Question 3.1.1: Figure 3.1-2 shows a cube of sides 1 m acted on by eight for...
Figure 3.1-2 shows a cube of sides 1 m acted on by eight forces and a moment as shown. Determine the magnitude and direction of the resultant force R and the coordinates (\overline{x},\overline{y} , \overline{z}) of the point at which R intersects the face OACB. Is there a resultant moment in addition to the resultant force R ?
similarly,
Then
\qquad \qquad R = √(2^2+2^2+4^2) =\underline{4.9 \ N}†See Appendix 1 of Ref. 1 for a more detailed discussion of this type of problem.
The angles \overline{\alpha }, \overline{\beta }, and \overline{\gamma } which R makes with the x-, y-, and z-axes are :
\overline{\alpha }=\cos ^{-1}\frac{\sum{P_{x}}}{R}=\cos ^{-1}\frac{2}{4.9}=\underline{66°}\\ \overline{\beta }=66°;\qquad \overline{\gamma} = \cos ^{-1}\frac{4}{4.9}=\underline{35.3°}Thus the unit vector in the direction of R is
\hspace{5em}e_{R} = 0.408i + 0.408j + 0.816k \ where \ 0.408 = cos \ \bar{\alpha } \ etc.Let M=\begin{bmatrix} M_{x} & M_{y} & M_{z} \end{bmatrix}^{T} be the resultant moment due to the given forces and moment.
Taking moments about the point 0,
Similarly,
That is, M=\begin{bmatrix} M_{x} & M_{y} & M_{z} \end{bmatrix}^{T} = i + 3j + 2k Nm
M can be considered as made up of two components: M_{R} parallel to R and M_{N} normal to R. M_{R} is of course simply the projection of M on R. That is,
And M_{N} must be equal to R x r as given by Eqn 3.1-4. That is
\hspace{5em}-0.33i + 1.67j -0.66k = \begin{vmatrix} i & j & k \\ 2 & 2 & 4 \\ -\bar{x} & -\bar{y} & 0 \end{vmatrix}giving \bar{x}= -0.418m, \ \bar{y} = -0.083m
Thus the given system of forces and moment is reduced to the single force R acting through the point (-0.418, -0.083, 0) plus the moment M_{R}. Note that M_{R} cannot be made to vanish.