09 Moments Couples Concept Overview
MOMENTS (COUPLES) | CONCEPT OVERVIEW The TOPIC OF MOMENTS (COUPLES) can be referenced on page 67 of the NCEES Supplied Reference Handbook, 9.4 version for Computer Based Testing.
CONCEPT INTRO: A force has two effects when acting on a solid object. The first is that the force makes the objectβs center of mass move, creating an acceleration and along or more planes of displacement. The second is that the force causes the object to rotate about an axis. A MOMENT is a name given to the tendency of a force to rotate, turn, or twist a rigid body about an actual or assumed pivot point. When acted upon by a moment, unrestrained bodies rotate. When a restrained body is acted upon by a moment, there is no rotation. The moment of a force about some point quantifies its tendency to rotate an object about that point. The magnitude of a moment identifies the axis of rotation associated with the rotational force. If the line of action of a force passes through the center of rotation, the net resultant moment on a body is zero. The formulas for the MOMENTS (COUPLES) OF A TWO-FORCE SYSTEM can be referenced under the topic 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 MOMENT (π) is defined as the cross product of the RADIUS VECTOR (π) and the FORCE (πΉ) from a point to the line of action of the force π = π Γ πΉ π' = π¦πΉ) β π§πΉ, , π, = π§πΉ' β π₯πΉ) , πππ π) = π₯πΉ, β π¦πΉ' A moment is always equal to the FORCE TIMES THE DISTANCE, where the distance is referred to as the LEVER ARM. A moment is commonly referred to as TORQUE in applications such as shafts and other power-transmitting machines. Because a moment is often a direct result of the action of a force vector, a moment is also a vector and must also have similar characteristics. Like any vector, a moment vector has a magnitude, point of application, and sense. The magnitude of the moment is a measure of the intensity of the rotational effect. Instead of having a unique point of application (location in space) or a defined line of action (line in space on which the vector is acting) as forces do, the moment actually rotates around an axis called the AXIS OF ROTATION.
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The sense is the direction of rotation about its axis of rotation. You usually represent it as a clockwise (negative) or counterclockwise (positive) behavior. An axis of rotation can either be within the object, which results in a rotational behavior known as pivoting, or outside an object, which results in a rotational behavior known as ORBITING. It is important to realize that you can define positive as either clockwise or counterclockwise, as long as you are consistent throughout your analysis of the problem. Assuming you correctly write the equations of equilibrium, in the end your answer will have a negative scalar or sign opposite to what you assumed, indicating that the moment is indeed acting in the opposite direction than what you initially assumed. The moment of a force is calculated as the cross product of the radius vector (π) and the force (πΉ) from a point to the line of action of the force. π = π Γ πΉ = π πΉ π πππ
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Where: β’ βπβ is a vector or distance from the point at which the moment is being calculated to the point at which the force is applied. This distance is given in units of meters, inches, feet, and other units of length. β’ βπΉβ is the force applied causing the moment. This given in units of pounds or Newtons. β’ π is the angle In two dimensions, the π₯- and π¦ β Cartesian axes are usually in the plane of the page, which results in the third axis, the z-axis, being out of the page because it must be perpendicular to the two-dimensional axes. A two-dimensional moment of a force located in the π₯π¦ β Cartesian plane is always about the π§ β ππ₯ππ (the axis of rotation is parallel to the π§ β ππ₯ππ ). In vector terms, this fact means that the moment in the π₯π¦ β plane has a unit vector direction of π (either positive, or negative depending on the sense of the moment). An object experiences a moment whenever a force is applied to it. Only when the line of action of the force passes through the center of rotation will the moment be zero. A force has two effects when acting on a solid object which are that it makes the objectβs center of mass move, creating an acceleration, and it causes the object to rotate. The moment of a force about some point quantifies its tendency to rotate an object about that point, the magnitude of the moment identifies the extent of the rotational force, and the direction of the moment identifies the axis of rotation associated with the rotational force.
