Wize University Statics Textbook (Master) > Force System Resultants (Moment, Couple, Dist. Load)

Moments about an axis are almost always computed using the triple cross product (also known as the triple mixed product). The advantage is these types of questions is that they’re usually very systematic.

Typical solution method
Determine the force vector in Cartesian vector form
Obtain the simplest distance vector from the axis to the line of action of the force
Use coordinates of points you already know
3) Find a unit vector along the axis
Typically obtained via a distance vector that is normalized to become a unit vector
4) Compute the magnitude of the moment as follows:

M⃗A=u⃗A⋅(r⃗×F⃗)=∣uaxuayuaxrxryrzFxFyFz∣\vec{M}_A=\vec{u}_A\cdot(\vec{r}\times{\vec{F}})=\begin{vmatrix} 
  u_{a_x}    &u_{a_y} & u_{a_x}  \\ 
  r_x&r_y& r_z\\F_x&F_y&F_z
\end{vmatrix}MA​=uA​⋅(r×F)=​uax​​rx​Fx​​uay​​ry​Fy​​uax​​rz​Fz​​​
5) The moment vector can found using:
M⃗A=MAu⃗A\vec{M}_A=M_A\vec{u}_AMA​=MA​uA​
Triple Cross Product Example
Compute the moment about an axis with the following properties:

u⃗A=0.3i−0.4j\vec{u}_A=0.3i-0.4juA​=0.3i−0.4j
r⃗=(6i+12j−7k)m\vec{r}=\left(6i+12j-7k\right)mr=(6i+12j−7k)m
F⃗=(150i−200j+70k)N\vec{F}=\left(150i-200j+70k\right)NF=(150i−200j+70k)N

Solution:
M=∣0.3−0.40612−7150−20070∣M=\begin{vmatrix} 
  0.3     & -0.4&0\\ 
  6 &12& -7\\150&-200&70 
\end{vmatrix}M=​0.36150​−0.412−200​0−770​​
      =420Nm=420Nm=420Nm

0:00 / 0:00

Determine the moment that the force causes about the hinged axis

u⃗=1i^\vec{u}=1\hat{i}u=1i^
r⃗=4k^\vec{r}=4\hat{k}r=4k^
A(3,0,4)A(3,0,4)A(3,0,4)
B(0,2.82,1.03)B(0,2.82,1.03)B(0,2.82,1.03)
r⃗AB=3i^−2.82j^+2.97k^\vec{r}_{AB}=3\hat{i}-2.82\hat{j}+2.97\hat{k}rAB​=3i^−2.82j^​+2.97k^            ∣∣r⃗AB∣∣=5.08||\vec{r}_{AB}||=5.08∣∣rAB​∣∣=5.08
 F⃗BA=11.8i^−11.1j^+11.7k^\vec{F}_{BA}=11.8\hat{i}-11.1\hat{j}+11.7\hat{k}FBA​=11.8i^−11.1j^​+11.7k^
M⃗=Mu⃗\vec{M}=M\vec{u}M=Mu

  M=∣10000411.8−11.111.7∣M=\begin{vmatrix} 
  1   &0&0\\ 
  0 & 0&4\\11.8&-11.1&11.7 
\end{vmatrix}M=​1011.8​00−11.1​0411.7​​
       =44.4lb.ft=44.4lb.ft=44.4lb.ft
        M⃗=(44.4i^)lb.ft\vec{M}=(44.4\hat{i}) lb.ftM=(44.4i^)lb.ft

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Determine the moment cause around the hinged axis if the tension in the cable is 300 N. Note: the hinged axis is along line AB; the cable CED is continuous; and point D is 0.5 m away from point A in the z-direction.

Answer

I don't know

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Determine the moment cause around the hinged axis if the tension in the cable is 300 N. Note: the hinged axis is along line AB; the cable CED is continuous; and point D is 0.5 m away from point A in the z-direction.

MAB = (-228.8i) Nm

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Some workers are looking to setup a sign on un-even ground. Cable EF has 300 N of tension. Determine the tension needed in cable CG to stablize the sign from tipping sideways (around line AD).  Note: The ground is un-even, making points F and D below ground level, point A is at ground level, and point G is above ground level. Neglect the weight of the sign.

TEG = 206 N

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Quiz: Computing Moments About on Axis

Determine the moment about line AC.

MAC = 14.4 lbft

Answer

I don't know