is said to be in three-dimensional static equilibrium with 10 forces acting on it at various angles and directions. The number of equations of equilibrium that we can write for this particle is most close to: A. 2 B. 3 C. 4 D. None of the above
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PROBLEM 6: A particle is said to be in three-dimensional static equilibrium with no external forces acting on it. The expression that best represents the static equilibrium of this particle is: A. βπΉA π + βπΉB π + βπΉC π = 0 B. βπΉ = 0 C. βπΉA = βπΉB = βπΉC = 0 D. All of the above
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PROBLEM 7: When given a force in a three dimensional space, without a defined direction or magnitude, the number of unknowns present corresponding to that force is most close to: A. One B. Two C. Three D. Four
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RESOLUTION OF A FORCE | SOLUTIONS SOLUTION 1: The GENERAL FORMULA for breaking down a THREE-DIMENSIONAL FORCE in to itβs COMPONENTS, knowing the MAGNITUDE, or RESULTANT, as well as the general GEOMETRY of the 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. If we are given a FORCE in a TWO or THREE DIMENSIONAL space, we are able to reverse engineer it, and resolve the FORCE in to itβs individual COMPONENTS if we know the MAGNITUDE, or RESULTANT, of the ORIGINAL FORCE, as well as the fully developed GEOMETRY of the FORCE, using the RELATIONSHIP:
π =
π₯N + π¦N + π§N =
πΉAN + πΉBN + πΉCN
Taking this FORMULA and laying it out for each INDIVIDUAL COMPONENT, we have: π₯ πΉ π π¦ πΉB = πΉ π π§ πΉC = πΉ π πΉA =
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In this problem, we are given that the wind vector is: πΉ = 3.5π β 1.5π + 2.0π We can calculate the MAGNITUDE, or RESULTANT, of this FORCE using the RELATIONSHIP established above along with the GEOMETRICAL COMPONENTS of the VECTOR, such that we get:
π =
3.5
N
+ β1.5
N
+ 2.0 N
Or: π = 4.3 With the MAGNITUDE of our FORCE defined, we now need to determine how to derive an ANGLE between it and the Y-AXISβ¦this will require knowledge of DIRECTIONAL COSINES. When given a FORCE VECTOR in a THREE DIMENSIONAL rectangular force space, such as in this situation we are currently working, we are able to RESOLVE the force into its components using DIRECTIONAL COSINES.
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DIRECTIONAL COSINES allows us to take a FORCE stemming from a single POINT OF APPLICATION, at any given ANGLE with the X, Y, and Z AXES, and break it up in to individual COMPONENTS that run PARALLEL to itβs respective AXIS, such that:
The GENERAL FORMULAS for the DIRECTIONAL COSINES can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Given a FORCE in a THREE DIMENSIONAL space, each COMPONENT can be defined using the GENERAL FORMULAS as outlined: β’ The X-COMPONENT, which runβs parallel with the X-AXIS and typically represents the HORIZONTAL COMPONENT of the FORCE, is: πΉA = πΉ cos πA
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β’ The Y-COMPONENT, which runβs parallel with the Y-AXIS and typically represents the VERTICAL COMPONENT of the FORCE, is: πΉB = πΉ cos πB β’ The Z-COMPONENT, which runβs parallel with the Z-AXIS and typically represents the DEPTH or ELEVATION COMPONENT of the FORCE, is: πΉC = πΉ cos πC In this set of GENERAL FORMULAS, πΉT is the represents the COMPONENT FORCE, F is the MAGNITUDE of the original FORCE, and πT is the ANGLE in which that FORCE makes with respect to the COMPONENT AXIS in question. We are only concerned with the ANGLE between the FORCE and the Y-AXIS, so we will hone in on this GENERAL FORMULA: πΉB = πΉ cos πB Now we already have the COMPONENTS defined for us, we have the MAGNITUDE, but we donβt have the ANGLES. Our GENERAL FORMULA requires that we have two of the three components defined to work, and since we do, we can do a little tweaking to isolate that ANGLE that we are after.