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The line of action of the moment vector is normal to the plane containing the force vector and the position vector. The sense of the moment is determined from the RIGHT-HAND RULE. The right-hand rule only works with the right hand! If you use your left hand by mistake, the direction of your z β axis will be backward. After aligning your fingers with the Cartesian system, if your moment is about one of the Cartesian π₯β, π¦β, ππ π§ β ππ₯ππ , you can determine the sense of the moment by looking at the end of the finger that is parallel to the axis of rotation of the moment. In three dimensions, a moment is positive about the π₯ β ππ₯ππ if itβs acting counterclockwise when youβre looking at the tip of your thumb. The same applies to the π¦ β ππ₯ππ and your forefinger and to the π§ β ππ₯ππ and your middle finger.
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Grab a pencil and pen, and letβs practice using the right-hand rule. Designate the left end of the pencil as the pivot, and hold it rigidly with your finger. β’ The pen is the force. β’ If the pen is an upward force, you will be applying the force below the pencil, but directed upwards. β’ If the pen is a downward force, you will be applying the force above the pencil, but directed downwards.
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For each of these scenarios, notice which direction the pencil is moving in: clockwise or counterclockwise? 1. Apply an upward force through the center of the pencil (CCW) 2. Apply an upward force through the pivot (No movement) 3. Apply an upward force through the edge farthest from the pivot (CCW) 4. Apply a downward force through the center of the pencil (CW) 5. Apply a downward force through the pivot (No movement) 6. Apply a downward force through the edge farthest from the pivot (CCW)
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Consider the beam below, that is pivoting at some point due to a force being applied at a distance "π" away.
When a force is applied to the beam, it will cause it to rotate about the pivot point. This is illustrated by the force "πΉ" at a distance "π" from the pivot point. This will cause the beam to rotate counter-clockwise. To put a stop to this rotation, a couple or opposing force must be placed on the opposite side of the pivot point at the same distance. This can be achieved by: β’ Adding a force "πΉ" at an equal distance "π" to the right of the pivot
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β’ Adding a force "2πΉ" at an equal distance "π/2" to the right of the pivot
β’ Adding a force "πΉ/2" at an equal distance "2π" to the right of the pivot
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UNITS: The topic of UNITS OF FORCE can be referenced under the topic of UNITS on page 1 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The moment commonly used unit for MOMENTS or TORQUE is πππ€π‘ππ β πππ‘πππ , which is the SI unit denoted by the symbol of π β π. Though, we are probably more familiar with and comfortable using SI units, we should be prepared to work problems that use US units as well because we can expect to see these on the FE Exam. The standard US unit for moments or torque is the ππππ‘ β πππ’ππ, which is denoted by the symbol of ππ‘ β ππJ . The formula for the CONVERSION FOR UNITS OF MOMENTS can be referenced under the topic of UNITS on page 1 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. To convert from SI units to US units, we use the following conversion factor: 1 ππ = 0.738 ππ‘ ππ; ππ 1 ππ‘ β ππ = 1.356 π β π It is important that we visualize practical examples of moments, or torques, in engineering systems, to better help us understand where they occur; a few examples are: β’ Moments can also emerge as reaction forces. One example of this is the resistance you encounter when turning the steering wheel of a car. The force you feel is created by moments acting on the wheels where they touch the ground.
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β’ Moments can also appear as internal forces in structural members or components. One example of this can be seen when a beam bends because of an internal moment with a direction transverse to the direction of the beam. An internal moment causes points to rotate relative to each other just as an internal force causes points in a solid to move relative to each other. MOMENT-COUPLE SYSTEMS: 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 exerts 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 having a zero resultant, as well as exerting the same resultant moment about all point. 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, as expressed: πS = πT ππ πΉS πS = πΉT πT 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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One example of a couple is the forces from your fingers on a doorknob. As you turn the knob, one finger is pushing up on the side of the doorknob and the other is pushing down. These two forces together cause the knob to rotate. Actually opening or closing the door requires that you push on the door after you have turned the knob, which is the scenario that I will discuss in the preceding section. Consider the figure below of an open door with a pushing force applied by someone opening the door:
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In a door assembly, doors typically hang from several hinges that are aligned along a single vertical axis of rotation. When you push on the door (apply a force), the resulting action is that the door begins to move. Regardless of where you apply the force on the door (at the top, on the handle, or by your foot on the bottom of the door), the same resulting behavior (a moment) occurs along the axis of rotation. This moment is what creates the rotation that results in the door swinging open or closed. CONCENTRATED MOMENTS: A couple can also be referred to as a PURE MOMENT or CONCENTRATED MOMENT. Concentrated moments are actually fairly easy to depict on a free-body diagram. Because moments are vectors, too, they must again display magnitude, sense, and point of application. For both two- and three- dimensional objects, you need to display the magnitude of a concentrated moment as the numerical value (if itβs known), or a label for the vector (if itβs unknown). The transmission of moments can occur through multiple objects. An example of this type of situation occurs in the drive train of your automobile. The engine of your car causes the transmission to rotate, which in turn causes the axles to rotate, which in turn cause the wheels to rotate. Now obviously, moving a car down the road requires many more factors, but the overall concept of the transmitted moment is still valid.
