10.1 Moments (Couples) Problem Set
MOMENTS (COUPLES) | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.
PROBLEM 1: The moment resulting from applying a force (πΉ) at a distance (π) from a particular point (π) is best represented by the expression: A. π% = π β πΉ B. π% = β β πΉ C. π% = π Γ πΉ D. π% = β Γ πΉ
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PROBLEM 2: Which of the following statements best describes when the moment due to an applied force is zero? A. The force is negative B. The force is through the origin C. The line of action passes through the center of rotation D. The force is a function of time
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PROBLEM 3: The magnitude of the moment about the base point βπβ created by the 300 ππ force is most close to:
A. 2,702 lb β ft (counterclockwise) B. 2,010 lb β ft (clockwise) C. 1,559 lb β ft (counterclockwise) D. 1,224 lb β ft (clockwise)
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PROBLEM 4: A 10 N and 5 N force act on a particular beam, as illustrated. The magnitude of the moment about the base located a point βπ΄β is most close to:
A. 20 N β m (clockwise) B. 20 N β m (counterclockwise) C. 40 N β m (clockwise) D. 40 N β m (clockwise)
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PROBLEM 5: The moment generated by a 12 N force acting at a perpendicular distance of 3 meters away from a pivot point is most close to: A. 3 N β m B. 7 N β m C. 12 N β m D. 36 N β m
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PROBLEM 6: A force represented by the vector, πΉ = 10 π π, acts at a distance represented by a second vector, π = 5 π π. The moment generated by the force is most close to: A. β50 π B. 50 π C. β50 π D. 50 π
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PROBLEM 7: A couple is defined as _________ separated by a perpendicular distance. The expression that makes this statement TRUE is: A. Two forces in the same direction B. Two forces of equal magnitude C. Two forces of equal magnitude acting in the same direction D. Two forces of equal magnitude acting in opposite directions
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PROBLEM 8: If three couples act on a body, the statement that best describes the result on the body is: A. The net force is not equal to 0 B. The net force and net moment are equal to 0 C. The net moment equals 0, but the net force is not necessarily equal to 0 D. The net force equals 0, but the net moment is not necessarily equal to 0
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MOMENTS (COUPLES) | SOLUTIONS SOLUTION 1: The GENERAL FORMULAS revolving around MOMENTS (COUPLES) of a TWO FORCE SYSTEM can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. In this problem, we are told that there is some FORCE, F, acting some DISTANCE, r, from a particular POINT Oβ¦we are then asked to define an expression for the MOMENT that this creates. A MOMENT, with a general notation as M, is defined as the CROSS PRODUCT of the RADIUS VECTOR and the FORCE from a POINT to the LINE OF ACTION of the that FORCE. The RADIUS VECTOR is the same as the DISTANCE from the LINE OF ACTION of the FORCE to POINT O, therefore, we have everything we need defined to derive our expression. In FORMULAIC TERMS, the MOMENT can be expressed as: π = π Γ πΉ The correct answer choice is C. ππ = π« Γ π
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SOLUTION 2: The TOPIC of MOMENTS (COUPLES) 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 MOMENT of a FORCE about some specified point QUANTIFIES the TENDENCY of that FORCE to ROTATE a RIGID BODY about some specified point. The MAGNITUDE of a MOMENT identifies the AXIS OF ROTATION associated with the rotational force. The MAGNITUDE for a MOMENT, which is generally denoted as M, is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ So as a FORCE extends further away from a particular POINT, the MOMENT increases about the CENTER OF ROTATION. However, if the LINE OF ACTION of the FORCE PASSES THROUGH the CENTER OF ROTATION, the NET RESULTANT MOMENT on the body will be ZERO, thus, there will be NO ROTATION. The correct answer choice is C.
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SOLUTION 3: The GENERAL FORMULAS revolving around MOMENTS (COUPLES) of a TWO FORCE SYSTEM can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A MOMENT, with a general notation as M, is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ However, when we are assessing a MOMENT in a TWO DIMENSIONAL space, the GENERAL FORMULA using a CROSS PRODUCT simplifies down to the PRODUCT of the FORCE times the DISTANCE to the AXIS parallel to the one in which the ROTATION is occurring, or in formulaic terms: π = ππΉ In classical mechanics, this DISTANCE is referred to as the LEVER ARM. Remember this fundamental definition, as it could be a money shot problem in your favor revolving around basic terminology come exam day. A MOMENT is often a DIRECT RESULT of the ACTION of a FORCE VECTOR, and because of this, a MOMENT is also a VECTOR defined by similar characteristics.
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A MOMENT will generally have the following CHARACTERISTICS, or the EQUIVALENT, defined: β’ MAGNITUDE β’ POINT OF APPLICATION β’ SENSE In this problem, we have the POINT OF APPLICATION, but need to define the MAGNITUDE as well as the SENSE acting at POINT O. The MAGNITUDE of a moment quantifies the INTENSITY of its ROTATIONAL EFFECT. The SENSE of a MOMENT is the DIRECTION OF ROTATION about its AXIS OF ROTATION. When we are dealing with a FORCE that is at an ANGLE relative to the COORDINATE SYSTEM in which it resides, we will need to use our knowledge of RESOLVING a FORCE in to COMPONENTS so that we can assess each one INDIVIDUALLY and how it affects the ROTATION about our AXIS OF ROTATION.