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Doing this we get:
cos πB =
πΉB πΉ
Or:
πB = cos UV
πΉB πΉ
At this point, we have defined: πΉ = π = 4.3 πΉB = β1.5 Plugging this data in to our formula, we get:
π = cos UV β
1.5 4.3
Or: π = 110.4Β°
The correct answer choice is D. πππΒ°
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SOLUTION 2: The GENERAL FORMULA for breaking down a THREE-DIMENSIONAL FORCE in to itβs COMPONENTS, knowing the MAGNITUDE, or RESULTANT, as well as the general GEOMETRY of the 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. If we are given a FORCE in a TWO or THREE DIMENSIONAL space, we are able to reverse engineer it, and resolve the FORCE in to itβs individual COMPONENTS if we know the MAGNITUDE, or RESULTANT, of the ORIGINAL FORCE, as well as the fully developed GEOMETRY of the FORCE, using the RELATIONSHIP:
π =
π₯N + π¦N + π§N =
πΉAN + πΉBN + πΉCN
We know the line of action of the force is acting on the bolt passing through points A and B, and that the force is directed from A to B. Knowing this, along with knowing the GEOMETRY of the assembly, we can define the INDIVIDUAL COMPONENTS of the VECTOR AB as: πA = β40 π πB = +80 π πC = +30 π
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With this information, we can define the FORCE for the cable in vector form as: πΉ = 2,500 β40π + 80π + 30π Or: πΉ = β100,000π + 200,000π + 75,000π Plugging in the value for each component, we can quantify the RESULTANT FORCE as:
π =
β100,000
N
+ 200,000
N
+ 75,000 N
Or: π = 235,849.53 The correct answer choice is A. πππ, πππ
SOLUTION 3: The GENERAL FORMULA for breaking down a THREE-DIMENSIONAL FORCE in to itβs COMPONENTS, knowing the MAGNITUDE, or RESULTANT, as well as the general GEOMETRY of the 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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If we are given a FORCE in a TWO or THREE DIMENSIONAL space, we are able to reverse engineer it, and resolve the FORCE in to itβs individual COMPONENTS if we know the MAGNITUDE, or RESULTANT, of the ORIGINAL FORCE, as well as the fully developed GEOMETRY of the FORCE, using the RELATIONSHIP:
π =
π₯N + π¦N + π§N =
πΉAN + πΉBN + πΉCN
Taking this FORMULA and laying it out for each INDIVIDUAL COMPONENT, we have: π₯ πΉ π π¦ πΉB = πΉ π π§ πΉC = πΉ π πΉA =
In this problem, we are given that the vector is: πΉ = β1060π + 2120π + 795π We can calculate the MAGNITUDE, or RESULTANT, of this FORCE using the RELATIONSHIP established above along with the GEOMETRICAL COMPONENTS of the VECTOR, such that we get:
π =
1060
N
+ 2120
N
+ 795 N
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Or: π = 2500 With the MAGNITUDE of our FORCE defined, we now need to determine how to derive an ANGLE between it and the X-AXISβ¦this will require knowledge of DIRECTIONAL COSINES. When given a FORCE VECTOR in a THREE DIMENSIONAL rectangular force space, such as in this situation we are currently working, we are able to RESOLVE the force into its components using DIRECTIONAL COSINES. DIRECTIONAL COSINES allows us to take a FORCE stemming from a single POINT OF APPLICATION, at any given ANGLE with the X, Y, and Z AXES, and break it up in to individual COMPONENTS that run PARALLEL to itβs respective AXIS, such that:
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The GENERAL FORMULAS for the DIRECTIONAL COSINES can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Given a FORCE in a THREE DIMENSIONAL space, each COMPONENT can be defined using the GENERAL FORMULAS as outlined: β’ The X-COMPONENT, which runβs parallel with the X-AXIS and typically represents the HORIZONTAL COMPONENT of the FORCE, is: πΉA = πΉ cos πA β’ The Y-COMPONENT, which runβs parallel with the Y-AXIS and typically represents the VERTICAL COMPONENT of the FORCE, is: πΉB = πΉ cos πB β’ The Z-COMPONENT, which runβs parallel with the Z-AXIS and typically represents the DEPTH or ELEVATION COMPONENT of the FORCE, is: πΉC = πΉ cos πC In this set of GENERAL FORMULAS, πΉT is the represents the COMPONENT FORCE, F is the MAGNITUDE of the original FORCE, and πT is the ANGLE in which that FORCE makes with respect to the COMPONENT AXIS in question.