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CONCEPT EXAMPLE: The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. Calculate the magnitude of the moment about the base point βπβ created by the 600 π force.
A. 2,610 π β π (πππ’ππ‘ππππππππ€ππ π) B. 2,610 π β π (ππππππ€ππ π) C. 2,980 π β π (πππ’ππ‘ππππππππ€ππ π) D. 2,980 π β π (ππππππ€ππ π)
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SOLUTION: In this problem, the 600 π force generated a moment about the base at point βπβ. At point βπβ, a moment will generate rotation in either the clockwise or counterclockwise direction. It is typical to assume the counterclockwise direction as positive (+).
When working problems that involve moment, it is important to determine the line of action of the force, and the perpendicular lever arm distance between the line of action and axis of rotation.
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The next step is to define a Coordinate System relative to the axis of rotation. This is helpful when determining the perpendicular distance between the axis of rotation and the line of action of the force.
A moment is comprised of two components, the perpendicular distance and the magnitude of the force. As we are given the force, we can calculate the horizontal and vertical components based on the provided geometry.
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Looking at the formula for a moment, we need the magnitude of the force and the lever arm distance. As we know the magnitude of the force, we need to solve for the lever arm distance in order to calculate the moment about point βπβ. The tricky part about this problem is realizing that you need to calculate a horizontal lever arm distance and a vertical lever arm distance, that correspond to the respective horizontal and vertical components.
We now have a clear representation of the horizontal and vertical perpendicular distances between the axis of rotation and the line of action for the force. The next step for us is to break the force into its horizontal and vertical components, which we can then use to calculate the moment generated by each component. The formulas for the COMPONENTS OF A VECTOR are not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
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The π₯ β ππ₯ππ usually represents the horizontal component of the force: π»ππππ§πππ‘ππ πΆππππππππ‘: πΉ' = πΉ cos π = 600 π (cos 40Β°) = 460 π The π¦ β ππ₯ππ usually represents the vertical component of the force: ππππ‘ππππ πΆππππππππ‘: πΉ, = πΉ sin π = (600 π)(sin 40Β°) = 386 π The formulas for the MOMENTS (COUPLES) OF A TWO-FORCE SYSTEM can be referenced under the topic 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 MOMENT (π) is defined as the cross product of the RADIUS VECTOR (π) and the force (πΉ) from a point to the line of action of the force: π = π Γ πΉ Now that we have the force for each component, we calculate the overall moment as the sum of the moment generated by each component such that: πe = πΉ' π' + πΉ, π,
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Plugging in the numbers provided in the problem, we can calculate the overall moment generated at point βπβ as: πe = 460 π 4.00 π + (386 π) 2.00 π = 2,610 π β π The last thing we need to worry about in this problem is determining the sense of the moment. Using the right hand rule of visually looking at the free body diagram, we can see that the force generates a clockwise moment.
