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6.1: Why is this thing turning?

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    81499
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    Anyone who has changed a flat tire, used a wrench, thrown a Frisbee, hung on a tree limb, or pushed a large box knows something about the tendency for objects to rotate or resist rotation. Angular momentum is the extensive physical property related to this phenomenon. Before we can develop the conservation of angular momentum relation, we must introduce several concepts for describing angular motion and the moment of a force.

    Consider:

    What's the purpose of a "cheater" bar applied to a wrench? (A "cheater" bar is a piece of pipe slipped over the end of a wrench to extend its length.)

    Where is the best place to push a filing cabinet on rollers? How did you judge best?

    To refresh your memory, imagine that your car has a flat tire and you must change it (See Figure \(\PageIndex{1}\)). To change the tire, you first raise the tire off the ground using a car jack. Next, you place the tire iron on a lug nut and try to loosen it. In the sketch, the force \(\mathbf{F}\) acts on the tire iron at point \(P\), the axle of the tire is at point \(O\), and the nut is located at point \(N\).

    A tire is held slightly off the ground by a wedge. A lug nut N on the tire is being loosened by a tire iron, which is held parallel to the ground. A downwards force F is applied to the free end of the tire iron, at point P located a distance of L from the tire center.

    Figure \(\PageIndex{1}\): Changing a tire.

    If you have been lucky in life, you may be a novice at tire changing, and will discover that the tire rotates when you push on the tire iron. A driver with less luck and more experience changing tires might only raise the tire partially or insert a wedge to prevent the tire from rotating. Experienced tire changers know that our ability to loosen the nut depends on both the force \(\mathbf{F}\) and its point of application (point \(P\)). They also know that the effectiveness of the applied force \(\mathbf{F}\) is less if it is applied at an angle to the tire iron.

    What would happen if the force was applied at point \(N\)? Would you be able to loosen the nut? Would the tire turn?

    6.1.1 Moment of a force about a point

    To quantify the ability of the force to rotate the tire we need a physical quantity that accounts for the magnitude and direction of force \(\mathbf{F}\) as well as its point of application. The quantity we desire is the moment of a force about a point.

    The moment of a force \(\mathbf{F}\) about a point \(\bf{O}\) is the vector (or cross) product of the position vector \(\mathbf{r}\) and the force \(\mathbf{F}\):

    \[ \mathbf{M}_O = \mathbf{r} \times \mathbf{F} \nonumber \]

    where \(\mathbf{r}\) is the position vector that extends from the point \(O\) to the point of application of the force (See Figure \(\PageIndex{2}\)).

    A three-dimensional coordinate system with x, y, and z axes contains a point P whose location relative to the origin O is given by the vector r. A force represented as a vector F is applied to point P. The smaller of the two angles made by the line of action of F with r is given by theta.

    Figure \(\PageIndex{2}\): Calculating moment of a force \(\mathbf{F}\) about a point \(O\).

    When the position vector \(\mathbf{r}\) and the force vector \(\mathbf{F}\) are written in terms of their components as \[\mathbf{r} = x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \quad \text { and } \quad \mathbf{F}=F_{x} \mathbf{i}+F_{y} \mathbf{j}+F_{z} \mathbf{k} \nonumber \] Eq. \((\PageIndex{1})\) is evaluated as the follows using standard cross-product operations: \[\begin{align} \mathbf{M}_{o} &=\mathbf{r} \times \mathbf{F}=\left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x & y & z \\ F_{x} & F_{y} & F_{z} \end{array}\right| \nonumber \\[4pt] &=\mathbf{i}\left|\begin{array}{cc} y & z \\ F_{y} & F_{z} \end{array}\right|-\mathbf{j}\left|\begin{array}{cc} x & z \\ F_{x} & F_{z} \end{array}\right|+\mathbf{k}\left|\begin{array}{cc} x & y \\[4pt] F_{x} & F_{y} \end{array}\right| \\ &=\left(y F_{z}-z F_{y}\right) \mathbf{i}-\left(x F_{z}-z F_{x}\right) \mathbf{j}+\left(x F_{y}-y F_{x}\right) \mathbf{k} \nonumber \end{align} \nonumber \]

    If we restrict ourselves to motion in the \(x \text{-} y\) plane (plane motion), the only component of \(\mathbf{M}_O\) with a non-zero value is the \(z\)-component — the last term on the right-hand side of Eq. \((\PageIndex{3})\).