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We are only concerned with the ANGLE between the FORCE and the X-AXIS, so we will hone in on this GENERAL FORMULA: πΉA = πΉ cos πA Now we already have the COMPONENTS defined for us, we have the MAGNITUDE, but we donβt have the ANGLES. Our GENERAL FORMULA requires that we have two of the three components defined to work, and since we do, we can do a little tweaking to isolate that ANGLE that we are after. Doing this we get:
cos πA =
πΉA πΉ
Or:
πA = cos UV
πΉA πΉ
At this point, we have defined: πΉ = π = 2500 πΉA = β1060
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Plugging this data in to our formula, we get:
π = cos UV β
1060 2500
Or: π = 115.1Β°
The correct answer choice is D. πππ. πΒ°
SOLUTION 4: The GENERAL FORMULA for breaking down a THREE-DIMENSIONAL FORCE in to itβs COMPONENTS, knowing the MAGNITUDE, or RESULTANT, as well as the general GEOMETRY of the 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. If we are given a FORCE in a TWO or THREE DIMENSIONAL space, we are able to reverse engineer it, and resolve the FORCE in to itβs individual COMPONENTS if we know the MAGNITUDE, or RESULTANT, of the ORIGINAL FORCE, as well as the fully developed GEOMETRY of the FORCE, using the RELATIONSHIP:
π =
π₯N + π¦N + π§N =
πΉAN + πΉBN + πΉCN
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Taking this FORMULA and laying it out for each INDIVIDUAL COMPONENT, we have: π₯ πΉ π π¦ πΉB = πΉ π π§ πΉC = πΉ π πΉA =
In this problem, we are given that the wind vector is: πΉ = β1060π + 2120π + 795π We can calculate the MAGNITUDE, or RESULTANT, of this FORCE using the RELATIONSHIP established above along with the GEOMETRICAL COMPONENTS of the VECTOR, such that we get:
π =
1060
N
+ 2120
N
+ 795 N
Or: π = 2500 With the MAGNITUDE of our FORCE defined, we now need to determine how to derive an ANGLE between it and the Y-AXISβ¦this will require knowledge of DIRECTIONAL COSINES.
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When given a FORCE VECTOR in a THREE DIMENSIONAL rectangular force space, such as in this situation we are currently working, we are able to RESOLVE the force into its components using DIRECTIONAL COSINES. DIRECTIONAL COSINES allows us to take a FORCE stemming from a single POINT OF APPLICATION, at any given ANGLE with the X, Y, and Z AXES, and break it up in to individual COMPONENTS that run PARALLEL to itβs respective AXIS, such that:
The GENERAL FORMULAS for the DIRECTIONAL COSINES can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Given a FORCE in a THREE DIMENSIONAL space, each COMPONENT can be defined using the GENERAL FORMULAS as outlined: β’ The X-COMPONENT, which runβs parallel with the X-AXIS and typically represents the HORIZONTAL COMPONENT of the FORCE, is:
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πΉA = πΉ cos πA β’ The Y-COMPONENT, which runβs parallel with the Y-AXIS and typically represents the VERTICAL COMPONENT of the FORCE, is: πΉB = πΉ cos πB β’ The Z-COMPONENT, which runβs parallel with the Z-AXIS and typically represents the DEPTH or ELEVATION COMPONENT of the FORCE, is: πΉC = πΉ cos πC In this set of GENERAL FORMULAS, πΉT is the represents the COMPONENT FORCE, F is the MAGNITUDE of the original FORCE, and πT is the ANGLE in which that FORCE makes with respect to the COMPONENT AXIS in question. We are only concerned with the ANGLE between the FORCE and the Y-AXIS, so we will hone in on this GENERAL FORMULA: πΉB = πΉ cos πB Now we already have the COMPONENTS defined for us, we have the MAGNITUDE, but we donβt have the ANGLES. Our GENERAL FORMULA requires that we have two of the three components defined to work, and since we do, we can do a little tweaking to isolate that ANGLE that we are after.
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Doing this we get:
cos πB =
πΉB πΉ
Or:
πB = cos UV
πΉB πΉ
At this point, we have defined: πΉ = π = 2500 πΉB = 2120 Plugging this data in to our formula, we get:
π = cos UV
2120 2500
Or: π = 32.0Β°
The correct answer choice is B. ππ. πΒ°
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SOLUTION 5: When working with problems in three-dimensions, we can only sum the forces around each axis that a force, or forces, is acting about. Therefore, as we are working in a THREE DIMENSIONAL space, there are three planesβ¦x, y, and z. We can add the FORCES that are acting in these THREE PLANES, therefore, the maximum number of equations of equilibrium that can be written is THREE...one for each PLANE. The correct answer choice is B. π
SOLUTION 6: When working with problems in three-dimensions, we can only sum the forces around each axis that a force, or forces, is acting about. Therefore, as we are working in a THREE DIMENSIONAL space, there are three planesβ¦x, y, and z. We can add the FORCES that are acting in these THREE PLANES, therefore, the maximum number of equations of equilibrium that can be written is THREE...one for each PLANE. As we are told that the particle is in STATIC EQUILIBRIUM, we know that the sum of the FORCES about any plane, or axis, will be EQUAL to ZERO.