Therefore, the correct answer choice is B. π, πππ π β π¦ (πππππππππ) HINT: REMEMBER these steps before starting any problem that involves finding reactions or moments. 1. Establish a coordinate axis system. Drawing a small coordinate system in the corner of the problem will help! 2. Establish whether clockwise, or counter-clockwise moment direction will be positive. Stick to whichever way you choose! Itβs important to stay consistent! 3. Always remember the following equations that you must solve for in order the find the reactions and moment of the given problem. Remember, in order for the structure or beam to be in equilibrium, everything must balance. Sum of the forces in the βπ₯β direction:
πΉ' = 0:
Sum of the forces in the βπ¦β direction:
πΉ, = 0:
Sum of the moment about a chosen point:
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CONCEPT INTRO: A force has two effects when acting on a solid object. The first is that the force makes the objectβs center of mass move, creating an acceleration and along or more planes of displacement. The second is that the force causes the object to rotate about an axis. A MOMENT is a name given to the tendency of a force to rotate, turn, or twist a rigid body about an actual or assumed pivot point. When acted upon by a moment, unrestrained bodies rotate. When a restrained body is acted upon by a moment, there is no rotation. The moment of a force about some point quantifies its tendency to rotate an object about that point. The magnitude of a moment identifies the axis of rotation associated with the rotational force. If the line of action of a force passes through the center of rotation, the net resultant moment on a body is zero. The formulas for the MOMENTS (COUPLES) OF A TWO-FORCE SYSTEM can be referenced under the topic 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 MOMENT (π) is defined as the cross product of the RADIUS VECTOR (π) and the FORCE (πΉ) from a point to the line of action of the force π = π Γ πΉ π' = π¦πΉ) β π§πΉ, , π, = π§πΉ' β π₯πΉ) , πππ π) = π₯πΉ, β π¦πΉ' A moment is always equal to the FORCE TIMES THE DISTANCE, where the distance is referred to as the LEVER ARM. A moment is commonly referred to as TORQUE in applications such as shafts and other power-transmitting machines. Because a moment is often a direct result of the action of a force vector, a moment is also a vector and must also have similar characteristics. Like any vector, a moment vector has a magnitude, point of application, and sense. The magnitude of the moment is a measure of the intensity of the rotational effect. Instead of having a unique point of application (location in space) or a defined line of action (line in space on which the vector is acting) as forces do, the moment actually rotates around an axis called the AXIS OF ROTATION.
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The sense is the direction of rotation about its axis of rotation. You usually represent it as a clockwise (negative) or counterclockwise (positive) behavior. An axis of rotation can either be within the object, which results in a rotational behavior known as pivoting, or outside an object, which results in a rotational behavior known as ORBITING. It is important to realize that you can define positive as either clockwise or counterclockwise, as long as you are consistent throughout your analysis of the problem. Assuming you correctly write the equations of equilibrium, in the end your answer will have a negative scalar or sign opposite to what you assumed, indicating that the moment is indeed acting in the opposite direction than what you initially assumed. The moment of a force is calculated as the cross product of the radius vector (π) and the force (πΉ) from a point to the line of action of the force. π = π Γ πΉ = π πΉ π πππ
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Where: β’ βπβ is a vector or distance from the point at which the moment is being calculated to the point at which the force is applied. This distance is given in units of meters, inches, feet, and other units of length. β’ βπΉβ is the force applied causing the moment. This given in units of pounds or Newtons. β’ π is the angle In two dimensions, the π₯- and π¦ β Cartesian axes are usually in the plane of the page, which results in the third axis, the z-axis, being out of the page because it must be perpendicular to the two-dimensional axes. A two-dimensional moment of a force located in the π₯π¦ β Cartesian plane is always about the π§ β ππ₯ππ (the axis of rotation is parallel to the π§ β ππ₯ππ ). In vector terms, this fact means that the moment in the π₯π¦ β plane has a unit vector direction of π (either positive, or negative depending on the sense of the moment). An object experiences a moment whenever a force is applied to it. Only when the line of action of the force passes through the center of rotation will the moment be zero. A force has two effects when acting on a solid object which are that it makes the objectβs center of mass move, creating an acceleration, and it causes the object to rotate. The moment of a force about some point quantifies its tendency to rotate an object about that point, the magnitude of the moment identifies the extent of the rotational force, and the direction of the moment identifies the axis of rotation associated with the rotational force.
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The line of action of the moment vector is normal to the plane containing the force vector and the position vector. The sense of the moment is determined from the RIGHT-HAND RULE. The right-hand rule only works with the right hand! If you use your left hand by mistake, the direction of your z β axis will be backward. After aligning your fingers with the Cartesian system, if your moment is about one of the Cartesian π₯β, π¦β, ππ π§ β ππ₯ππ , you can determine the sense of the moment by looking at the end of the finger that is parallel to the axis of rotation of the moment. In three dimensions, a moment is positive about the π₯ β ππ₯ππ if itβs acting counterclockwise when youβre looking at the tip of your thumb. The same applies to the π¦ β ππ₯ππ and your forefinger and to the π§ β ππ₯ππ and your middle finger.
