3D Rigid Body Equilibrium

Solution:

1) Draw Free Body Diagram, notice the hinge (ball and socket support) at B has been modified to allow for axial translation along the y-axis (axis of rotation for the hinge) so there is NO By reaction.

2) Determine all forces and moments in cartesian vector form

Cable force CD,

$\overrightarrow{T_{C D}}=\left|T_{C D}\right| u_{C D}=\left|T_{C D}\right|\left(\frac{r_{C D}}{\left|r_{C D}\right|}\right)$

$r_{CD}=D-C=(0,-0.57,1.14)-(0.6,-1.14,0)\rightarrow r_{CD}=-0.6i+0.57j+1.14k$

$\left|r_{CD}\right|=\sqrt{(0.6)^2+(0.57)^2+(1.14)^2}=1.409$

$\overrightarrow{T_{CD}}=\left|T_{CD}\right|\left(\frac{-0.6i+0.57j+1.14k}{1.409}\right)=T_{CD}(-0.426i+0.405j+0.809k)$

$\overrightarrow{T_{CD}}=-0.426T_{CD}i+0.405T_{CD}j+0.809T_{CD}k$

Weight,

$\vec{W}=-(20 k g)\left(9.81 \frac{m}{s^{2}}\right) k \rightarrow \vec{W}=(0 i+0 j-196.2 k) N$

Reaction forces,

$\begin{array}{l}{\overrightarrow{F_{A}}=A_{x} i+A_{y} j+A_{z} k} \\ {\overrightarrow{F_{B}}=B_{x} i+0 j+B_{z} k}\end{array}$

3) Apply 3D Force Equations of Equilibrium and Equate i, j, and k components from the Forces above,

$\Sigma F_x=0:-0.426T_{CD}+A_x+B_x=0\ \ \ \ (1)\ i-components$

$\Sigma F_{y}=0: 0.405 T_{C D}+A_{y}=0\ \ \ \ (2 ) \ j-components$

$\Sigma F_{z}=0: 0.809 T_{C D}-196.2 k+A_{z}+B_{z}=0\ \ \ \ (3 ) \ k-components$

Notice we couldn’t solve for anything using the equations of equilibrium from above. We need to take the moment about a point where we can eliminate the most unknowns. This is point A!

4) Moment equation of equilibrium about point A,

$\Sigma M_A=0:r_{AD}\times\overrightarrow{T_{CD}}+r_{AB}\times\overrightarrow{F_B}+r_{AG}\times\vec{W}=0$

We need the position vectors $r_{AD},\ \ r_{AB},\ \ r_{AG}$

$\overrightarrow{r_{AD}}=D-A=(0,-0.57,1.14)-(0,-0.12,0.54)\rightarrow\overrightarrow{r_{AD}}=0i-0.45j+0.6k$

$\overrightarrow{r_{AB}}=B-A=(0,-1.02,0.54)-(0,-0.12,0.54)\rightarrow\overrightarrow{r_{AB}}=0i-0.9j+0k$

$\overrightarrow{r_{AG}}=G-A=(0.3,-0.57,0.27)-(0,-0.12,0.54)\rightarrow\overrightarrow{r_{AG}}=0.3i-0.45j-0.27k$

Therefore,

$r_{A D} \times \overrightarrow{T_{C D}}+r_{A B} \times \overrightarrow{F_{B}}+r_{A G} \times \vec{W}=0$

$\overrightarrow{T_{C D}}\left|\begin{array}{ccc}{i} & {j} & {k} \\ {0} & {-0.45} & {0.6} \\ {-0.426} & {0.405} & {0.809}\end{array}\right|+\left|\begin{array}{ccc}{i} & {j} & {k} \\ {0} & {-0.9} & {0} \\ {B_{x}} & {0} & {B_{Z}}\end{array}\right|+\left|\begin{array}{ccc}{i} & {j} & {k} \\ {0.3} & {-0.45} & {-0.27} \\ {0} & {0} & {-196.2}\end{array}\right|=0$

