Couple Moment A85A78

Question: Can you convert the text numbers for it to be easier to read Problem 1 — Resultant Couple Moment We are given two couples acting on the beam. Horizontal forces: $(2,\text{kN})$ Vertical forces: $(F)$ Horizontal distance between the vertical forces: $(0.9,\text{m})$ Vertical distance between the horizontal forces: $(0.3,\text{m})$ Desired resultant: $(1.5,\text{kN}\cdot\text{m})$ clockwise The standard couple relation is $$ M=Fd $$ and the two couple moments are added algebraically. This problem is also listed as Hibbeler F4–21, with the result $(F=2.33,\text{kN})$. Step 1: Moment from the $2$-kN forces The two $(2,\text{kN})$ forces are separated vertically by $$ d=0.3,\text{m} $$ Therefore, $$ M_{2\text{ kN}}=(2)(0.3) $$ $$ \boxed{M_{2\text{ kN}}=0.6,\text{kN}\cdot\text{m}} $$ Looking at the directions, the upper $(2,\text{kN})$ force points left and the lower one points right, producing a counterclockwise moment. Take counterclockwise as positive: $$ M_{2\text{ kN}}=+0.6,\text{kN}\cdot\text{m} $$ Step 2: Moment from the $(F)$ forces The vertical forces are separated horizontally by $$ d=0.9,\text{m} $$ Thus, $$ M_F=F(0.9) $$ The force $(F)$ produces a clockwise couple, so $$ M_F=-0.9F $$ Step 3: Add the couple moments The desired resultant is $(1.5,\text{kN}\cdot\text{m})$ clockwise. Therefore, with CCW positive, $$ M_R=-1.5,\text{kN}\cdot\text{m} $$ So, $$ 0.6-0.9F=-1.5 $$ Move $(0.6)$ to the right: $$ -0.9F=-2.1 $$ Therefore, $$ F=\frac{2.1}{0.9} $$ $$ F=2.333\ldots $$ Answer $$ \boxed{\boxed{F=2.33,\text{kN}}} $$ The direction shown in the diagram is upward at A and downward at B. Problem 2 — Couple Moment in Cartesian Vector Form We have two equal and opposite $(450,\text{N})$ forces, so they form a couple. For a couple, $$ \boxed{\mathbf M=\mathbf r\times\mathbf F} $$ The geometry and force components correspond to Hibbeler F4–24. Step 1: Resolve the $450$-N force The $(3-4-5)$ triangle tells us the force has components based on $$ \frac{4}{5},\qquad \frac{3}{5} $$ From the diagram, the force at $(B)$ points in the $(+y)$ and $(-z)$ directions. Therefore, $$ \mathbf F_B = 450\left(\frac45\mathbf j-\frac35\mathbf k\right) $$ Calculate each component: (y)-component $$ F_{By}=450\left(\frac45\right) $$ $$ F_{By}=360,\text{N} $$ (z)-component $$ F_{Bz}=450\left(-\frac35\right) $$ $$ F_{Bz}=-270,\text{N} $$ So, $$ \boxed{\mathbf F_B=360\mathbf j-270\mathbf k\ \text{N}} $$ The force at $(A)$ is equal and opposite: $$ \mathbf F_A=-360\mathbf j+270\mathbf k\ \text{N} $$ Step 2: Find the position vector The pipe from $(A)$ to $(B)$ is $(0.4,\text{m})$ in the $(+x)$ direction. Thus, $$ \boxed{\mathbf r_{AB}=0.4\mathbf i\ \text{m}} $$ Notice that the $(0.3,\text{m})$ vertical dimension does not appear here. Why? Because $(A)$ and $(B)$ are at the same height. The separation between the two force lines is represented by $$ \mathbf r_{AB}=0.4\mathbf i $$ The $(0.3,\text{m})$ dimension locates the pipe relative to $(O)$, but the moment of a couple does not depend on the reference point. Step 3: Calculate the couple moment Use $$ \mathbf M=\mathbf r{AB}\times\mathbf FB $$ Substitute: $$ \mathbf M = (0.4\mathbf i) \times (360\mathbf j-270\mathbf k) $$ Expand: $$ \mathbf M = (0.4)(360)(\mathbf i\times\mathbf j) + (0.4)(-270)(\mathbf i\times\mathbf k) $$ Recall the cross-product rules: $$ \mathbf i\times\mathbf j=\mathbf k $$ and $$ \mathbf i\times\mathbf k=-\mathbf j $$ Therefore, $$ \mathbf M = 144\mathbf k + 108\mathbf j $$ So, $$ \boxed{\mathbf M=108\mathbf j+144\mathbf k\ \text{N}\cdot\text{m}} $$ Step 4: Same calculation using a determinant This is often the format your professor will expect: $$ \mathbf M= \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\ 0.4&0&0\ 0&360&-270 \end{vmatrix} $$ Expand: $$ \mathbf M = \mathbf i[0(-270)-0(360)] -\mathbf j[0.4(-270)-0(0)] +\mathbf k[0.4(360)-0(0)] $$ First component: $$ M_x=0 $$ Second component: $$ M_y=-[-108]=108 $$ Third component: $$ M_z=144 $$ Therefore, $$ \boxed{\mathbf M=(0\mathbf i+108\mathbf j+144\mathbf k)\ \text{N}\cdot\text{m}} $$ or simply $$ \boxed{\boxed{\mathbf M=108\mathbf j+144\mathbf k\ \text{N}\cdot\text{m}}} $$ This matches the published solution for the same F4–24 problem. Final Answers 1. Beam $$ \boxed{F=2.33\ \text{kN}} $$ 2. Pipe assembly $$ \boxed{\mathbf M=108\mathbf j+144\mathbf k\ \text{N}\cdot\text{m}} $$ Key formulas to remember For a couple: $$ \boxed{M=Fd} $$ For a 3D couple: $$ \boxed{\mathbf M=\mathbf r\times\mathbf F} $$ And for the cross product: $$ \boxed{ \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\ rx&ry&rz\ Fx&Fy&Fz \end{vmatrix}} $$