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Cycling through the answer options, we see that all three are different forms of expressing the equations of equilibrium of THREE DIMENSIONAL particle in STATIC EQUILIBRIUM.
Therefore, the correct answer choice is D. ππ₯π₯ π¨π ππ‘π πππ¨π―π SOLUTION 7: When working with problems in THREE DIMENSIONS, we can only sum the forces around each axis that a force, or forces, is acting about. Therefore, as we are working in a THREE DIMENSIONAL space, there are three planesβ¦x, y, and z. We can add the FORCES that are acting in these THREE PLANES, therefore, the maximum number of equations of equilibrium that can be written is THREE...one for each PLANE. As we have a SINGLE FORCE acting in a THREE DIMENSIONAL space, there are THREE COMPONENTS that are UNDEFINEDβ¦this is because we have no information for the MAGNITUDE or DIRECTION. Therefore, these THREE COMPONENTS must be defined, all of which are originally UNKNOWN. The correct answer choice is C. ππ‘π«ππ
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PROBLEM 1: A hurricane wind blowing at 300 mph is acting on a building in Sunrise, FL with a defined force vector πΉ = 3.5π β 1.5π + 2.0π. The angle by which the force makes with the positive y-axis is most close to: A. 20.4Β° B. 66.4Β° C. 69.6Β° D. 110Β°
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PROBLEM 2: A transmission tower is supported by a guy wire that anchored to the ground at point A and attached to the top of the tower at point B. If the cable has a tension of 2,500, the resultant of the tensile force is most close to:
A. 235,850 B. 243,280 C. 246,800 D. 268,000
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PROBLEM 3: A force is represented by the force vector πΉ = β1060π + 2120π + 795π, the angle by which the force makes with the positive x-axis is most close to: A. 20.4Β° B. 32.0Β° C. 71.5Β° D. 115.1Β°
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PROBLEM 4: A force is represented by the force vector πΉ = β1060π + 2120π + 795π, the angle by which the force makes with the positive y-axis is most close to: A. 20.4Β° B. 32.0Β° C. 71.5 D. 115.1Β°
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PROBLEM 5: A particle is said to be in three-dimensional static equilibrium with 10 forces acting on it at various angles and directions. The number of equations of equilibrium that we can write for this particle is most close to: A. 2 B. 3 C. 4 D. None of the above
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PROBLEM 6: A particle is said to be in three-dimensional static equilibrium with no external forces acting on it. The expression that best represents the static equilibrium of this particle is: A. βπΉA π + βπΉB π + βπΉC π = 0 B. βπΉ = 0 C. βπΉA = βπΉB = βπΉC = 0 D. All of the above
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PROBLEM 7: When given a force in a three dimensional space, without a defined direction or magnitude, the number of unknowns present corresponding to that force is most close to: A. One B. Two C. Three D. Four
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RESOLUTION OF A FORCE | SOLUTIONS SOLUTION 1: The GENERAL FORMULA for breaking down a THREE-DIMENSIONAL FORCE in to itβs COMPONENTS, knowing the MAGNITUDE, or RESULTANT, as well as the general GEOMETRY of the 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. If we are given a FORCE in a TWO or THREE DIMENSIONAL space, we are able to reverse engineer it, and resolve the FORCE in to itβs individual COMPONENTS if we know the MAGNITUDE, or RESULTANT, of the ORIGINAL FORCE, as well as the fully developed GEOMETRY of the FORCE, using the RELATIONSHIP:
π =
π₯N + π¦N + π§N =
πΉAN + πΉBN + πΉCN
Taking this FORMULA and laying it out for each INDIVIDUAL COMPONENT, we have: π₯ πΉ π π¦ πΉB = πΉ π π§ πΉC = πΉ π πΉA =
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In this problem, we are given that the wind vector is: πΉ = 3.5π β 1.5π + 2.0π We can calculate the MAGNITUDE, or RESULTANT, of this FORCE using the RELATIONSHIP established above along with the GEOMETRICAL COMPONENTS of the VECTOR, such that we get:
π =
3.5
N
+ β1.5
N
+ 2.0 N
Or: π = 4.3 With the MAGNITUDE of our FORCE defined, we now need to determine how to derive an ANGLE between it and the Y-AXISβ¦this will require knowledge of DIRECTIONAL COSINES. When given a FORCE VECTOR in a THREE DIMENSIONAL rectangular force space, such as in this situation we are currently working, we are able to RESOLVE the force into its components using DIRECTIONAL COSINES.
