Moment of a Force (PDF)

## 4.5 Moment of a Force about a Specified Axis Sometimes, the moment produced by a force about a specified axis must be determined. For example, suppose the lug nut at *O* on the car tire in Fig. 4-20a needs to be loosened. The force applied to the wrench will create a tendency for the wrench and...

## 4.5 Moment of a Force about a Specified Axis Sometimes, the moment produced by a force about a specified axis must be determined. For example, suppose the lug nut at *O* on the car tire in Fig. 4-20a needs to be loosened. The force applied to the wrench will create a tendency for the wrench and the nut to rotate about the moment axis passing through *O*; however, the nut can only rotate about the *y* axis. Therefore, to determine the turning effect, only the *y* component of the moment is needed, and the total moment produced is not important. To determine this component, we can use either a scalar or vector analysis. ### Scalar Analysis To use a scalar analysis in the case of the lug nut in Fig. 4-20a, the moment arm, or perpendicular distance from the axis to the line of action of the force, is $d_1 = d cos θ$. Thus the moment of *F* about the *y* axis is $M_y = Fd_1 = F(d cos θ)$. According to the right-hand rule, $M_y$ is directed along the positive *y* axis as shown in the figure. In general, for any axis *a*, the moment is $M_a = Fd_a$ ### Vector Analysis To find the moment of force *F* in Fig. 4-20b about the *y* axis using a vector analysis, we must first determine the moment of the force about any point *O* on the *y* axis by applying Eq. 4-7, $M_o = r × F$. The component $M_y$ along the *y* axis is the projection of $M_o$ onto the *y* axis. It can be found using the dot product discussed in Chapter 2, so that $M_y = j • M_o = j• (r × F)$, where *j* is the unit vector for the *y* axis. We can generalize this approach by letting *u* be the unit vector that specifies the direction of the *a* axis shown in Fig. 4-21. Then the moment of *F* about a point *O* on the axis is $M_o = r × F$, and the projection of this moment onto the *a* axis is $M_a = u_a· (r × F)$. This combination is referred to as the scalar triple product. If the vectors are written in Cartesian form, we have $M_a = [u_{ax}i + u_{ay}j + u_{az}k]· \begin{vmatrix} i & j & k \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix}$ $ = u_{ax}(r_yF_z - r_zF_y) – u_{ay}(r_xF_z - r_zF_x) + u_{az}(r_xF_y – r_yF_x)$ This result can also be written in the form of a determinant, making it easier to memorize. $M_a = u_a (r × F) = \begin{vmatrix} u_{ax} & u_{ay} & u_{az} \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix}$ Where: * $u_a$, $u_{ay}$, $u_{az}$ represent the *x*, *y*, *z* components of the unit vector defining the direction of the *a* axis * $r_x$, $r_y$, $r_z$ represent the *x*, *y*, *z* components of the position vector extended from any point *O* on the *a* axis to any point *A* on the line of action of the force * $F_x$, $F_y$, $F_z$ represent the *x*, *y*, *z* components of the force vector. When $M_a$ is evaluated from Eq. 4-11, it will yield a positive or negative scalar. The sign of this scalar indicates the sense of direction of $M_a$ along the *a* axis. If it is positive, then $M_a$ will have the same sense as $u_a$, whereas if it is negative, then $M_a$ will act opposite to $u_a$. Once the *a* axis is established, point your right-hand thumb in the direction of $M_a$, and the curl of your fingers will indicate the sense of twist about the axis, Fig. 4-21. Provided $M_a$ is determined, we can then express $M_a$ as a Cartesian vector, namely, $M_a = M_{a}u_a$ ## Important Points * The moment of a force about a specified axis can be determined provided the perpendicular distance $d_a$ from the force line of action to the axis can be determined. $M_a = Fd_a$ * If vector