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In this problem, we are given the illustration:
The 300 ππ force generates a MOMENT about the base at point βπβ. At point βπβ, this MOMENT will have a SENSE in either the clockwise or counterclockwise direction. We will typically represent the SENSE as a CLOCKWISE (NEGATIVE) or COUNTERCLOCKWISE (POSITIVE) behavior. The DIRECTION you take in your analysis of MOMENTS is not set in stone, itβs just a suggested route, and an ASSUMPTION of the resulting MOMENT in the end. Whether you define clockwise or counterclockwise rotation as positive is inconsequential, itβs that you remain consistent throughout your analysis of the problem that determines whether or not you arrive at a valid conclusion. Assuming you correctly write the EQUATIONS OF EQUILIBRIUM, in the end, your answer may result in a NEGATIVE SCALAR, or SIGN OPPOSITE to what you assumed
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originally as negative or positive. This indicates that the MOMENT is indeed acting in the OPPOSITE DIRECTION than what you initially assumed. This is not a bad thing, it just allows you to adjust your moment at the end of your analysis to be properly represented and distributed within your FORCE system. We will assume the COUNTERCLOCKWISE direction as POSITIVE. In this problem, we will break the FORCE VECTOR in to its COMPONENTS and define a PAIR of MOMENTS to determine the CUMULATIVE AFFECT at POINT O. Illustrating the FORCE COMPONENTS of the 300 lb FORCE, we have:
Letβs put some definition around these 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.
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Knowing where our ANGLE of 30Β° resides in relation to the HORIZONTAL X-AXIS, we can identify the X-AXIS as our OPPOSITE side, Y-AXIS as our ADJACENT 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: πΉ^ = πΉ_ sin π β’ The VERTICAL COMPONENT of the force: πΉa = πΉ_ cos π Letβs start with calculating the HORIZONTAL COMPONENT. Pulling the data we have and plugging it in to the general formula established, we get: πΉ^bcc = (300 ππ) sin(30Β°) Or: πΉ^bcc = 150 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: πΉabcc = (300 ππ) cos(30Β°)
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Or: πΉabcc = 260 ππ With this, we now have:
With each FORCE COMPONENT now defined, we can identify the LINE OF ACTION and associated LEVER ARM back to the AXIS OF ROTATION for each, such that:
Remember that the LEVER ARM must be PERPINDICULAR from the LINE OF ACTION back to the AXIS OF ROTATIONβ¦this is IMPORTANT.
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Many will attempt to find the DISTANCE from the AXIS OF ROTATION back to the POINT OF APPLICATION, such that:
This will result in a WRONG ANSWER, and trust us, the NCEES knows that this is a SIMPLE yet COMMON error we as engineers make:
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By Observation, we can see that our HORIZONTAL COMPONENT FORCE will cause a MOMENT about POINT O in the CLOCKWISE direction, or rather, a NEGATIVE MOMENT:
In contrast to that, the VERTICAL COMPONENT FORCE will cause a MOMENT about POINT O in the COUNTERCLOCKWISE direction, or rather, a POSITIVE MOMENT:
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The CUMULATIVE effects of each MOMENT when SUMMED about POINT O will determine the ultimate results, or in FORMULAIC TERMS: π% = βπΉ^ π^ + πΉa πa It is important to note the SENSE of each component and whether they create a clockwise or counterclockwise moment and account for this when combining the two MOMENTS. At this point, we have defined: πΉ^ = 150 ππ πΉa = 260 ππ We need to QUANTIFY the LEVER ARM form the AXIS OF ROTATION to the LINE OF ACTION of each individual COMPONENT FORCE.
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To do this, we will use the GEOMETRY we are given in the original illustrationβ¦and more specifically, we will break the system in to two separate RIGHT TRIANGLES as illustrated:
We see that breaking the SYSTEM up like this gives us a lot of information, and we can almost define both LEVER ARMS, minus one little portion of the HORIZONTAL LEVER ARM, dx, which is highlighted here in RED:
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However, using the GEOMETRY to define the BASE of the TRIANGLE that allows us to determine the VERTICAL LEVER ARM, we will have everything we needβ¦so letβs start here, we have:
Using simple TRIGNOMETRIC RELATIONSHIPS, we can calculate the VERTICAL LEVER ARM as: πa = 6 ππ‘ cos 30Β° = 5.2 ππ‘ This also allows us to determine the BASE, giving us the missing portion of the HORIZONTAL LEVER ARM, as: π^ = (6 ππ‘) sin 30Β° = 3 ππ‘
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Therefore, already having been given 6 ft of the HORIZONTAL LEVER ARM:
We can add this calculated horizontal distance to fully quantify our HORIZONTAL LEVER ARM as: π^ = 3 + 6 = 9 ππ‘ Letβs now revisit our CUMULATIVE MOMENT as we previously developed in FORMULAIC TERMS: π% = βπΉ^ π^ + πΉa πa We are now at a point where we can just plug and chug. Plugging in the data that we have defined up to this point, we can calculate the CUMULATIVE MOMENT generated about point βπβ as: π% = 260 ππ 9 ππ‘ β (150 ππ) 5.2 ππ‘
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Or: π% = 1,559 ππ β ππ‘ The last thing we need to do is determine the SENSE of the MOMENT about POINT O. We originally assumed that the COUNTERCLOCKWISE direction represented a POSITIVE MOMENT. In the end, if our MOMENT resulted in a NEGATIVE SCALAR, then that would indicate that the MOMENT actually acts in the OPPOSITE DIRECTION than what we initially assumed. Our resultant MOMENT is POSITIVE, which tells us that our initial ASSUMPTION that the COUNTERCLOCKWISE direction represented a POSITIVE MOMENT was correct. The correct answer choice is C. π, πππ π₯π β ππ (ππππππππππππππππ)
SOLUTION 4: The GENERAL FORMULAS to use when working with MOMENTS (COUPLES) in a TWO FORCE SYSTEM 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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A MOMENT is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ However, when we are assessing a MOMENT in a TWO DIMENSIONAL space, the GENERAL FORMULA using a CROSS PRODUCT simplifies down to the PRODUCT of the FORCE times the DISTANCE to the AXIS parallel to the one in which the ROTATION is occurring, or in formulaic terms: π = ππΉ In classical mechanics, this DISTANCE is referred to as the LEVER ARM. In this problem, we are given a beam loaded as:
The beam is being acted upon by two forces on either side of a dimensioned PIVOT POINT at A.