$\Sigma M_{A}=0:$

$\overrightarrow{T_{C D}}(+(-0.45(0.809)-0.6(0.405) i)-(0(0.809)-0.6(-0.426) j)+(0(0.405)-(-0.45(-0.426) k)$

$+\left(-0.9\left(B_{z}\right)-0(0)\right) i-\left(0\left(B_{z}\right)-0\left(B_{x}\right)\right) j+\left(0(0)-\left(-0.9\left(B_{x}\right)\right) k\right.$

$+(-0.45(-196.2)-(-0.27(0)) i-(0.3(-196.2)-(-0.27(0)) j+(0.3(0)-(-0.45(0))) k=0$

$\Sigma M_{A}=0:$

$-0.607 T_{C D} i-0.256 T_{C D} j-0.192 T_{C D} k-0.9 B_{Z} i+0.9 B_{x} k+88.29 i+58.86 j=0$

5) Equate i j and k Components,

$-0.607 T_{C D}-0.9 B_{z}+88.29=0\ \ \ \ (4 ) \ i-components$

$-0.256 T_{C D}+58.86=0\ \ \ \ (5 ) \ j-components$

$-0.192 T_{C D}+0.9 B_{x}=0\ \ \ \ (6 ) \ k-components$

6) We can find TCD using the equation (5),

$\begin{array}{l}{-0.256 T_{C D}+58.86=0} \\ {T_{C D}=230 N(\text { ans })}\end{array}$

Plug TCD into equation (6) and solve for Bx,

$\begin{array}{l}{-0.192(230 N)+0.9 B_{x}=0} \\ {B_{x}=49.1 \mathrm{N}(\text {ans})}\end{array}$

Plug TCD into equation (4) and solve for Bz,

$\begin{array}{l}{-0.607(230 N)-0.9 B_{z}+88.29=0} \\ {B_{Z}=-57.0 N(\text { ans })}\end{array}$

7) Solve for the reactions at A, using our initial equilibrium equations (1), (2), (3)

$\begin{array}{l}{-0.426 T_{C D}+A_{x}+B_{x}=0} \\ {-0.426(230 N)+A_{x}+(49.1 N)=0} \\ {A_{x}=48.9 N(\text {ans})}\end{array}$

$\begin{array}{l}{0.405(230 N)+A_{y}=0} \\ {A_{y}=-93.2 N(\text {ans})}\end{array}$

$\begin{array}{l}{0.809(230 N)-196.2+A_{z}+-57.0 N=0} \\ {A_{z}=67.1 \mathrm{N}(\text {ans})}\end{array}$

Solution:

1) Draw Free Body Diagram

2) Determine all forces and moments in cartesian vector form

Cable force CD,

$\overrightarrow{T_{C D}}=\left|T_{C D}\right| u_{C D}=\left|T_{C D}\right|\left(\frac{r_{C D}}{\left|r_{C D}\right|}\right)$

$r_{C D}=D-C=(0,-0.45,1.14)-(0.6,-1.14,0) \rightarrow r_{C D}=-0.6 i+0.69 j+1.14 k$

$\left|r_{C D}\right|=\sqrt{(0.6)^{2}+(0.69)^{2}+(1.14)^{2}}=1.461$

$\overrightarrow{T_{C D}}=\left|T_{C D}\right|\left(\frac{-0.6 i+0.69 j+1.14 k}{1.461}\right)=T_{C D}(-0.411 i+0.472 j+0.780 k)$

$\overrightarrow{T_{CD}}=-0.411T_{CD}i+0.472T_{CD}j+0.780T_{CD}k$

Weight,

$\vec{W}=-(20 k g)\left(9.81 \frac{m}{s^{2}}\right) k \rightarrow \vec{W}=(0 i+0 j-196.2 k) N$

Reaction forces,

$\begin{array}{l}{\overrightarrow{F_{A}}=A_{x} i+A_{y} j+A_{z} k} \\ {\overrightarrow{F_{B}}=B_{x} i+0 j+B_{z} k}\end{array}$