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DIRECTIONAL COSINES allows us to take a FORCE stemming from a single POINT OF APPLICATION, at any given ANGLE with the X, Y, and Z AXES, and break it up in to individual COMPONENTS that run PARALLEL to itβs respective AXIS, such that:
The GENERAL FORMULAS for the DIRECTIONAL COSINES can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Given a FORCE in a THREE DIMENSIONAL space, each COMPONENT can be defined using the GENERAL FORMULAS as outlined: β’ The X-COMPONENT, which runβs parallel with the X-AXIS and typically represents the HORIZONTAL COMPONENT of the FORCE, is: πΉA = πΉ cos πA
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β’ The Y-COMPONENT, which runβs parallel with the Y-AXIS and typically represents the VERTICAL COMPONENT of the FORCE, is: πΉB = πΉ cos πB β’ The Z-COMPONENT, which runβs parallel with the Z-AXIS and typically represents the DEPTH or ELEVATION COMPONENT of the FORCE, is: πΉC = πΉ cos πC In this set of GENERAL FORMULAS, πΉT is the represents the COMPONENT FORCE, F is the MAGNITUDE of the original FORCE, and πT is the ANGLE in which that FORCE makes with respect to the COMPONENT AXIS in question. We are only concerned with the ANGLE between the FORCE and the Y-AXIS, so we will hone in on this GENERAL FORMULA: πΉB = πΉ cos πB Now we already have the COMPONENTS defined for us, we have the MAGNITUDE, but we donβt have the ANGLES. Our GENERAL FORMULA requires that we have two of the three components defined to work, and since we do, we can do a little tweaking to isolate that ANGLE that we are after.
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Doing this we get:
cos πB =
πΉB πΉ
Or:
πB = cos UV
πΉB πΉ
At this point, we have defined: πΉ = π = 4.3 πΉB = β1.5 Plugging this data in to our formula, we get:
π = cos UV β
1.5 4.3
Or: π = 110.4Β°
The correct answer choice is D. πππΒ°
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SOLUTION 2: The GENERAL FORMULA for breaking down a THREE-DIMENSIONAL FORCE in to itβs COMPONENTS, knowing the MAGNITUDE, or RESULTANT, as well as the general GEOMETRY of the 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. If we are given a FORCE in a TWO or THREE DIMENSIONAL space, we are able to reverse engineer it, and resolve the FORCE in to itβs individual COMPONENTS if we know the MAGNITUDE, or RESULTANT, of the ORIGINAL FORCE, as well as the fully developed GEOMETRY of the FORCE, using the RELATIONSHIP:
π =
π₯N + π¦N + π§N =
πΉAN + πΉBN + πΉCN
We know the line of action of the force is acting on the bolt passing through points A and B, and that the force is directed from A to B. Knowing this, along with knowing the GEOMETRY of the assembly, we can define the INDIVIDUAL COMPONENTS of the VECTOR AB as: πA = β40 π πB = +80 π πC = +30 π
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With this information, we can define the FORCE for the cable in vector form as: πΉ = 2,500 β40π + 80π + 30π Or: πΉ = β100,000π + 200,000π + 75,000π Plugging in the value for each component, we can quantify the RESULTANT FORCE as:
π =
β100,000
N
+ 200,000
N
+ 75,000 N
Or: π = 235,849.53 The correct answer choice is A. πππ, πππ
SOLUTION 3: The GENERAL FORMULA for breaking down a THREE-DIMENSIONAL FORCE in to itβs COMPONENTS, knowing the MAGNITUDE, or RESULTANT, as well as the general GEOMETRY of the 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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by Prepineer | Prepineer.com
If we are given a FORCE in a TWO or THREE DIMENSIONAL space, we are able to reverse engineer it, and resolve the FORCE in to itβs individual COMPONENTS if we know the MAGNITUDE, or RESULTANT, of the ORIGINAL FORCE, as well as the fully developed GEOMETRY of the FORCE, using the RELATIONSHIP:
π =
π₯N + π¦N + π§N =
πΉAN + πΉBN + πΉCN