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Each FORCE applied on this beam will create a tendency for it to ROTATE about the PIVOT POINT at A. This is otherwise known as the MOMENT. We are able to quantify the MAGNITUDE of the MOMENT that each FORCE creates individually, then SUM them to determine the CUMULATIVE AFFECT about POINT A. Generally speaking, in ENGINEERING, and specifically ENGINEERING MECHANICS: STATICS, we donβt want ROTATION to occur in a way that is not controlled. Therefore, after analyzing the CUMULATIVE AFFECTS of these two forces, we will have the data necessary to install a third component somewhere on the beam to balance the beam and establish the EQUILIBRIUM we are often after. This can be done in one of two ways, we either: β’ CREATE a COUPLE at some LOCATION on this BEAM that is EQUAL and OPPOSITE in MAGNITUDE. β’ ADD a FORCE at some LOCATION on this BEAM that creates a MOMENT that is EQUAL and OPPOSITE in MAGNITUDE. However, that is not what we are seeking at this point, so letβs move on and get in to our problem. The first thing we should do is establish a COORDINATE SYSTEM to work within.
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We could do this mentally in our head, which is the recommended route, or we can quickly scratch it out on the paper we are working our solution. In this problem, we will assume that we are working in a standard CARTESIAN COORDINATE SYSTEM and that any MOMENTS acting about POINT A will be POSITIVE if they are acting with a SENSE in the COUNTERCLOCKWISE direction and NEGATIVE if they are acting with a SENSE in the CLOCKWISE direction. The initial assumptions we build around the DIRECTION of the MOMENT at the front end of our analysis is not set in stone, itβs just a suggested route, and an ASSUMPTION of the resulting MOMENT in the end. Whether you define CLOCKWISE or COUNTERCLOCKWISE rotation as POSITIVE is inconsequential, itβs that you remain consistent throughout your analysis of the problem that determines whether or not you arrive at a valid conclusion. Assuming you correctly write the EQUATIONS, in the end, your answer may result in a NEGATIVE SCALAR to what you assumed originally as negative or positive. This indicates that the MOMENT is indeed acting in the OPPOSITE DIRECTION than what you initially assumed. This is not a bad thing, it just allows you to adjust your MOMENT at the end of your analysis to be properly represented and distributed within your FORCE system. So letβs continue with our assumption that the COUNTERCLOCKWISE direction is POSITIVE.
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Based on this convention, we can conclude generically up front that the FORCE acting to the LEFT of the PIVOT POINT, and with a DOWNWARD direction, will create a ROTATION, or MOMENT, about the PIVOT POINT at A in the COUNTERCLOCKWISE direction. The FORCE acting to the RIGHT of the PIVOT POINT, and with a DOWNWARD direction, will create a ROTATION, or MOMENT, about the PIVOT POINT at A in the CLOCKWISE direction. Therefore, our GENERAL FORMULA to determine the CUMULATIVE AFFECTS of these FORCES at POINT A can be written as: π| = πΉ}~ π}~ β πΉβ’ πβ’ We are given everything we need to carry forward with a plug and chug sprinkled in with a light deployment of our calculator. We are given: πΉ}~ = 10 π πΉβ’ = 5 π π}~ = 3 π πβ’ = 2 π Plugging these values in, we get: π| = 10 π 3 π β 5 π 2 π
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Or: π| = 20 π β π The last thing we need to confirm is the SENSE of this MOMENT. Originally, we had assumed that a MOMENT about POINT A would be POSITIVE if it was acting in the COUNTERCLOCKWISE direction. We then went on to state that it didnβt matter what our original assumption was, it could be anything, and that the final result would tell us whether we were correct or not. If we werenβt correct in our original assumptions, then our result would play out as a NEGATIVE value, telling us that our MOMENT was acting in the OPPOSITE DIRECTION. However, in this problem, our resulting MOMENT is a POSITIVE VALUE, confirming that our original assumption that a POSITIVE MOMENT would act in a COUNTERCLOCKWISE direction. Therefore, our resulting MOMENT too will act in the COUNTERCLOCKWISE direction. To further solidify this point, had our value came back as: π| = β20 π β π
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Then we would have concluded that the MOMENT, with the same MAGNITUDE, was acting in a CLOCKWISE direction about POINT A. The correct answer choice is B. ππ π β π¦ (ππππππππππππππππ)
SOLUTION 5: The GENERAL FORMULAS to use when working with MOMENTS (COUPLES) in a TWO FORCE SYSTEM can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A MOMENT is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ However, when we are assessing a MOMENT in a TWO DIMENSIONAL space, the GENERAL FORMULA using a CROSS PRODUCT simplifies down to the PRODUCT of the FORCE times the DISTANCE to the AXIS parallel to the one in which the ROTATION is occurring, or in FORMULAIC TERMS: π = ππΉ In classical mechanics, this DISTANCE is referred to as the LEVER ARM.
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In this problem, we are given all the information we need to solve the problem right off the bat: πΉ = 12 π π = 3 π Plugging these values in to the GENERAL FORMULA for a MOMENT in a TWO DIMENSIONAL space, we get: π = 12 π 3 π Or: π = 36 π β π Generally speaking, in ENGINEERING, and specifically ENGINEERING MECHANICS: STATICS, we donβt want ROTATION to occur in a way that is not controlledβ¦we seek EQUILIBRIUM. However, in this problem, we donβt know enough about the FORCE to determine the SENSE of the MOMENT created, so we canβt go any further in our analysis to conclude where we should implement an additional component to establish this EQUILIBRIUM.