3) Apply 3D Force Equations of Equilibrium and Equate i, j, and k components from the Forces above,

$\Sigma F_{x}=0:-0.411 T_{C D}+A_{x}+B_{x}=0 \ \ \ \ (1 ) \ i-components$

$\Sigma F_{y}=0: 0.472 T_{C D}+A_{y}=0\ \ \ \ (2 ) \ j-components$

$\Sigma F_{z}=0: 0.780 T_{C D}-196.2 k+A_{z}+B_{z}=0\ \ \ \ (3 ) \ k-components$

Notice we couldn’t solve for anything using the equations of equilibrium from above. We need to take the moment about a point where we can eliminate the most unknowns. This is point A!

4) Moment equation of equilibrium about point A,

$\Sigma M_A=0:r_{AD}\times\overrightarrow{T_{CD}}+r_{AB}\times\overrightarrow{F_B}+r_{AG}\times\vec{W}=0$

We need the position vectors $r_{AD},\ \ r_{AB},\ \ r_{AG}$

$\overrightarrow{r_{AD}}=D-A=(0,-0.57,1.14)-(0,-0.12,0.54)\rightarrow\overrightarrow{r_{AD}}=0i-0.45j+0.6k$

$\overrightarrow{r_{AB}}=B-A=(0,-1.02,0.54)-(0,-0.12,0.54)\rightarrow\overrightarrow{r_{AB}}=0i-0.9j+0k$

$\overrightarrow{r_{AG}}=G-A=(0.3,-0.57,0.27)-(0,-0.12,0.54)\rightarrow\overrightarrow{r_{AG}}=0.3i-0.45j-0.27k$

Therefore,

$r_{A D} \times \overrightarrow{T_{C D}}+r_{A B} \times \overrightarrow{F_{B}}+r_{A G} \times \vec{W}=0$

$\overrightarrow{T_{C D}}\left|\begin{array}{ccc}{i} & {j} & {k} \\ {0} & {-0.45} & {0.6} \\ {-0.411} & {0.472} & {0.780}\end{array}\right|+\left|\begin{array}{ccc}{i} & {j} & {k} \\ {0} & {-0.9} & {0} \\ {B_{x}} & {0} & {B_{Z}}\end{array}\right|+\left|\begin{array}{ccc}{i} & {j} & {k} \\ {0.3} & {-0.45} & {-0.27} \\ {0} & {0} & {-196.2}\end{array}\right|=0$

$\Sigma M_{A}=0:$

$\overrightarrow{T_{C D}}(+(-0.45(0.780)-0.6(0.472) i)-(0(0.780)-0.6(-0.411) j)+(0(0.472)-(-0.45(-0.411) k)$

$+\left(-0.9\left(B_{z}\right)-0(0)\right) i-\left(0\left(B_{z}\right)-0\left(B_{x}\right)\right) j+\left(0(0)-\left(-0.9\left(B_{x}\right)\right) k\right.$

$+(-0.45(-196.2)-(-0.27(0)) i-(0.3(-196.2)-(-0.27(0)) j+(0.3(0)-(-0.45(0))) k=0$

$\Sigma M_{A}=0:$

$-0.6342 T_{C D} i-0.2466 T_{C D} j+0.185 T_{C D} k-0.9 B_{Z} i+0.9 B_{x} k+88.29 i+58.86 j=0$

5) Equate i j and k Components,

$-0.6342 T_{C D}-0.9 B_{z}+88.29=0\ \ \ \ (4 ) \ i-components$

$-0.2466 T_{C D}+58.86=0\ \ \ \ (5 ) \ j-components$

$0.185 T_{C D}+0.9 B_{x}=0\ \ \ \ (6 ) \ k-components$

6) We can find TCD using the equation (5),

$\begin{array}{l}{-0.2466 T_{C D}+58.86=0} \\ {T_{C D}=239 N(\text { ans })}\end{array}$

Plug TCD into equation (6) and solve for Bx,

$\begin{array}{l}{0.185(239 N)+0.9 B_{x}=0} \\ {B_{x}=-49.1 \mathrm{N}(\text {ans})}\end{array}$