    From our knowledge of vectors, we can say several things about this result:

    • The position vector \(\mathbf{r}\) and the force vector \(\mathbf{F}\) lie in a plane.
    • The moment \(\mathbf{M}_O\) of the force \(\mathbf{F}\) about point \(O\) is a vector (see Eq. \((\PageIndex{3})\)).
    • The line of action of the moment vector \(\mathbf{M}_{O}\) is normal (perpendicular) to the plane that contains both \(\mathbf{r}\) and \(\mathbf{F}\).
    • The point of application of the moment vector \(\mathbf{M}_{O}\) is at point \(O\), the point about which we are taking the moment.
    • The sense of the direction of the moment \(\mathbf{M}_{O}\) and the sense of the rotation it could impart can be described by the right-hand rule in several ways (See Figure \(\PageIndex{3}\)):
      Vectors r, pointing to the right and out of the page, and F, pointing to the right and into the page, lie in a plane parallel to the ground. The tail of r is located at point O and the tail of F is placed at the head of r. A parallelogram formed by 2 instances each of r and F placed head-to-tail is shaded in, with the smaller of the two angle between an r vector and an adjacent F designated theta. A vector M_O, indicating the moment about point O, points upwards and into the page.
      Figure \(\PageIndex{3}\): Sense of direction of \(\mathbf{M}_O\) using the right-hand rule.
      • Align the fingers of your right hand in the direction of the position vector \(\mathbf{r}\) and curl them in the direction of the force \(\mathbf{F}\). Your thumb now points in the direction of the moment \(\mathbf{M}_{O}\) and your fingers curl to show the sense of rotation for the moment \(\mathbf{M}_{O}\).
      • Imagine sliding vector \(\mathbf{r}\) or \(\mathbf{F}\) until they are tail-to-tail. Align the fingers of your right hand in the direction of the position vector \(\mathbf{r}\) and curl your fingers in the direction of force \(\mathbf{F}\). Your curled fingers now indicate the sense of rotation for the moment \(\mathbf{M}_{O}\) and your thumb points in the direction of moment \(\mathbf{M}_{\text {O}}\).
      • Curl the fingers of your right hand and point the thumb of your right hand in the direction of moment \(\mathbf{M}_{O}\). The direction your fingers curl is the sense of rotation for the moment \(\mathbf{M}_{O}\).
    • The magnitude of the moment vector \(\mathbf{M}_{O}\) is the square root of the dot product of \(\mathbf{M}_{O}\) with itself and is calculated as follows: \[\begin{array} M M_{O} &=\left|\mathbf{M}_{o}\right|=\left(\mathbf{M}_{O} \cdot \mathbf{M}_{O}\right)^{1 / 2} \\ &=\left[\left(y F_{z}-z F_{y}\right)^{2}+\left(x F_{z}-z F_{x}\right)^{2}+\left(x F_{y}-y F_{x}\right)^{2}\right]^{1 / 2} \end{array} \nonumber \] It can also be calculated using the relationship \[M_{o}=\left|\mathbf{M}_{o}\right|=|\mathbf{r}||\mathbf{F}| \sin \theta=r F(\sin \theta) \nonumber \] where \(\theta\) is the angle between the lines-of-action of vectors \(\mathbf{r}\) and \(\mathbf{F}\) and is measured as though \(\mathbf{r}\) and \(\mathbf{F}\) were placed tail-to-tail. A positive angle satisfies the right-hand rule as described above. Eq. \(( \PageIndex{5})\) can be interpreted as the area of a parallelogram formed by the vectors \(\mathbf{r}\) and \(\mathbf{F}\) (See Figure \(\PageIndex{4}\)).
      Vector r, with its tail at point O, points to the right. Vector F, with its tail at the head of vector r, points up and to the right. A parallelogram is formed by two instances each of r and F, placed head-to-tail. The smaller of the two angles of the parallelogram is theta. The parallelogram area is equivalent to the area of a rectangle of base r and height F(sin theta), or a rectangle of base d = r(sin theta) and height F.
      Figure \(\PageIndex{4}\): Interpreting the magnitude of \(\mathbf{M}_O\) as an area.

