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Wize University Statics Textbook (Master) > Equilibrium of Rigid Body

3D Rigid Body Equilibrium

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In 3D, we can use up to a maximum of 6 equations.

ΣFx=0ΣFy=0ΣFz=0\begin{array}{l}{\Sigma F_{x}=0} \\ {\Sigma F_{y}=0} \\ {\Sigma F_{z}=0}\end{array}

ΣMx=0ΣMy=0ΣMz=0\begin{array}{l}{\Sigma M_{x}=0} \\ {\Sigma M_{y}=0} \\ {\Sigma M_{z}=0}\end{array}

Bodies are usually attached to a fixed reference (wall, ground, etc.) by some connection. These connections are typically called reaction forces, or support reactions, and are common to be asked on a question. As you may notice, there are many types of support reactions.

Wize Concept
The key to knowing whether they provide a reaction or not is to ask yourself whether the 3D body would move in the x-direction, y-direction, z-direction and whether it would rotate about the x-axis, y-axis, and z-axis.

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A brief review of these reactions follows:



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As always, to solve a system of equations, you can have a maximum number of unknowns equal to the number of equations – up to 6 unknowns in 3D. Therefore, we sometimes simplify the unknown reaction forces to reduce the number of unknowns by:

Wize Tip
Choosing the direction of an axis for moment summation (ΣMx=0ΣMy=0ΣMz=0\begin{array}{l}{\Sigma M_{x}=0} \\ {\Sigma M_{y}=0} \\ {\Sigma M_{z}=0}\end{array} ),
such that it intersects the lines of action of as many unknown forces as possible. Realize that the moments of forces passing through points on this axis and the moments of forces which are parallel to the axis will then be zero.








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Lets Draw The FREE BODY DIAGRAMS for the following cases:














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Example

The mass for the window below is 20 kg and it acts at the geometric center of the window, at G. The window is attached at A and B by two hinges which can be idealized as ball and socket supports. Determine all the forces acting on the window when it is held open in the position shown by the rope attached at C. The hinge at B has been modified to allow for translation along its own axis of rotation.




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,

TCD→=∣TCD∣uCD=∣TCD∣(rCD∣rCD∣)\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)

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

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

TCD→=∣TCD∣(−0.6i+0.57j+1.14k1.409)=TCD(−0.426i+0.405j+0.809k)\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)

TCD→=−0.426TCDi+0.405TCDj+0.809TCDk\overrightarrow{T_{CD}}=-0.426T_{CD}i+0.405T_{CD}j+0.809T_{CD}k

Weight,

W⃗=−(20kg)(9.81ms2)k→W⃗=(0i+0j−196.2k)N\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,

FA→=Axi+Ayj+AzkFB→=Bxi+0j+Bzk\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,

ΣFx=0:−0.426TCD+Ax+Bx=0    (1) i−components\Sigma F_x=0:-0.426T_{CD}+A_x+B_x=0\ \ \ \ (1)\ i-components

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

ΣFz=0:0.809TCD−196.2k+Az+Bz=0    (3) k−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,

ΣMA=0:rAD×TCD→+rAB×FB→+rAG×W⃗=0\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 rAD,  rAB,  rAGr_{AD},\ \ r_{AB},\ \ r_{AG}




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

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

rAG→=G−A=(0.3,−0.57,0.27)−(0,−0.12,0.54)→rAG→=0.3i−0.45j−0.27k\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,

rAD×TCD→+rAB×FB→+rAG×W⃗=0r_{A D} \times \overrightarrow{T_{C D}}+r_{A B} \times \overrightarrow{F_{B}}+r_{A G} \times \vec{W}=0

TCD→∣ijk0−0.450.6−0.4260.4050.809∣+∣ijk0−0.90Bx0BZ∣+∣ijk0.3−0.45−0.2700−196.2∣=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

ΣMA=0:\Sigma M_{A}=0:

TCD→(+(−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)\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)

+(−0.9(Bz)−0(0))i−(0(Bz)−0(Bx))j+(0(0)−(−0.9(Bx))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+(-0.45(-196.2)-(-0.27(0)) i-(0.3(-196.2)-(-0.27(0)) j+(0.3(0)-(-0.45(0))) k=0


ΣMA=0:\Sigma M_{A}=0:

−0.607TCDi−0.256TCDj−0.192TCDk−0.9BZi+0.9Bxk+88.29i+58.86j=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,

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

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

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

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

−0.256TCD+58.86=0TCD=230N( ans )\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,

−0.192(230N)+0.9Bx=0Bx=49.1N(ans)\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,

−0.607(230N)−0.9Bz+88.29=0BZ=−57.0N( ans )\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)

−0.426TCD+Ax+Bx=0−0.426(230N)+Ax+(49.1N)=0Ax=48.9N(ans)\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}

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

0.809(230N)−196.2+Az+−57.0N=0Az=67.1N(ans)\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}

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Example

The mass for the window below is 20 kg and it acts at the geometric center of the window, at G. The window is attached at A and B by two hinges which can be idealized as ball and socket supports. Determine all the forces acting on the window when it is held open in the position shown by the rope attached at C. The hinge at B has been modified to allow for translation along its own axis of rotation.
Solution:
1) Draw Free Body Diagram

2) Determine all forces and moments in cartesian vector form

Cable force CD,

TCD→=∣TCD∣uCD=∣TCD∣(rCD∣rCD∣)\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)

rCD=D−C=(0,−0.45,1.14)−(0.6,−1.14,0)→rCD=−0.6i+0.69j+1.14kr_{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

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

TCD→=∣TCD∣(−0.6i+0.69j+1.14k1.461)=TCD(−0.411i+0.472j+0.780k)\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)

TCD→=−0.411TCDi+0.472TCDj+0.780TCDk\overrightarrow{T_{CD}}=-0.411T_{CD}i+0.472T_{CD}j+0.780T_{CD}k

Weight,

W⃗=−(20kg)(9.81ms2)k→W⃗=(0i+0j−196.2k)N\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,

FA→=Axi+Ayj+AzkFB→=Bxi+0j+Bzk\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,

ΣFx=0:−0.411TCD+Ax+Bx=0    (1) i−components\Sigma F_{x}=0:-0.411 T_{C D}+A_{x}+B_{x}=0 \ \ \ \ (1 ) \ i-components

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

ΣFz=0:0.780TCD−196.2k+Az+Bz=0    (3) k−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,

ΣMA=0:rAD×TCD→+rAB×FB→+rAG×W⃗=0\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 rAD,  rAB,  rAGr_{AD},\ \ r_{AB},\ \ r_{AG}




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

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

rAG→=G−A=(0.3,−0.57,0.27)−(0,−0.12,0.54)→rAG→=0.3i−0.45j−0.27k\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,

rAD×TCD→+rAB×FB→+rAG×W⃗=0r_{A D} \times \overrightarrow{T_{C D}}+r_{A B} \times \overrightarrow{F_{B}}+r_{A G} \times \vec{W}=0

TCD→∣ijk0−0.450.6−0.4110.4720.780∣+∣ijk0−0.90Bx0BZ∣+∣ijk0.3−0.45−0.2700−196.2∣=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


ΣMA=0:\Sigma M_{A}=0:

TCD→(+(−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)\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)

+(−0.9(Bz)−0(0))i−(0(Bz)−0(Bx))j+(0(0)−(−0.9(Bx))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+(-0.45(-196.2)-(-0.27(0)) i-(0.3(-196.2)-(-0.27(0)) j+(0.3(0)-(-0.45(0))) k=0


ΣMA=0:\Sigma M_{A}=0:

−0.6342TCDi−0.2466TCDj+0.185TCDk−0.9BZi+0.9Bxk+88.29i+58.86j=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,