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Grab a pencil and pen, and letβs practice using the right-hand rule. Designate the left end of the pencil as the pivot, and hold it rigidly with your finger. β’ The pen is the force. β’ If the pen is an upward force, you will be applying the force below the pencil, but directed upwards. β’ If the pen is a downward force, you will be applying the force above the pencil, but directed downwards.
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For each of these scenarios, notice which direction the pencil is moving in: clockwise or counterclockwise? 1. Apply an upward force through the center of the pencil (CCW) 2. Apply an upward force through the pivot (No movement) 3. Apply an upward force through the edge farthest from the pivot (CCW) 4. Apply a downward force through the center of the pencil (CW) 5. Apply a downward force through the pivot (No movement) 6. Apply a downward force through the edge farthest from the pivot (CCW)
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Consider the beam below, that is pivoting at some point due to a force being applied at a distance "π" away.
When a force is applied to the beam, it will cause it to rotate about the pivot point. This is illustrated by the force "πΉ" at a distance "π" from the pivot point. This will cause the beam to rotate counter-clockwise. To put a stop to this rotation, a couple or opposing force must be placed on the opposite side of the pivot point at the same distance. This can be achieved by: β’ Adding a force "πΉ" at an equal distance "π" to the right of the pivot
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β’ Adding a force "2πΉ" at an equal distance "π/2" to the right of the pivot
β’ Adding a force "πΉ/2" at an equal distance "2π" to the right of the pivot
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UNITS: The topic of UNITS OF FORCE can be referenced under the topic of UNITS on page 1 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. The moment commonly used unit for MOMENTS or TORQUE is πππ€π‘ππ β πππ‘πππ , which is the SI unit denoted by the symbol of π β π. Though, we are probably more familiar with and comfortable using SI units, we should be prepared to work problems that use US units as well because we can expect to see these on the FE Exam. The standard US unit for moments or torque is the ππππ‘ β πππ’ππ, which is denoted by the symbol of ππ‘ β ππJ . The formula for the CONVERSION FOR UNITS OF MOMENTS can be referenced under the topic of UNITS on page 1 of the NCEES Supplied Reference Handbook, 9.4 Version for Computer Based Testing. To convert from SI units to US units, we use the following conversion factor: 1 ππ = 0.738 ππ‘ ππ; ππ 1 ππ‘ β ππ = 1.356 π β π It is important that we visualize practical examples of moments, or torques, in engineering systems, to better help us understand where they occur; a few examples are: β’ Moments can also emerge as reaction forces. One example of this is the resistance you encounter when turning the steering wheel of a car. The force you feel is created by moments acting on the wheels where they touch the ground.
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β’ Moments can also appear as internal forces in structural members or components. One example of this can be seen when a beam bends because of an internal moment with a direction transverse to the direction of the beam. An internal moment causes points to rotate relative to each other just as an internal force causes points in a solid to move relative to each other. MOMENT-COUPLE SYSTEMS: 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 exerts 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 having a zero resultant, as well as exerting the same resultant moment about all point. 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, as expressed: πS = πT ππ πΉS πS = πΉT πT 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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One example of a couple is the forces from your fingers on a doorknob. As you turn the knob, one finger is pushing up on the side of the doorknob and the other is pushing down. These two forces together cause the knob to rotate. Actually opening or closing the door requires that you push on the door after you have turned the knob, which is the scenario that I will discuss in the preceding section. Consider the figure below of an open door with a pushing force applied by someone opening the door:
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by Prepineer | Prepineer.com
In a door assembly, doors typically hang from several hinges that are aligned along a single vertical axis of rotation. When you push on the door (apply a force), the resulting action is that the door begins to move. Regardless of where you apply the force on the door (at the top, on the handle, or by your foot on the bottom of the door), the same resulting behavior (a moment) occurs along the axis of rotation. This moment is what creates the rotation that results in the door swinging open or closed. CONCENTRATED MOMENTS: A couple can also be referred to as a PURE MOMENT or CONCENTRATED MOMENT. Concentrated moments are actually fairly easy to depict on a free-body diagram. Because moments are vectors, too, they must again display magnitude, sense, and point of application. For both two- and three- dimensional objects, you need to display the magnitude of a concentrated moment as the numerical value (if itβs known), or a label for the vector (if itβs unknown). The transmission of moments can occur through multiple objects. An example of this type of situation occurs in the drive train of your automobile. The engine of your car causes the transmission to rotate, which in turn causes the axles to rotate, which in turn cause the wheels to rotate. Now obviously, moving a car down the road requires many more factors, but the overall concept of the transmitted moment is still valid.