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Had we known another characteristic of this FORCE, such as whether the FORCE acts to the RIGHT or LEFT of the PIVOT POINT, then we could establish EQUILIBIRUM in one of two ways, we either: β’ CREATE a COUPLE at some LOCATION on this BEAM that is EQUAL and OPPOSITE in MAGNITUDE. β’ ADD a FORCE at some LOCATION on this BEAM that creates a MOMENT that is EQUAL and OPPOSITE in MAGNITUDE. The correct answer choice is D. ππ π β π¦
SOLUTION 6: The GENERAL FORMULAS to use when working with MOMENTS (COUPLES) in a TWO FORCE SYSTEM can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A MOMENT is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ However, when we are assessing a MOMENT in a TWO DIMENSIONAL space, the GENERAL FORMULA using a CROSS PRODUCT simplifies down to the PRODUCT of
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the FORCE times the DISTANCE to the AXIS parallel to the one in which the ROTATION is occurring, or in FORMULAIC TERMS: π = ππΉ In classical mechanics, this DISTANCE is referred to as the LEVER ARM. This problem is in TWO DIMENSIONS, and if we are quick to recognize the CARTESIAN COMPONENTS, then in our head, we can conclude that we have a FORCE that is acting UPDWARDS at a DISTANCE, or with a LEVER ARM, of 5 m. However, we are given VECTORS, so letβs move forward through this problem using the CROSS PRODUCT. The GENERAL FORMULA for the CROSS PRODUCT can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The CROSS PRODUCT is one of two OPERATIONS that can be deployed using VECTOR MULTIPLICATION, the other operation being the DOT PRODUCT. In itβs most fundamental form, the CROSS PRODUCT takes two vectors, runs them through a standard process, and creates a third vector that is ORTHOGONAL (or PERPENDICULAR) to each of the original vectors.
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In this problem, our two vectors are given as: πΉ = 10 π π π = 5 π π Or otherwise written as: πΉ =< 0, 0, 10 > π π =< 0, 5, 0 > π There are two routes we can take in solving this problem, the first will be using the GENERAL FORMULA that we are explicitly given in the NCEES Reference Handbook, which is written as: π Γ πΉ =< πa πΉΛ β πΛ πΉa , πΛ πΉ^ β π^ πΉΛ , π^ πΉa β πa πΉ^ >= πΉ π π πππ We have all the ELEMENTS defined for each of our VECTORS, but we donβt have an ANGLE between them. This is not a problem, the first half of this GENERAL FORMULA is enough to define our resulting VECTORβ¦or MOMENT in this case. Plugging in the ELEMENTS where outlined, we get: π Γ πΉ =< 5(10) β 0(0),0(0) β 0(10), 0(0) β 5(0) > This simplifies down to: π Γ πΉ =< 5(10) >
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Which results in a new VECTOR: π Γ πΉ =< 50, 0, 0 > Or in VECTOR NOTATION: π Γ πΉ = 50π β 0π + 0π Which tells us that the resulting MOMENT in VECTOR NOTATION is: π = 50π π β π
A QUICK HACK TO THIS PROBLEM: You can use your CALCULATOR to hack this problem in a fraction of the time. Check out the CALCULATOR WORKSHOPS page within PREPINEER to study just how we suggest you move forward with these types of problems. The correct answer choice is B. ππ π
SOLUTION 7: The topic of MOMENTS (COUPLES) 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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A system of TWO FORCES that are EQUAL in MAGNITUDE, OPPOSITE IN DIRECTION, and PARALLEL to each other is called a COUPLE. A COUPLE is equivalent to a single moment vector exerting a resultant moment, but no resultant force. The moment vector of the couple is independent of the choice of the origin of the coordinate axes. It is a FREE VECTOR that can be applied at any point with the same effect. A COUPLE is characterized as having a ZERO RESULTANT, as well as exerting the same resultant moment about all points. The NET EFFECT is ROTATION WITHOUT TRANSLATION. A force couple system is said to have moments equal to each such that they can be expressed as equal and opposite of each other, expressed as: π} = πβΉ Or: πΉ} π} = πΉβΉ πβΉ A COUPLE can be COUNTERACTED ONLY by another COUPLE. A COUPLE can be MOVED to any location WITHOUT AFFECTING THE EQUILIBRIUM requirements.
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Therefore, the correct answer choice is D.
SOLUTION 8: The topic of MOMENTS (COUPLES) can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A system of TWO FORCES that are EQUAL in MAGNITUDE, OPPOSITE IN DIRECTION, and PARALLEL to each other is called a COUPLE. A COUPLE is equivalent to a single moment vector exerting a resultant moment, but no resultant force. The moment vector of the couple is independent of the choice of the origin of the coordinate axes. It is a FREE VECTOR that can be applied at any point with the same effect. A COUPLE is characterized as having a ZERO RESULTANT, as well as exerting the same resultant moment about all points. The NET EFFECT is ROTATION WITHOUT TRANSLATION. A force couple system is said to have moments equal to each such that they can be expressed as equal and opposite of each other, expressed as: π} = πβΉ
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Or: πΉ} π} = πΉβΉ πβΉ A COUPLE can be COUNTERACTED ONLY by another COUPLE. A COUPLE can be MOVED to any location WITHOUT AFFECTING THE EQUILIBRIUM requirements. The correct answer choice is D.