Taking this FORMULA and laying it out for each INDIVIDUAL COMPONENT, we have: π₯ πΉ π π¦ πΉB = πΉ π π§ πΉC = πΉ π πΉA =
In this problem, we are given that the vector is: πΉ = β1060π + 2120π + 795π We can calculate the MAGNITUDE, or RESULTANT, of this FORCE using the RELATIONSHIP established above along with the GEOMETRICAL COMPONENTS of the VECTOR, such that we get:
π =
1060
N
+ 2120
N
+ 795 N
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Or: π = 2500 With the MAGNITUDE of our FORCE defined, we now need to determine how to derive an ANGLE between it and the X-AXISβ¦this will require knowledge of DIRECTIONAL COSINES. When given a FORCE VECTOR in a THREE DIMENSIONAL rectangular force space, such as in this situation we are currently working, we are able to RESOLVE the force into its components using DIRECTIONAL COSINES. DIRECTIONAL COSINES allows us to take a FORCE stemming from a single POINT OF APPLICATION, at any given ANGLE with the X, Y, and Z AXES, and break it up in to individual COMPONENTS that run PARALLEL to itβs respective AXIS, such that:
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The GENERAL FORMULAS for the DIRECTIONAL COSINES can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Given a FORCE in a THREE DIMENSIONAL space, each COMPONENT can be defined using the GENERAL FORMULAS as outlined: β’ The X-COMPONENT, which runβs parallel with the X-AXIS and typically represents the HORIZONTAL COMPONENT of the FORCE, is: πΉA = πΉ cos πA β’ The Y-COMPONENT, which runβs parallel with the Y-AXIS and typically represents the VERTICAL COMPONENT of the FORCE, is: πΉB = πΉ cos πB β’ The Z-COMPONENT, which runβs parallel with the Z-AXIS and typically represents the DEPTH or ELEVATION COMPONENT of the FORCE, is: πΉC = πΉ cos πC In this set of GENERAL FORMULAS, πΉT is the represents the COMPONENT FORCE, F is the MAGNITUDE of the original FORCE, and πT is the ANGLE in which that FORCE makes with respect to the COMPONENT AXIS in question.
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We are only concerned with the ANGLE between the FORCE and the X-AXIS, so we will hone in on this GENERAL FORMULA: πΉA = πΉ cos πA Now we already have the COMPONENTS defined for us, we have the MAGNITUDE, but we donβt have the ANGLES. Our GENERAL FORMULA requires that we have two of the three components defined to work, and since we do, we can do a little tweaking to isolate that ANGLE that we are after. Doing this we get:
cos πA =
πΉA πΉ
Or:
πA = cos UV
πΉA πΉ
At this point, we have defined: πΉ = π = 2500 πΉA = β1060
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Plugging this data in to our formula, we get:
π = cos UV β
1060 2500
Or: π = 115.1Β°
The correct answer choice is D. πππ. πΒ°
SOLUTION 4: The GENERAL FORMULA for breaking down a THREE-DIMENSIONAL FORCE in to itβs COMPONENTS, knowing the MAGNITUDE, or RESULTANT, as well as the general GEOMETRY of the 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. If we are given a FORCE in a TWO or THREE DIMENSIONAL space, we are able to reverse engineer it, and resolve the FORCE in to itβs individual COMPONENTS if we know the MAGNITUDE, or RESULTANT, of the ORIGINAL FORCE, as well as the fully developed GEOMETRY of the FORCE, using the RELATIONSHIP:
π =
π₯N + π¦N + π§N =
πΉAN + πΉBN + πΉCN
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Taking this FORMULA and laying it out for each INDIVIDUAL COMPONENT, we have: π₯ πΉ π π¦ πΉB = πΉ π π§ πΉC = πΉ π πΉA =
In this problem, we are given that the wind vector is: πΉ = β1060π + 2120π + 795π We can calculate the MAGNITUDE, or RESULTANT, of this FORCE using the RELATIONSHIP established above along with the GEOMETRICAL COMPONENTS of the VECTOR, such that we get:
π =
1060
N
+ 2120
N
+ 795 N
Or: π = 2500 With the MAGNITUDE of our FORCE defined, we now need to determine how to derive an ANGLE between it and the Y-AXISβ¦this will require knowledge of DIRECTIONAL COSINES.