Plug TCD into equation (4) and solve for Bz,

$\begin{array}{l}{-0.6342(239 N)-0.9 B_{z}+88.29=0} \\ {B_{Z}=-70.3 N(\text { ans })}\end{array}$

7) Solve for the reactions at A, using our initial equilibrium equations (1), (2), (3)

$\begin{array}{l}{-0.411 T_{C D}+A_{x}+B_{x}=0} \\ {-0.411(239 N)+A_{x}+(-49.1 N)=0} \\ {A_{x}=147 N(\text {ans})}\end{array}$

$\begin{array}{l}{0.472(239 N)+A_{y}=0} \\ {A_{y}=113 N(\text {ans})}\end{array}$

$\begin{array}{l}{0.780(239 N)-196.2+A_{z}+-70.3 N=0} \\ {A_{z}=80.1 \mathrm{N}(\text {ans})}\end{array}$

Solution:

1) Draw Free Body Diagram

2) Determine all forces and moments in cartesian vector form

Cable force CD,

$\overrightarrow{T_{C D}}=\left|T_{C D}\right| u_{C D}=\left|T_{C D}\right|\left(\frac{r_{C D}}{\left|r_{C D}\right|}\right)$

$r_{C D}=D-C=(0,-0.45,1.14)-(0.6,-1.14,0) \rightarrow r_{C D}=-0.6 i+0.69 j+1.14 k$

$\left|r_{C D}\right|=\sqrt{(0.6)^{2}+(0.69)^{2}+(1.14)^{2}}=1.461$

$\overrightarrow{T_{C D}}=\left|T_{C D}\right|\left(\frac{-0.6 i+0.69 j+1.14 k}{1.461}\right)=T_{C D}(-0.411 i+0.472 j+0.780 k)$

$\overrightarrow{T_{CD}}=-0.411T_{CD}i+0.472T_{CD}j+0.780T_{CD}k$

Weight,

$\vec{W}=-(20 k g)\left(9.81 \frac{m}{s^{2}}\right) k \rightarrow \vec{W}=(0 i+0 j-196.2 k) N$

Reaction forces,

$\begin{array}{l}{\overrightarrow{F_{A}}=A_{x} i+A_{y} j+A_{z} k} \\ {\overrightarrow{F_{B}}=B_{x} i+0 j+B_{z} k}\end{array}$

3) Apply 3D Force Equations of Equilibrium and Equate i, j, and k components from the Forces above,

$\Sigma F_{x}=0:-0.411 T_{C D}+A_{x}+B_{x}=0 \ \ \ \ (1 ) \ i-components$

$\Sigma F_{y}=0: 0.472 T_{C D}+A_{y}=0\ \ \ \ (2 ) \ j-components$

$\Sigma F_{z}=0: 0.780 T_{C D}-196.2 k+A_{z}+B_{z}=0\ \ \ \ (3 ) \ k-components$

Notice we couldn’t solve for anything using the equations of equilibrium from above. We need to take the moment about a point where we can eliminate the most unknowns. This is point A!

4) Moment equation of equilibrium about point A,

$\Sigma M_A=0:r_{AD}\times\overrightarrow{T_{CD}}+r_{AB}\times\overrightarrow{F_B}+r_{AG}\times\vec{W}=0$

We need the position vectors $r_{AD},\ \ r_{AB},\ \ r_{AG}$

$\overrightarrow{r_{AD}}=D-A=(0,-0.57,1.14)-(0,-0.12,0.54)\rightarrow\overrightarrow{r_{AD}}=0i-0.45j+0.6k$

$\overrightarrow{r_{AB}}=B-A=(0,-1.02,0.54)-(0,-0.12,0.54)\rightarrow\overrightarrow{r_{AB}}=0i-0.9j+0k$

$\overrightarrow{r_{AG}}=G-A=(0.3,-0.57,0.27)-(0,-0.12,0.54)\rightarrow\overrightarrow{r_{AG}}=0.3i-0.45j-0.27k$