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PROBLEM 1: The moment resulting from applying a force (πΉ) at a distance (π) from a particular point (π) is best represented by the expression: A. π% = π β πΉ B. π% = β β πΉ C. π% = π Γ πΉ D. π% = β Γ πΉ
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PROBLEM 2: Which of the following statements best describes when the moment due to an applied force is zero? A. The force is negative B. The force is through the origin C. The line of action passes through the center of rotation D. The force is a function of time
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PROBLEM 3: The magnitude of the moment about the base point βπβ created by the 300 ππ force is most close to:
A. 2,702 lb β ft (counterclockwise) B. 2,010 lb β ft (clockwise) C. 1,559 lb β ft (counterclockwise) D. 1,224 lb β ft (clockwise)
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PROBLEM 4: A 10 N and 5 N force act on a particular beam, as illustrated. The magnitude of the moment about the base located a point βπ΄β is most close to:
A. 20 N β m (clockwise) B. 20 N β m (counterclockwise) C. 40 N β m (clockwise) D. 40 N β m (clockwise)
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PROBLEM 5: The moment generated by a 12 N force acting at a perpendicular distance of 3 meters away from a pivot point is most close to: A. 3 N β m B. 7 N β m C. 12 N β m D. 36 N β m
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PROBLEM 6: A force represented by the vector, πΉ = 10 π π, acts at a distance represented by a second vector, π = 5 π π. The moment generated by the force is most close to: A. β50 π B. 50 π C. β50 π D. 50 π
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PROBLEM 7: A couple is defined as _________ separated by a perpendicular distance. The expression that makes this statement TRUE is: A. Two forces in the same direction B. Two forces of equal magnitude C. Two forces of equal magnitude acting in the same direction D. Two forces of equal magnitude acting in opposite directions
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PROBLEM 8: If three couples act on a body, the statement that best describes the result on the body is: A. The net force is not equal to 0 B. The net force and net moment are equal to 0 C. The net moment equals 0, but the net force is not necessarily equal to 0 D. The net force equals 0, but the net moment is not necessarily equal to 0
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MOMENTS (COUPLES) | SOLUTIONS SOLUTION 1: The GENERAL FORMULAS revolving around MOMENTS (COUPLES) of a TWO FORCE SYSTEM can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. In this problem, we are told that there is some FORCE, F, acting some DISTANCE, r, from a particular POINT Oβ¦we are then asked to define an expression for the MOMENT that this creates. A MOMENT, with a general notation as M, is defined as the CROSS PRODUCT of the RADIUS VECTOR and the FORCE from a POINT to the LINE OF ACTION of the that FORCE. The RADIUS VECTOR is the same as the DISTANCE from the LINE OF ACTION of the FORCE to POINT O, therefore, we have everything we need defined to derive our expression. In FORMULAIC TERMS, the MOMENT can be expressed as: π = π Γ πΉ The correct answer choice is C. ππ = π« Γ π
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SOLUTION 2: The TOPIC of MOMENTS (COUPLES) 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 MOMENT of a FORCE about some specified point QUANTIFIES the TENDENCY of that FORCE to ROTATE a RIGID BODY about some specified point. The MAGNITUDE of a MOMENT identifies the AXIS OF ROTATION associated with the rotational force. The MAGNITUDE for a MOMENT, which is generally denoted as M, is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ So as a FORCE extends further away from a particular POINT, the MOMENT increases about the CENTER OF ROTATION. However, if the LINE OF ACTION of the FORCE PASSES THROUGH the CENTER OF ROTATION, the NET RESULTANT MOMENT on the body will be ZERO, thus, there will be NO ROTATION. The correct answer choice is C.
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SOLUTION 3: The GENERAL FORMULAS revolving around MOMENTS (COUPLES) of a TWO FORCE SYSTEM can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A MOMENT, with a general notation as M, is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ However, when we are assessing a MOMENT in a TWO DIMENSIONAL space, the GENERAL FORMULA using a CROSS PRODUCT simplifies down to the PRODUCT of the FORCE times the DISTANCE to the AXIS parallel to the one in which the ROTATION is occurring, or in formulaic terms: π = ππΉ In classical mechanics, this DISTANCE is referred to as the LEVER ARM. Remember this fundamental definition, as it could be a money shot problem in your favor revolving around basic terminology come exam day. A MOMENT is often a DIRECT RESULT of the ACTION of a FORCE VECTOR, and because of this, a MOMENT is also a VECTOR defined by similar characteristics.
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A MOMENT will generally have the following CHARACTERISTICS, or the EQUIVALENT, defined: β’ MAGNITUDE β’ POINT OF APPLICATION β’ SENSE In this problem, we have the POINT OF APPLICATION, but need to define the MAGNITUDE as well as the SENSE acting at POINT O. The MAGNITUDE of a moment quantifies the INTENSITY of its ROTATIONAL EFFECT. The SENSE of a MOMENT is the DIRECTION OF ROTATION about its AXIS OF ROTATION. When we are dealing with a FORCE that is at an ANGLE relative to the COORDINATE SYSTEM in which it resides, we will need to use our knowledge of RESOLVING a FORCE in to COMPONENTS so that we can assess each one INDIVIDUALLY and how it affects the ROTATION about our AXIS OF ROTATION.
