Question 3.10: Three cables are attached to a bracket as shown. Replace the......

Three cables are attached to a bracket as shown. Replace the forces exerted by the cables with an equivalent force-couple system at A .

3.10.1
The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

We first determine the relative position vectors drawn from point A to the points of application of the various forces and resolve the forces into rectangular components. Observing that { F}_{B}\,=\,(700\mathrm{~N})L_{B E} where

L_{B E}=\frac{{{\overrightarrow{{B E}} }}}{B E}=\frac{75\mathrm{i}-\,150\mathrm{j}+50\mathrm{k}}{175}

we have, using meters and newtons,

{r}_{R/A}={\overrightarrow{{ AB}} }=0.075\mathrm{i}+{\mathrm{0.050k}}\qquad{F}_{B}=300\mathrm{i}-{\mathrm{600j}}+200\mathrm{k} \\ \\ {r}_{C/ A}= \overrightarrow{A C} = 0.075{\mathrm{i}}-0.050\mathrm{k}\qquad{F}_{C}=707 {\mathrm{i}}\qquad\qquad-\,{ {7}}07 {\mathrm{k}} \\ \\ {r}_{D/A}=\overrightarrow{A D} = 0.100\mathrm{i}-0.100\mathrm{j}\qquad {F}_{D}=600\mathrm{i}+\mathrm{1039j}

The force-couple system at A equivalent to the given forces consists of a force R~=~\Sigma {F} and a couple M_A^R~=~\Sigma {(r\times F)} . The force R is readily obtained by adding respectively the x, y , and z components of the forces:

{ R}\,=\,\Sigma{ F}\,=\,(1607~{ N})\mathrm{i}\,+\,(439~{\ N})\mathrm{j}\,-\,(507~{ N})\mathrm{k}

The computation of {M}_{\mathrm{A}}^{R} will be facilitated if we express the moments of the forces in the form of determinants (Sec. 3.8):

{r}_{B/A}\times{F}_{B}= \begin{vmatrix} \mathrm i & \mathrm j & \mathrm k \\ 0.075 & 0 & 0.050 \\ 300 & -600 & 200 \end{vmatrix} =\,30\mathrm{i}\qquad\qquad\qquad-45\mathrm{k} \\ \\ {r}_{C/A}\times{F}_{C}= \begin{vmatrix} \mathrm i & \mathrm j & \mathrm k \\ 0.075 & 0 & -0.050 \\ 707 & 0 & -707 \end{vmatrix} =\qquad\qquad17.68 \mathrm j \\ \\ {r}_{D/A}\times{F}_{D}= \begin{vmatrix} \mathrm i & \mathrm j & \mathrm k \\ 0.100 & -0.100 & 0 \\ 600 & 1039 & 0 \end{vmatrix} =~~~\qquad\qquad\qquad163.9\mathrm{k}

Adding the expressions obtained, we have

{M}_{\Lambda}^{R}=\Sigma(\mathrm{r}\times\mathrm{F})\,=\,(30\ \mathrm{N}\cdot\mathrm{m})\mathrm{i}\,+\,(17.68\ \mathrm {N}\cdot\mathrm{m})\mathrm{j}\,+\,(118.9\ \mathrm{N}\cdot\mathrm{m})\mathrm{k}

The rectangular components of the force R and the couple {M}_{A}^{R} are shown in the adjoining sketch.

3.10.2
Loading more images...

Related Answered Questions

Question: 3.6

Verified Answer:

Our computations will be simplified if we attach t...
Question: 3.3

Verified Answer:

The force is replaced by two components, one compo...
Question: 3.2

Verified Answer:

The moment {M}_{B} of the force F a...