Therefore,

$r_{A D} \times \overrightarrow{T_{C D}}+r_{A B} \times \overrightarrow{F_{B}}+r_{A G} \times \vec{W}=0$

$\overrightarrow{T_{C D}}\left|\begin{array}{ccc}{i} & {j} & {k} \\ {0} & {-0.45} & {0.6} \\ {-0.411} & {0.472} & {0.780}\end{array}\right|+\left|\begin{array}{ccc}{i} & {j} & {k} \\ {0} & {-0.9} & {0} \\ {B_{x}} & {0} & {B_{Z}}\end{array}\right|+\left|\begin{array}{ccc}{i} & {j} & {k} \\ {0.3} & {-0.45} & {-0.27} \\ {0} & {0} & {-196.2}\end{array}\right|=0$

$\Sigma M_{A}=0:$

$\overrightarrow{T_{C D}}(+(-0.45(0.780)-0.6(0.472) i)-(0(0.780)-0.6(-0.411) j)+(0(0.472)-(-0.45(-0.411) k)$

$+\left(-0.9\left(B_{z}\right)-0(0)\right) i-\left(0\left(B_{z}\right)-0\left(B_{x}\right)\right) j+\left(0(0)-\left(-0.9\left(B_{x}\right)\right) k\right.$

$+(-0.45(-196.2)-(-0.27(0)) i-(0.3(-196.2)-(-0.27(0)) j+(0.3(0)-(-0.45(0))) k=0$

$\Sigma M_{A}=0:$

$-0.6342 T_{C D} i-0.2466 T_{C D} j+0.185 T_{C D} k-0.9 B_{Z} i+0.9 B_{x} k+88.29 i+58.86 j=0$

5) Equate i j and k Components,

$-0.6342 T_{C D}-0.9 B_{z}+88.29=0\ \ \ \ (4 ) \ i-components$

$-0.2466 T_{C D}+58.86=0\ \ \ \ (5 ) \ j-components$

$0.185 T_{C D}+0.9 B_{x}=0\ \ \ \ (6 ) \ k-components$

6) We can find TCD using the equation (5),

$\begin{array}{l}{-0.2466 T_{C D}+58.86=0} \\ {T_{C D}=239 N(\text { ans })}\end{array}$

Plug TCD into equation (6) and solve for Bx,

$\begin{array}{l}{0.185(239 N)+0.9 B_{x}=0} \\ {B_{x}=-49.1 \mathrm{N}(\text {ans})}\end{array}$

Plug TCD into equation (4) and solve for Bz,

$\begin{array}{l}{-0.6342(239 N)-0.9 B_{z}+88.29=0} \\ {B_{Z}=-70.3 N(\text { ans })}\end{array}$

7) Solve for the reactions at A, using our initial equilibrium equations (1), (2), (3)

$\begin{array}{l}{-0.411 T_{C D}+A_{x}+B_{x}=0} \\ {-0.411(239 N)+A_{x}+(-49.1 N)=0} \\ {A_{x}=147 N(\text {ans})}\end{array}$

$\begin{array}{l}{0.472(239 N)+A_{y}=0} \\ {A_{y}=113 N(\text {ans})}\end{array}$

$\begin{array}{l}{0.780(239 N)-196.2+A_{z}+-70.3 N=0} \\ {A_{z}=80.1 \mathrm{N}(\text {ans})}\end{array}$

Solution:

1) Draw Free Body Diagram.