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In this problem, we are given the illustration:
The 300 ππ force generates a MOMENT about the base at point βπβ. At point βπβ, this MOMENT will have a SENSE in either the clockwise or counterclockwise direction. We will typically represent the SENSE as a CLOCKWISE (NEGATIVE) or COUNTERCLOCKWISE (POSITIVE) behavior. The DIRECTION you take in your analysis of MOMENTS is not set in stone, itβs just a suggested route, and an ASSUMPTION of the resulting MOMENT in the end. Whether you define clockwise or counterclockwise rotation as positive is inconsequential, itβs that you remain consistent throughout your analysis of the problem that determines whether or not you arrive at a valid conclusion. Assuming you correctly write the EQUATIONS OF EQUILIBRIUM, in the end, your answer may result in a NEGATIVE SCALAR, or SIGN OPPOSITE to what you assumed
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originally as negative or positive. This indicates that the MOMENT is indeed acting in the OPPOSITE DIRECTION than what you initially assumed. This is not a bad thing, it just allows you to adjust your moment at the end of your analysis to be properly represented and distributed within your FORCE system. We will assume the COUNTERCLOCKWISE direction as POSITIVE. In this problem, we will break the FORCE VECTOR in to its COMPONENTS and define a PAIR of MOMENTS to determine the CUMULATIVE AFFECT at POINT O. Illustrating the FORCE COMPONENTS of the 300 lb FORCE, we have:
Letβs put some definition around these 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.
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Knowing where our ANGLE of 30Β° resides in relation to the HORIZONTAL X-AXIS, we can identify the X-AXIS as our OPPOSITE side, Y-AXIS as our ADJACENT 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: πΉ^ = πΉ_ sin π β’ The VERTICAL COMPONENT of the force: πΉa = πΉ_ cos π Letβs start with calculating the HORIZONTAL COMPONENT. Pulling the data we have and plugging it in to the general formula established, we get: πΉ^bcc = (300 ππ) sin(30Β°) Or: πΉ^bcc = 150 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: πΉabcc = (300 ππ) cos(30Β°)
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Or: πΉabcc = 260 ππ With this, we now have:
With each FORCE COMPONENT now defined, we can identify the LINE OF ACTION and associated LEVER ARM back to the AXIS OF ROTATION for each, such that:
Remember that the LEVER ARM must be PERPINDICULAR from the LINE OF ACTION back to the AXIS OF ROTATIONβ¦this is IMPORTANT.
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Many will attempt to find the DISTANCE from the AXIS OF ROTATION back to the POINT OF APPLICATION, such that:
This will result in a WRONG ANSWER, and trust us, the NCEES knows that this is a SIMPLE yet COMMON error we as engineers make:
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By Observation, we can see that our HORIZONTAL COMPONENT FORCE will cause a MOMENT about POINT O in the CLOCKWISE direction, or rather, a NEGATIVE MOMENT:
In contrast to that, the VERTICAL COMPONENT FORCE will cause a MOMENT about POINT O in the COUNTERCLOCKWISE direction, or rather, a POSITIVE MOMENT:
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The CUMULATIVE effects of each MOMENT when SUMMED about POINT O will determine the ultimate results, or in FORMULAIC TERMS: π% = βπΉ^ π^ + πΉa πa It is important to note the SENSE of each component and whether they create a clockwise or counterclockwise moment and account for this when combining the two MOMENTS. At this point, we have defined: πΉ^ = 150 ππ πΉa = 260 ππ We need to QUANTIFY the LEVER ARM form the AXIS OF ROTATION to the LINE OF ACTION of each individual COMPONENT FORCE.
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To do this, we will use the GEOMETRY we are given in the original illustrationβ¦and more specifically, we will break the system in to two separate RIGHT TRIANGLES as illustrated:
We see that breaking the SYSTEM up like this gives us a lot of information, and we can almost define both LEVER ARMS, minus one little portion of the HORIZONTAL LEVER ARM, dx, which is highlighted here in RED:
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However, using the GEOMETRY to define the BASE of the TRIANGLE that allows us to determine the VERTICAL LEVER ARM, we will have everything we needβ¦so letβs start here, we have:
Using simple TRIGNOMETRIC RELATIONSHIPS, we can calculate the VERTICAL LEVER ARM as: πa = 6 ππ‘ cos 30Β° = 5.2 ππ‘ This also allows us to determine the BASE, giving us the missing portion of the HORIZONTAL LEVER ARM, as: π^ = (6 ππ‘) sin 30Β° = 3 ππ‘
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Therefore, already having been given 6 ft of the HORIZONTAL LEVER ARM:
We can add this calculated horizontal distance to fully quantify our HORIZONTAL LEVER ARM as: π^ = 3 + 6 = 9 ππ‘ Letβs now revisit our CUMULATIVE MOMENT as we previously developed in FORMULAIC TERMS: π% = βπΉ^ π^ + πΉa πa We are now at a point where we can just plug and chug. Plugging in the data that we have defined up to this point, we can calculate the CUMULATIVE MOMENT generated about point βπβ as: π% = 260 ππ 9 ππ‘ β (150 ππ) 5.2 ππ‘
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Or: π% = 1,559 ππ β ππ‘ The last thing we need to do is determine the SENSE of the MOMENT about POINT O. We originally assumed that the COUNTERCLOCKWISE direction represented a POSITIVE MOMENT. In the end, if our MOMENT resulted in a NEGATIVE SCALAR, then that would indicate that the MOMENT actually acts in the OPPOSITE DIRECTION than what we initially assumed. Our resultant MOMENT is POSITIVE, which tells us that our initial ASSUMPTION that the COUNTERCLOCKWISE direction represented a POSITIVE MOMENT was correct. The correct answer choice is C. π, πππ π₯π β ππ (ππππππππππππππππ)
SOLUTION 4: The GENERAL FORMULAS to use when working with MOMENTS (COUPLES) in a TWO FORCE SYSTEM 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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A MOMENT is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ However, when we are assessing a MOMENT in a TWO DIMENSIONAL space, the GENERAL FORMULA using a CROSS PRODUCT simplifies down to the PRODUCT of the FORCE times the DISTANCE to the AXIS parallel to the one in which the ROTATION is occurring, or in formulaic terms: π = ππΉ In classical mechanics, this DISTANCE is referred to as the LEVER ARM. In this problem, we are given a beam loaded as:
The beam is being acted upon by two forces on either side of a dimensioned PIVOT POINT at A.