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

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

0.185TCD+0.9Bx=0    (6) k−components0.185 T_{C D}+0.9 B_{x}=0\ \ \ \ (6 ) \ k-components

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

−0.2466TCD+58.86=0TCD=239N( ans )\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,

0.185(239N)+0.9Bx=0Bx=−49.1N(ans)\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,

−0.6342(239N)−0.9Bz+88.29=0BZ=−70.3N( ans )\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)

−0.411TCD+Ax+Bx=0−0.411(239N)+Ax+(−49.1N)=0Ax=147N(ans)\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}

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

0.780(239N)−196.2+Az+−70.3N=0Az=80.1N(ans)\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}


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Example

The mass for the window below is 20 kg and it acts at the geometric center of the window, at G. The window is attached at A and B by two hinges which can be idealized as ball and socket supports. Determine all the forces acting on the window when it is held open in the position shown by the rope attached at C. The hinge at B has been modified to allow for translation along its own axis of rotation.
Solution:
1) Draw Free Body Diagram

2) Determine all forces and moments in cartesian vector form

Cable force CD,

TCD→=∣TCD∣uCD=∣TCD∣(rCD∣rCD∣)\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)

rCD=D−C=(0,−0.45,1.14)−(0.6,−1.14,0)→rCD=−0.6i+0.69j+1.14kr_{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

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

TCD→=∣TCD∣(−0.6i+0.69j+1.14k1.461)=TCD(−0.411i+0.472j+0.780k)\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)

TCD→=−0.411TCDi+0.472TCDj+0.780TCDk\overrightarrow{T_{CD}}=-0.411T_{CD}i+0.472T_{CD}j+0.780T_{CD}k

Weight,

W⃗=−(20kg)(9.81ms2)k→W⃗=(0i+0j−196.2k)N\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,

FA→=Axi+Ayj+AzkFB→=Bxi+0j+Bzk\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,

ΣFx=0:−0.411TCD+Ax+Bx=0    (1) i−components\Sigma F_{x}=0:-0.411 T_{C D}+A_{x}+B_{x}=0 \ \ \ \ (1 ) \ i-components

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

ΣFz=0:0.780TCD−196.2k+Az+Bz=0    (3) k−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,

ΣMA=0:rAD×TCD→+rAB×FB→+rAG×W⃗=0\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 rAD,  rAB,  rAGr_{AD},\ \ r_{AB},\ \ r_{AG}




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

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

rAG→=G−A=(0.3,−0.57,0.27)−(0,−0.12,0.54)→rAG→=0.3i−0.45j−0.27k\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,

rAD×TCD→+rAB×FB→+rAG×W⃗=0r_{A D} \times \overrightarrow{T_{C D}}+r_{A B} \times \overrightarrow{F_{B}}+r_{A G} \times \vec{W}=0

TCD→∣ijk0−0.450.6−0.4110.4720.780∣+∣ijk0−0.90Bx0BZ∣+∣ijk0.3−0.45−0.2700−196.2∣=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


ΣMA=0:\Sigma M_{A}=0:

TCD→(+(−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)\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)

+(−0.9(Bz)−0(0))i−(0(Bz)−0(Bx))j+(0(0)−(−0.9(Bx))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+(-0.45(-196.2)-(-0.27(0)) i-(0.3(-196.2)-(-0.27(0)) j+(0.3(0)-(-0.45(0))) k=0


ΣMA=0:\Sigma M_{A}=0:

−0.6342TCDi−0.2466TCDj+0.185TCDk−0.9BZi+0.9Bxk+88.29i+58.86j=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,

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

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

0.185TCD+0.9Bx=0    (6) k−components0.185 T_{C D}+0.9 B_{x}=0\ \ \ \ (6 ) \ k-components

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

−0.2466TCD+58.86=0TCD=239N( ans )\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,

0.185(239N)+0.9Bx=0Bx=−49.1N(ans)\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,

−0.6342(239N)−0.9Bz+88.29=0BZ=−70.3N( ans )\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)

−0.411TCD+Ax+Bx=0−0.411(239N)+Ax+(−49.1N)=0Ax=147N(ans)\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}

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

0.780(239N)−196.2+Az+−70.3N=0Az=80.1N(ans)\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}


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Disclaimer: At 16:15 of video it should read "Ay=-800N".

