Chapter 5, Problem 5.46P

The 80-lb homogeneous plate is supported by a ball-and-socket joint at A, a slider hearing at B, and the cable CE. A 120-lb vertical force is applied to the corner D. Determine the force in the cable and the magnitude of the bearing reaction at B.

To determine

Force in cable and the magnitude of bearing reaction at B.

Answer to Problem 5.46P

  $T_{CE}=1400\text{ lb},B=1138.8\text{ lb}\text{.}$

Explanation of Solution

Given Information:

The 80-lb homogeneous plate is supported by a ball-socket joint at A, a slider bearing at B, and the cable CE. See figure below.

  

Calculation:

Draw free body diagram of as shown in following figure,

  

Forces and corresponding position vectors are listed in the following table,

Force	Position vector
$\overrightarrow{A}=A_x\widehat{i}+A_y\widehat{j}+A_z\widehat{k}$	${\overrightarrow{r}}_A=\overrightarrow{0}$
$\overrightarrow{B}=B_x\widehat{i}+B_z\widehat{k}$	${\overrightarrow{r}}_B=18\widehat{i}+28\widehat{j}$
${\overrightarrow{T}}_{CE}=T_{CE}\left[\frac{-18\widehat{i}+12\widehat{k}}{\sqrt{{(-18)}^2+{12}^2}}\right]$	${\overrightarrow{r}}_C=18\widehat{i}+24\widehat{j}$
$\overrightarrow{D}=-120\widehat{k}$	${\overrightarrow{r}}_D=18\widehat{i}+4\widehat{j}$
$\overrightarrow{W}=-80\widehat{k}$	${\overrightarrow{r}}_W=9\widehat{i}+14\widehat{j}$

Take equilibrium of moments about origin as,

  $\begin{array}{l}\sum \overrightarrow{M}=0\\ \Rightarrow {\overrightarrow{r}}_A\times \overrightarrow{A}+{\overrightarrow{r}}_B\times \overrightarrow{B}+{\overrightarrow{r}}_C\times {\overrightarrow{T}}_{CE}+{\overrightarrow{r}}_D\times \overrightarrow{D}+{\overrightarrow{r}}_W\times \overrightarrow{W}=0\\ \Rightarrow \left(\overrightarrow{0}\right)\times \left(A_x\widehat{i}+A_y\widehat{j}+A_z\widehat{k}\right)+\left(18\widehat{i}+28\widehat{j}\right)\times \left(B_x\widehat{i}+B_z\widehat{k}\right)\\ +\left(18\widehat{i}+24\widehat{j}\right)\times \left(T_{CE}\left[\frac{-18\widehat{i}+12\widehat{k}}{\sqrt{{(-18)}^2+{12}^2}}\right]\right)+\left(18\widehat{i}+4\widehat{j}\right)\times \left(-120\widehat{k}\right)+\left(9\widehat{i}+14\widehat{j}\right)\times \left(-80\widehat{k}\right)=0\end{array}$

Above vector equations gives following scalars equations,

  $\begin{array}{c}28B_z+13T_{CE}-1600=0\mathrm{...}\text{(1)}\\ -18B_z-9.7T_{CE}+2880=0\mathrm{...}\text{(2)}\\ -28B_x+20T_{CE}=0\mathrm{...}\text{(3)}\end{array}$

Take equilibrium of forces as,

  $\begin{array}{l}\sum \overrightarrow{F}=0\\ \Rightarrow \overrightarrow{A}+\overrightarrow{B}+{\overrightarrow{T}}_{CE}+\overrightarrow{D}+\overrightarrow{W}=0\\ \Rightarrow \left(A_x\widehat{i}+A_y\widehat{j}+A_z\widehat{k}\right)+\left(B_x\widehat{i}+B_z\widehat{k}\right)+\left(T_{CE}\left[\frac{-18\widehat{i}+12\widehat{k}}{\sqrt{{(-18)}^2+{12}^2}}\right]\right)+\left(-120\widehat{k}\right)+\left(-80\widehat{k}\right)=0\end{array}$

Above vector equations gives following scalars equations,

  $\begin{array}{c}{A}_x+B_x-0.83T_{CE}=0\mathrm{...}\text{(4)}\\ A_y=0\mathrm{...}\text{(5)}\\ A_z+B_z+0.54T_{CE}-200=0\mathrm{...}\text{(6)}\end{array}$

Solve equations (1-6) to get

  $A_x=160\text{ lb, }{A}_y=0\text{, }{A}_z=38\text{ lb, }{B}_x=980\text{ lb, }{B}_z=-580\text{ lb, }{T}_{CE}=1400\text{ lb}\text{.}$

The magnitude of reaction at B is $B=\sqrt{B_x{}^2+B_z{}^2}=\sqrt{{980}^2+{(-580)}^2}=1138.8\text{ lb}\text{.}$

Conclusion:

Therefore, $T_{CE}=1400\text{ lb},B=1138.8\text{ lb}\text{.}$