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Each FORCE applied on this beam will create a tendency for it to ROTATE about the PIVOT POINT at A. This is otherwise known as the MOMENT. We are able to quantify the MAGNITUDE of the MOMENT that each FORCE creates individually, then SUM them to determine the CUMULATIVE AFFECT about POINT A. Generally speaking, in ENGINEERING, and specifically ENGINEERING MECHANICS: STATICS, we donβt want ROTATION to occur in a way that is not controlled. Therefore, after analyzing the CUMULATIVE AFFECTS of these two forces, we will have the data necessary to install a third component somewhere on the beam to balance the beam and establish the EQUILIBRIUM we are often after. This can be done in one of two ways, we either: β’ CREATE a COUPLE at some LOCATION on this BEAM that is EQUAL and OPPOSITE in MAGNITUDE. β’ ADD a FORCE at some LOCATION on this BEAM that creates a MOMENT that is EQUAL and OPPOSITE in MAGNITUDE. However, that is not what we are seeking at this point, so letβs move on and get in to our problem. The first thing we should do is establish a COORDINATE SYSTEM to work within.
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We could do this mentally in our head, which is the recommended route, or we can quickly scratch it out on the paper we are working our solution. In this problem, we will assume that we are working in a standard CARTESIAN COORDINATE SYSTEM and that any MOMENTS acting about POINT A will be POSITIVE if they are acting with a SENSE in the COUNTERCLOCKWISE direction and NEGATIVE if they are acting with a SENSE in the CLOCKWISE direction. The initial assumptions we build around the DIRECTION of the MOMENT at the front end of our analysis is not set in stone, itβs just a suggested route, and an ASSUMPTION of the resulting MOMENT in the end. Whether you define CLOCKWISE or COUNTERCLOCKWISE rotation as POSITIVE is inconsequential, itβs that you remain consistent throughout your analysis of the problem that determines whether or not you arrive at a valid conclusion. Assuming you correctly write the EQUATIONS, in the end, your answer may result in a NEGATIVE SCALAR to what you assumed originally as negative or positive. This indicates that the MOMENT is indeed acting in the OPPOSITE DIRECTION than what you initially assumed. This is not a bad thing, it just allows you to adjust your MOMENT at the end of your analysis to be properly represented and distributed within your FORCE system. So letβs continue with our assumption that the COUNTERCLOCKWISE direction is POSITIVE.
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Based on this convention, we can conclude generically up front that the FORCE acting to the LEFT of the PIVOT POINT, and with a DOWNWARD direction, will create a ROTATION, or MOMENT, about the PIVOT POINT at A in the COUNTERCLOCKWISE direction. The FORCE acting to the RIGHT of the PIVOT POINT, and with a DOWNWARD direction, will create a ROTATION, or MOMENT, about the PIVOT POINT at A in the CLOCKWISE direction. Therefore, our GENERAL FORMULA to determine the CUMULATIVE AFFECTS of these FORCES at POINT A can be written as: π| = πΉ}~ π}~ β πΉβ’ πβ’ We are given everything we need to carry forward with a plug and chug sprinkled in with a light deployment of our calculator. We are given: πΉ}~ = 10 π πΉβ’ = 5 π π}~ = 3 π πβ’ = 2 π Plugging these values in, we get: π| = 10 π 3 π β 5 π 2 π
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Or: π| = 20 π β π The last thing we need to confirm is the SENSE of this MOMENT. Originally, we had assumed that a MOMENT about POINT A would be POSITIVE if it was acting in the COUNTERCLOCKWISE direction. We then went on to state that it didnβt matter what our original assumption was, it could be anything, and that the final result would tell us whether we were correct or not. If we werenβt correct in our original assumptions, then our result would play out as a NEGATIVE value, telling us that our MOMENT was acting in the OPPOSITE DIRECTION. However, in this problem, our resulting MOMENT is a POSITIVE VALUE, confirming that our original assumption that a POSITIVE MOMENT would act in a COUNTERCLOCKWISE direction. Therefore, our resulting MOMENT too will act in the COUNTERCLOCKWISE direction. To further solidify this point, had our value came back as: π| = β20 π β π
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Then we would have concluded that the MOMENT, with the same MAGNITUDE, was acting in a CLOCKWISE direction about POINT A. The correct answer choice is B. ππ π β π¦ (ππππππππππππππππ)
SOLUTION 5: The GENERAL FORMULAS to use when working with MOMENTS (COUPLES) in a TWO FORCE SYSTEM can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A MOMENT is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ However, when we are assessing a MOMENT in a TWO DIMENSIONAL space, the GENERAL FORMULA using a CROSS PRODUCT simplifies down to the PRODUCT of the FORCE times the DISTANCE to the AXIS parallel to the one in which the ROTATION is occurring, or in FORMULAIC TERMS: π = ππΉ In classical mechanics, this DISTANCE is referred to as the LEVER ARM.
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In this problem, we are given all the information we need to solve the problem right off the bat: πΉ = 12 π π = 3 π Plugging these values in to the GENERAL FORMULA for a MOMENT in a TWO DIMENSIONAL space, we get: π = 12 π 3 π Or: π = 36 π β π Generally speaking, in ENGINEERING, and specifically ENGINEERING MECHANICS: STATICS, we donβt want ROTATION to occur in a way that is not controlledβ¦we seek EQUILIBRIUM. However, in this problem, we donβt know enough about the FORCE to determine the SENSE of the MOMENT created, so we canβt go any further in our analysis to conclude where we should implement an additional component to establish this EQUILIBRIUM.