      Two additional rectangles shown in Figure \(\PageIndex{4}\) have areas equal in magnitude to the shaded area. The rectangle of length \(d_{\perp}\) and width \(F\) has an area of \[\begin{align} M_{O} &=(r \sin \theta) F \nonumber \\ &=d_{\perp} \times F \quad=\underbrace{\left[\begin{array}{c} \text { Shortest distance } \\ \text { between point } O \\ \text { and the line-of-action } \\ \text { of force } \mathbf{F} \end{array}\right]}_{\text {Often called the "lever arm" }} \times\left[\begin{array}{c} \text { Magnitude } \\ \text { of } \\ \text { force } \mathbf{F} \end{array}\right] \end{align} \nonumber \] The rectangle of length \(r\) and width \(F \sin \theta\) has an area of \[\begin{align} M_{O} &=r(F \sin \theta) \nonumber \\ &=r \times F_{\perp}=\left[\begin{array}{c} \text { Magnitude } \\ \text { of the } \\ \text { position vector } \mathbf{r} \end{array}\right] \times\left[\begin{array}{c} \text { Component of force } \mathbf{F} \\ \text { that is } \perp \text { to } \\ \text { the line-of-action of } \mathbf{r} \end{array}\right] \end{align} \nonumber \] For most cases with two-dimensional (plane) motion, you will find that using Eqs. \((\PageIndex{6})\) or \((\PageIndex{7})\) is much simpler than using the formal cross product mathematics.

    Test your understanding

    For each of the following two-dimensional cases, determine the magnitude and direction [Clockwise (CW) or counter-clockwise (CCW)] of the moment of force \(\mathbf{F}\) about point \(O\):

    a)

    A horizontal 4-meter-long bar has point O at its left end. A force of 50 N directed upwards and to the right, at 20 degrees from the vertical, is applied at the right end of the bar.

    Figure \(\PageIndex{5}\): A force is applied at an angle to one end of a bar.

    b)

    A horizontal 4-meter bar has point O at its left endpoint. A force of 50 N directed downwards and to the left at 80 degrees from the vertical is applied to the right end of the bar.

    Figure \(\PageIndex{6}\): A force is applied at an angle to one end of a bar.

    c)

    An object is composed of two horizontal bars with the right end of the upper bar connected to the left end of the lower bar. Point O is the leftmost point of the assembly. A force of 50 N directed upwards and to the right at 45 degrees from the vertical is applied to a point 4 meters to the right and 1 meter below O.

    Figure \(\PageIndex{7}\): A force is applied at an angle to one end of a rigid body.

    Answer

    a) \(187.9 \ \mathrm{N} \cdot \mathrm{m CCW}\)

    b) \(34.73 \ \mathrm{N} \cdot \mathrm{m CW}\)

    c) \(176.8 \ \mathrm{N} \cdot \mathrm{m CCW}\)

    6.1.2 Moment of a force couple

    There is a special type of external force system called a force couple. A force couple consists of two external forces that have equal magnitude, parallel lines-of-action, and opposite sense (See Figure \(\PageIndex{8}\)). Thus, a force couple results in a zero net force on a system. Or stating this another way, a force couple transfers zero linear momentum to a system. However, a quick examination of the figure shows that the force couple does in fact attempt to turn or rotate the system.

    An irregularly curved body contains three points O, O', and O''. Force vector F1 represents a force applied to some point of the body, with magnitude F. Force vector F2 represents a force applied to some other point of the body, also of magnitude F and along the same line of action as F1 while pointing in the opposite direction. The distance between the forces' lines of action is given by d.

    Figure \(\PageIndex{8}\): A force couple.