analysis is used, $M_a = u_a· (r × F)$, where *u* defines the direction of the axis and *r* is extended from any point on the axis to any point on the line of action of the force. * If $M_a$ is calculated as a negative scalar, then the sense of direction of $M_a$ is opposite to $u_a$. * The moment $M_a$ expressed as a Cartesian vector is determined from $M_a = M_{a}u_a$. ## Example 4.7 Determine the resultant moment of the three forces in Fig. 4-22 about the x axis, the y axis, and the z axis. **Solution:** A force that is parallel to a coordinate axis or has a line of action that passes through the axis does not produce any moment or tendency for turning about that axis. Therefore, defining the positive direction of the moment of a force according to the right-hand rule, as shown in the figure, we have: $M_x = (60 lb)(2ft) + (50 lb)(2ft) + 0 = 220 lbft$ $M_y = 0 -(50 lb)(3 ft) - (40 lb)(2ft) = -230 lb.ft $ $M_z = 0 + 0 - (40 lb)(2 ft) = -80 lb.ft$ The negative signs indicate that $M_y$ and $M_z$ act in the *-y* and *-z* directions, respectively. ## Example 4.8 Determine the moment $M_{AB}$ produced by the force *F* in Fig. 4-23a, which tends to rotate the rod about the *AB* axis. **Solution:** A vector analysis using $M_{AB} = u_{AB} · (r × F)$ will be considered for the solution rather than trying to find the moment arm or perpendicular distance from the line of action of *F* to the *AB* axis. Each of the terms in the equation will now be identified. $u_{AB}$ defines the direction of the *AB* axis of the rod, Fig. 4-23b, where: $u_{AB} = \frac{\{0.4i + 0.2j\} m}{\sqrt{(0.4 m)² + (0.2 m)²}} = 0.8944i + 0.4472j$ *r* is directed from any point on the *AB* axis to any point on the line of action of the force. For example, position vectors $r_C$ and $r_D$ are suitable, Fig. 4-23b. (Although not shown, $r_{BC}$ or $r_{BD}$ can also be used.) For simplicity, we choose $r_D$, where: $r_D = \{0.6i\} m$ $F = \{-300k\} N$ Substituting these vectors into the determinant form and expanding, we have: $M_{AB} = u_{AB} (r_D × F) = \begin{vmatrix} 0.8944 & 0.4472 & 0 \\ 0.6 & 0 & 0 \\ 0 & 0 & -300 \end{vmatrix}$ $ = 0.8944[0(-300) - 0(0)] - 0.4472[0.6(-300) - 0(0)] + 0[0.6(0) - 0(0)]$ $= 80.50 N·m$ This positive result indicates that the sense of $M_{AB}$ is in the same direction as $u_{AB}$. Expressing $M_{AB}$ in Fig. 4-23b as a Cartesian vector yields: $M_{AB} = M_{AB}u_{AB} = (80.50 N·m)(0.8944i + 0.4472j)$ $= \{72.0i + 36.0j\} N·m$ **Note:** If axis *AB* is defined using a unit vector directed from *B* toward *A*, then in the above formulation $-u_{AB}$ would have to be used. This would lead to $M_{AB} = -80.50 Nm$. Consequently, $M_{AB} = M_{AB}(-u_{AB})$, and the same result would be obtained. ## Example 4.9 Determine the magnitude of the moment of force *F* about segment *OA* of the pipe assembly in Fig. 4-24a. **Solution:** The moment of *F* about the *OA* axis is determined from: $M_{OA} = u_{OA} (r × F)$, where *r* is a position vector extending from any point on the *OA* axis to any point on the line of action of *F*. As indicated in Fig. 4-24b, either $r_{OD}$, $r_{OC}$, $r_{AD}$, or $r_{AC}$ can be used; however, $r_{OD}$ will be considered since it will simplify the calculation. The unit vector $u_{OA}$, which specifies the direction of the *OA* axis, is: $u_{OA} = \frac{\{0.3i + 0.4j\} m}{\sqrt{(0.3 m)² + (0.4 m)²}} = 0.6i + 0.8j$ the position vector $r_{OD}$ is: $r_{OD} = \{0.5i + 0.5k\} m$ The force *F* expressed as a Cartesian vector is: $F = F(\frac{F_{CD}}{CD})$ $ = (300 N)\frac{\{0.4i -0.4j + 0.2k\} m}{\sqrt{(0.4 m)² + (-0.4 m)² + (0.2 m)²}}$ $ = \{200i - 200j + 100k\} N$ Therefore, $M_{OA} = u_{OA} (r_{OD} × F) = \begin{vmatrix} 0.6 & 0.8 & 0 \\ 0.5 & 0 & 0.5 \\ 200 & -200 & 100 \end{vmatrix} $ $ = 0.6[0(100) - (0.5)(-200)] - 0.8[0.5(100) - (0.5)(200)] + 0$ $ = 100 N·m$