2) Determine all forces and moments in cartesian vector form

Cable force CD,

$\mathbf{T}_{B C}=\mathbf{T}_{B C}\left(\frac{\mathbf{r}_{B C}}{\mathbf{r}_{B C}}\right)$

$r_{BC}=C-D=(0,1.5,0)-(3,0,-1)\rightarrow r_{CD}=-3i+1.5j+1k$

$\left|r_{CD}\right|=\sqrt{(-3)^2+(1.5)^2+(1)^2}=3.5$

$\overrightarrow{T_{CD}}=\left|T_{CD}\right|\left(\frac{-3i+1.5j+1k}{3.5}\right)$

$\overrightarrow{T_{CD}}=-\frac{6}{7}T_{BC}\mathbf{i}+\frac{3}{7}T_{BC}\mathbf{j}+\frac{2}{7}T_{BC}\mathbf{k}$

Applied forces,

$\mathbf{F}=\{200 \mathbf{j}-400 \mathbf{k}\} \mathbf{N}$

Reaction forces,

$F_A=A_x\mathbf{i}+A_y\mathbf{j}$

Reaction moment,

$\mathbf{M}_A=\left(M_A\right)_x\mathbf{i}+\left(M_A\right)_y\mathbf{j}+\left(M_A\right)_z\mathbf{k}$

3) Apply 3D Force Equations of Equilibrium and Equate i, j, and k components from the Forces above,

$\Sigma F_x=0:-\frac{6}{7}T_{BC}+A_x=0\ \ \ \ (1)\ i-components$

$\Sigma F_y=0:\frac{3}{7}T_{BC}+200+A_y=0\ \ \ \ (2)\ j-components$

$\Sigma F_z=0:\frac{2}{7}T_{BC}-400=0\ \ \ \ (3)\ k-components$

From Eqs (1) to (3), we get,

$\begin{aligned} T_{B C} &=1400 \mathrm{N}=1.40 \mathrm{kN} \\ A_{y} &=-800 \mathrm{N} \\ A_{x} &=1200 \mathrm{N}=1.20 \mathrm{kN} \end{aligned}$

4) Moment equation of equilibrium about point A,

$\Sigma M_A=\mathbf{0};\quad\mathbf{M}_A+\mathbf{r}_1\times\mathbf{F}+r_2\times\mathbf{T}_{BC}=0$

We need the position vectors $r_{1},\ \ r_{2}$

$\mathbf{r}_{1}=\{3 \mathbf{i}\} \mathrm{m} \quad r_{2}=\{1.5 \mathbf{j}\} \mathrm{m}$

Therefore,

$\Sigma M_A=\mathbf{0};\quad\mathbf{M}_A+\mathbf{r}_1\times\mathbf{F}+r_2\times\mathbf{T}_{BC}=0$

$\left(M_{A}\right)_{x} \mathbf{i}+\left(M_{A}\right)_{y} \mathbf{j}+\left(M_{A}\right)_{z} \mathbf{k}+\left|\begin{array}{ccc}{\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \\ {3} & {0} & {0} \\ {0} & {200} & {-400}\end{array}\right|+\left|\begin{array}{ccc}{\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \\ {0} & {1.5} & {0} \\ {-\frac{6}{7} T_{B C}} & {\frac{3}{7} T_{B C}} & {\frac{2}{7} T_{B C}}\end{array}\right|=0$

$\left[\left(M_A\right)_x+\frac{3}{7}T_{BC}\right]\mathbf{i}+\left[\left(M_A\right)_y+1200\right]\mathbf{j}+\left[\left(M_A\right)_z+\frac{9}{7}T_{BC}+600\right]\mathbf{k}=0$

5) Equate i j and k Components,

$\left(M_A\right)_x+\frac{3}{7}T_{BC}=0\ \ \ (4)\ i-components$

$\left(M_A\right)_y+1200=0\ \ \ \ (5)\ j-components$

$\left(M_A\right)_z+\frac{9}{7}T_{BC}+600=0\ \ \ (6)\ k-components$

Plug TBC=1400 N into equations (4), (5), and (6) and solve for the moment reactions,

$\begin{array}{l}{\left(M_{A}\right)_{x}=600 \mathrm{N} \cdot \mathrm{m}} \\ {\left(M_{A}\right)_{y}=-1200 \mathrm{N} \cdot \mathrm{m}=1.20 \mathrm{kN} \cdot \mathrm{m}} \\ {\left(M_{A}\right)_{z}=-2400 \mathrm{N} \cdot \mathrm{m}=2.40 \mathrm{kN} \cdot \mathrm{m}}\end{array}$