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When given a FORCE VECTOR in a THREE DIMENSIONAL rectangular force space, such as in this situation we are currently working, we are able to RESOLVE the force into its components using DIRECTIONAL COSINES. DIRECTIONAL COSINES allows us to take a FORCE stemming from a single POINT OF APPLICATION, at any given ANGLE with the X, Y, and Z AXES, and break it up in to individual COMPONENTS that run PARALLEL to itβs respective AXIS, such that:
The GENERAL FORMULAS for the DIRECTIONAL COSINES can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. Given a FORCE in a THREE DIMENSIONAL space, each COMPONENT can be defined using the GENERAL FORMULAS as outlined: β’ The X-COMPONENT, which runβs parallel with the X-AXIS and typically represents the HORIZONTAL COMPONENT of the FORCE, is:
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πΉA = πΉ cos πA β’ The Y-COMPONENT, which runβs parallel with the Y-AXIS and typically represents the VERTICAL COMPONENT of the FORCE, is: πΉB = πΉ cos πB β’ The Z-COMPONENT, which runβs parallel with the Z-AXIS and typically represents the DEPTH or ELEVATION COMPONENT of the FORCE, is: πΉC = πΉ cos πC In this set of GENERAL FORMULAS, πΉT is the represents the COMPONENT FORCE, F is the MAGNITUDE of the original FORCE, and πT is the ANGLE in which that FORCE makes with respect to the COMPONENT AXIS in question. We are only concerned with the ANGLE between the FORCE and the Y-AXIS, so we will hone in on this GENERAL FORMULA: πΉB = πΉ cos πB Now we already have the COMPONENTS defined for us, we have the MAGNITUDE, but we donβt have the ANGLES. Our GENERAL FORMULA requires that we have two of the three components defined to work, and since we do, we can do a little tweaking to isolate that ANGLE that we are after.
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Doing this we get:
cos πB =
πΉB πΉ
Or:
πB = cos UV
πΉB πΉ
At this point, we have defined: πΉ = π = 2500 πΉB = 2120 Plugging this data in to our formula, we get:
π = cos UV
2120 2500
Or: π = 32.0Β°
The correct answer choice is B. ππ. πΒ°
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SOLUTION 5: When working with problems in three-dimensions, we can only sum the forces around each axis that a force, or forces, is acting about. Therefore, as we are working in a THREE DIMENSIONAL space, there are three planesβ¦x, y, and z. We can add the FORCES that are acting in these THREE PLANES, therefore, the maximum number of equations of equilibrium that can be written is THREE...one for each PLANE. The correct answer choice is B. π
SOLUTION 6: When working with problems in three-dimensions, we can only sum the forces around each axis that a force, or forces, is acting about. Therefore, as we are working in a THREE DIMENSIONAL space, there are three planesβ¦x, y, and z. We can add the FORCES that are acting in these THREE PLANES, therefore, the maximum number of equations of equilibrium that can be written is THREE...one for each PLANE. As we are told that the particle is in STATIC EQUILIBRIUM, we know that the sum of the FORCES about any plane, or axis, will be EQUAL to ZERO.
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Cycling through the answer options, we see that all three are different forms of expressing the equations of equilibrium of THREE DIMENSIONAL particle in STATIC EQUILIBRIUM.
Therefore, the correct answer choice is D. ππ₯π₯ π¨π ππ‘π πππ¨π―π SOLUTION 7: When working with problems in THREE DIMENSIONS, we can only sum the forces around each axis that a force, or forces, is acting about. Therefore, as we are working in a THREE DIMENSIONAL space, there are three planesβ¦x, y, and z. We can add the FORCES that are acting in these THREE PLANES, therefore, the maximum number of equations of equilibrium that can be written is THREE...one for each PLANE. As we have a SINGLE FORCE acting in a THREE DIMENSIONAL space, there are THREE COMPONENTS that are UNDEFINEDβ¦this is because we have no information for the MAGNITUDE or DIRECTION. Therefore, these THREE COMPONENTS must be defined, all of which are originally UNKNOWN. The correct answer choice is C. ππ‘π«ππ
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