    The moment of a force couple about point \(O\) can be calculated with reference to Figure \(\PageIndex{8}\) as follows: \[ \begin{align} \mathbf{M}_{O, \text { couple }} &= \mathbf{M}_{O, \ F_{1}} + \mathbf{M}_{O, \ F_{2}} \nonumber \\ &= \left( \mathbf{r}_{1} \times \mathbf{F}_{1} \right) + \left( \mathbf{r}_{2} \times \mathbf{F}_{2}\right) \quad\quad\quad \text { where } \left| \mathbf{F}_{1} \right| = \left|\mathbf{F}_{2}\right| = F \nonumber \\ &= \left[ \left( d_{\perp, 1} \times F_{1}\right) \,\, \mathrm{CCW} \right] + \left[ \left( d_{\perp, 2} \times F_{2}\right) \,\, \mathrm{CW} \right] \\ & = \left[ -\left( d_{\perp, 1} \times F_{1}\right) \quad + \quad \left(d_{\perp, 2} \times F_{2}\right) \right] \quad \text { CW } \nonumber \\ &= d \times F \quad \text { CW } \quad\quad\quad\quad\quad\quad\quad \text { where } d=\left| d_{\perp, 1}-d_{\perp, 2}\right| \nonumber \\ {} \nonumber \\ \mathbf{M}_{O, \text { couple }} &= d \times F \text{ in a CW direction} \nonumber \end{align} \nonumber \]

    Note that if we had calculated the moment of the force couple shown in Figure \(\PageIndex{8}\) about point \(O'\) or point \(O''\) we would have found exactly the same result. Thus, the moment of a couple does not seem to be tied to a particular point, unlike what we found with the moment of a force about a point.

    We can say the following about a force couple with force \(\mathbf{F}\) and a distance \(d\) separating the forces' lines of action:

    • The lines of action of the two forces in a force couple lie in the same plane.
    • A force couple transfers no net linear momentum to a system.
    • The moment produced by a force couple is a vector \(\mathbf{M}_{\text {couple}}\).
    • The line of action of the moment vector \(\mathbf{M}_{\text {couple}}\) is normal (perpendicular) to the plane that contains the force couple.
    • The point of application of the moment vector \(\mathbf{M}_{\text {couple}}\) can be at any point in the plane of the couple. Because the point of application is free to move, this type of vector is sometimes called a free vector.
    • The sense of the moment vector \(\mathbf{M}_{\text {couple can be obtained from ob. }}\) serving the direction of rotation the force couple could impart. The direction of the moment vector can be found by curling the fingers of your right hand to match the sense of the force couple and aligning your thumb along the line of action. Your thumb now points in direction for the moment vector \(\mathbf{M}_{\text {couple. }}\).
    • The magnitude of the moment vector \(\mathbf{M}_{\text {couple is equal to the product }}\) of the magnitude of one of the force vectors and the shortest distance between the lines of action, \(d\): \[M_{\text {couple}} = \left| \mathbf{M}_{\text {couple}}\right| = d \times |\mathbf{F}| = d \times F \nonumber \]

    We will frequently encounter systems where the force distribution on some portion of the boundary looks like that shown in Part (a) of Figure \(\PageIndex{9}\). On the upper half of the boundary, the forces are compressive (they push on the system), and on the lower half of the boundary, the forces are tensile (they pull on the system). If every force on the upper half has a mirror image of opposite sense on the lower half, this distribution represents a couple and the net force on the boundary is zero; however there is a net moment due to the couple as shown in Part (b) of Figure \(\PageIndex{9}\). There will also be occasions where the force distribution on the boundary is actually the sum of a couple and a net force.

    A block experiences a distributed couple on its right side. For the upper half of the block the distributed force is directed to the left, and for the lower half the distributed force, distributed symmetrically to the top, is directed to the right.a) Distributed couple acting on the boundary
    Block from part A of figure has its distributed couple replaced by a single curved arrow representing the equivalent moment.

    b) Equivalent moment due to distributed couple

    Figure \(\PageIndex{9}\): Moment of a distributed couple.
    Test your understanding

    Calculate the magnitude and the direction (CW or CCW) for the net moment about the point \(O\) :

    Point O is located on an irregularly shaped body. 4 feet above O, a 100-lbf force is applied towards the right. 4 feet to the right of O, a 50-lbf force is applied upwards. Some distance to the right of O, a force couple with magnitude 75 lbf and a distance of 2 feet separating the forces is applied to the body.

    Figure \(\PageIndex{10}\): Point forces and a force couple are applied to a curved shape.


    This page titled 6.1: Why is this thing turning? is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Donald E. Richards (Rose-Hulman Scholar) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.