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Had we known another characteristic of this FORCE, such as whether the FORCE acts to the RIGHT or LEFT of the PIVOT POINT, then we could establish EQUILIBIRUM in one of two ways, we either: β’ CREATE a COUPLE at some LOCATION on this BEAM that is EQUAL and OPPOSITE in MAGNITUDE. β’ ADD a FORCE at some LOCATION on this BEAM that creates a MOMENT that is EQUAL and OPPOSITE in MAGNITUDE. The correct answer choice is D. ππ π β π¦
SOLUTION 6: The GENERAL FORMULAS to use when working with MOMENTS (COUPLES) in a TWO FORCE SYSTEM can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A MOMENT is defined as the CROSS PRODUCT of the RADIUS VECTOR, r, and the FORCE, F, from a POINT to the LINE OF ACTION of the that FORCE, or in FORMULAIC TERMS: π = π Γ πΉ However, when we are assessing a MOMENT in a TWO DIMENSIONAL space, the GENERAL FORMULA using a CROSS PRODUCT simplifies down to the PRODUCT of
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the FORCE times the DISTANCE to the AXIS parallel to the one in which the ROTATION is occurring, or in FORMULAIC TERMS: π = ππΉ In classical mechanics, this DISTANCE is referred to as the LEVER ARM. This problem is in TWO DIMENSIONS, and if we are quick to recognize the CARTESIAN COMPONENTS, then in our head, we can conclude that we have a FORCE that is acting UPDWARDS at a DISTANCE, or with a LEVER ARM, of 5 m. However, we are given VECTORS, so letβs move forward through this problem using the CROSS PRODUCT. The GENERAL FORMULA for the CROSS PRODUCT can be referenced under the subject of MATHEMATICS on page 35 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The CROSS PRODUCT is one of two OPERATIONS that can be deployed using VECTOR MULTIPLICATION, the other operation being the DOT PRODUCT. In itβs most fundamental form, the CROSS PRODUCT takes two vectors, runs them through a standard process, and creates a third vector that is ORTHOGONAL (or PERPENDICULAR) to each of the original vectors.
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In this problem, our two vectors are given as: πΉ = 10 π π π = 5 π π Or otherwise written as: πΉ =< 0, 0, 10 > π π =< 0, 5, 0 > π There are two routes we can take in solving this problem, the first will be using the GENERAL FORMULA that we are explicitly given in the NCEES Reference Handbook, which is written as: π Γ πΉ =< πa πΉΛ β πΛ πΉa , πΛ πΉ^ β π^ πΉΛ , π^ πΉa β πa πΉ^ >= πΉ π π πππ We have all the ELEMENTS defined for each of our VECTORS, but we donβt have an ANGLE between them. This is not a problem, the first half of this GENERAL FORMULA is enough to define our resulting VECTORβ¦or MOMENT in this case. Plugging in the ELEMENTS where outlined, we get: π Γ πΉ =< 5(10) β 0(0),0(0) β 0(10), 0(0) β 5(0) > This simplifies down to: π Γ πΉ =< 5(10) >
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Which results in a new VECTOR: π Γ πΉ =< 50, 0, 0 > Or in VECTOR NOTATION: π Γ πΉ = 50π β 0π + 0π Which tells us that the resulting MOMENT in VECTOR NOTATION is: π = 50π π β π
A QUICK HACK TO THIS PROBLEM: You can use your CALCULATOR to hack this problem in a fraction of the time. Check out the CALCULATOR WORKSHOPS page within PREPINEER to study just how we suggest you move forward with these types of problems. The correct answer choice is B. ππ π
SOLUTION 7: The topic of MOMENTS (COUPLES) 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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A system of TWO FORCES that are EQUAL in MAGNITUDE, OPPOSITE IN DIRECTION, and PARALLEL to each other is called a COUPLE. A COUPLE is equivalent to a single moment vector exerting a resultant moment, but no resultant force. The moment vector of the couple is independent of the choice of the origin of the coordinate axes. It is a FREE VECTOR that can be applied at any point with the same effect. A COUPLE is characterized as having a ZERO RESULTANT, as well as exerting the same resultant moment about all points. The NET EFFECT is ROTATION WITHOUT TRANSLATION. A force couple system is said to have moments equal to each such that they can be expressed as equal and opposite of each other, expressed as: π} = πβΉ Or: πΉ} π} = πΉβΉ πβΉ A COUPLE can be COUNTERACTED ONLY by another COUPLE. A COUPLE can be MOVED to any location WITHOUT AFFECTING THE EQUILIBRIUM requirements.
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Therefore, the correct answer choice is D.
SOLUTION 8: The topic of MOMENTS (COUPLES) can be referenced under the subject of STATICS on page 67 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. A system of TWO FORCES that are EQUAL in MAGNITUDE, OPPOSITE IN DIRECTION, and PARALLEL to each other is called a COUPLE. A COUPLE is equivalent to a single moment vector exerting a resultant moment, but no resultant force. The moment vector of the couple is independent of the choice of the origin of the coordinate axes. It is a FREE VECTOR that can be applied at any point with the same effect. A COUPLE is characterized as having a ZERO RESULTANT, as well as exerting the same resultant moment about all points. The NET EFFECT is ROTATION WITHOUT TRANSLATION. A force couple system is said to have moments equal to each such that they can be expressed as equal and opposite of each other, expressed as: π} = πβΉ
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Or: πΉ} π} = πΉβΉ πβΉ A COUPLE can be COUNTERACTED ONLY by another COUPLE. A COUPLE can be MOVED to any location WITHOUT AFFECTING THE EQUILIBRIUM requirements. The correct answer choice is D.
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