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CONCEPT EXAMPLE: The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. Calculate the magnitude of the moment about the base point βπβ created by the 600 π force.
A. 2,610 π β π (πππ’ππ‘ππππππππ€ππ π) B. 2,610 π β π (ππππππ€ππ π) C. 2,980 π β π (πππ’ππ‘ππππππππ€ππ π) D. 2,980 π β π (ππππππ€ππ π)
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SOLUTION: In this problem, the 600 π force generated a moment about the base at point βπβ. At point βπβ, a moment will generate rotation in either the clockwise or counterclockwise direction. It is typical to assume the counterclockwise direction as positive (+).
When working problems that involve moment, it is important to determine the line of action of the force, and the perpendicular lever arm distance between the line of action and axis of rotation.
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The next step is to define a Coordinate System relative to the axis of rotation. This is helpful when determining the perpendicular distance between the axis of rotation and the line of action of the force.
A moment is comprised of two components, the perpendicular distance and the magnitude of the force. As we are given the force, we can calculate the horizontal and vertical components based on the provided geometry.
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by Prepineer | Prepineer.com
Looking at the formula for a moment, we need the magnitude of the force and the lever arm distance. As we know the magnitude of the force, we need to solve for the lever arm distance in order to calculate the moment about point βπβ. The tricky part about this problem is realizing that you need to calculate a horizontal lever arm distance and a vertical lever arm distance, that correspond to the respective horizontal and vertical components.
We now have a clear representation of the horizontal and vertical perpendicular distances between the axis of rotation and the line of action for the force. The next step for us is to break the force into its horizontal and vertical components, which we can then use to calculate the moment generated by each component. The formulas for the COMPONENTS OF A VECTOR are not provided in the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. We must memorize this formula and understand its application independent of the NCEES Supplied Reference Handbook.
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The π₯ β ππ₯ππ usually represents the horizontal component of the force: π»ππππ§πππ‘ππ πΆππππππππ‘: πΉ' = πΉ cos π = 600 π (cos 40Β°) = 460 π The π¦ β ππ₯ππ usually represents the vertical component of the force: ππππ‘ππππ πΆππππππππ‘: πΉ, = πΉ sin π = (600 π)(sin 40Β°) = 386 π The formulas for the MOMENTS (COUPLES) OF A TWO-FORCE SYSTEM can be referenced under the topic 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 MOMENT (π) is defined as the cross product of the RADIUS VECTOR (π) and the force (πΉ) from a point to the line of action of the force: π = π Γ πΉ Now that we have the force for each component, we calculate the overall moment as the sum of the moment generated by each component such that: πe = πΉ' π' + πΉ, π,
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Plugging in the numbers provided in the problem, we can calculate the overall moment generated at point βπβ as: πe = 460 π 4.00 π + (386 π) 2.00 π = 2,610 π β π The last thing we need to worry about in this problem is determining the sense of the moment. Using the right hand rule of visually looking at the free body diagram, we can see that the force generates a clockwise moment.
Therefore, the correct answer choice is B. π, πππ π β π¦ (πππππππππ) HINT: REMEMBER these steps before starting any problem that involves finding reactions or moments. 1. Establish a coordinate axis system. Drawing a small coordinate system in the corner of the problem will help! 2. Establish whether clockwise, or counter-clockwise moment direction will be positive. Stick to whichever way you choose! Itβs important to stay consistent! 3. Always remember the following equations that you must solve for in order the find the reactions and moment of the given problem. Remember, in order for the structure or beam to be in equilibrium, everything must balance. Sum of the forces in the βπ₯β direction:
πΉ' = 0:
Sum of the forces in the βπ¦β direction:
πΉ, = 0:
Sum of the moment about a chosen point:
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πtuvwx = 0:
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