Member AB is supported by a cable BC and at A by a square rod which fits loosely through the square hole in the
collar fixed to the member as shown.

(a) (3 marks) Write the cable force T as a vector component.
(b) (8 marks) Determine the reactions at A.



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Solution:
1) Draw Free Body Diagram.


2) Determine all forces and moments in cartesian vector form

Cable force CD,

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

rBC=C−D=(0,1.5,0)−(3,0,−1)→rCD=−3i+1.5j+1kr_{BC}=C-D=(0,1.5,0)-(3,0,-1)\rightarrow r_{CD}=-3i+1.5j+1k

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

TCD→=∣TCD∣(−3i+1.5j+1k3.5)\overrightarrow{T_{CD}}=\left|T_{CD}\right|\left(\frac{-3i+1.5j+1k}{3.5}\right)

TCD→=−67TBCi+37TBCj+27TBCk\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,

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

Reaction forces,

FA=Axi+AyjF_A=A_x\mathbf{i}+A_y\mathbf{j}

Reaction moment,
MA=(MA)xi+(MA)yj+(MA)zk\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,

ΣFx=0:−67TBC+Ax=0    (1) i−components\Sigma F_x=0:-\frac{6}{7}T_{BC}+A_x=0\ \ \ \ (1)\ i-components

ΣFy=0:37TBC+200+Ay=0    (2) j−components\Sigma F_y=0:\frac{3}{7}T_{BC}+200+A_y=0\ \ \ \ (2)\ j-components

ΣFz=0:27TBC−400=0    (3) k−components\Sigma F_z=0:\frac{2}{7}T_{BC}-400=0\ \ \ \ (3)\ k-components

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

TBC=1400N=1.40kNAy=−800NAx=1200N=1.20kN\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,

ΣMA=0;MA+r1×F+r2×TBC=0\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 r1,  r2r_{1},\ \ r_{2}

r1={3i}mr2={1.5j}m\mathbf{r}_{1}=\{3 \mathbf{i}\} \mathrm{m} \quad r_{2}=\{1.5 \mathbf{j}\} \mathrm{m}

Therefore,

ΣMA=0;MA+r1×F+r2×TBC=0\Sigma M_A=\mathbf{0};\quad\mathbf{M}_A+\mathbf{r}_1\times\mathbf{F}+r_2\times\mathbf{T}_{BC}=0

(MA)xi+(MA)yj+(MA)zk+∣ijk3000200−400∣+∣ijk01.50−67TBC37TBC27TBC∣=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

[(MA)x+37TBC]i+[(MA)y+1200]j+[(MA)z+97TBC+600]k=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,

(MA)x+37TBC=0   (4) i−components\left(M_A\right)_x+\frac{3}{7}T_{BC}=0\ \ \ (4)\ i-components

(MA)y+1200=0    (5) j−components\left(M_A\right)_y+1200=0\ \ \ \ (5)\ j-components

(MA)z+97TBC+600=0   (6) k−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,

(MA)x=600N⋅m(MA)y=−1200N⋅m=1.20kN⋅m(MA)z=−2400N⋅m=2.40kN⋅m\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}

The bent bar illustrated has a negligible weight and is supported by a ball and socket joint at O, a cable connected between A and E, and a sliding bearing at D. The sliding bearing is allowed to slide freely in the x-direction. The bar is acted upon for a moment C and a force P, both are vectors parallel to the z-axis. Determine the magnitude of the total reaction force at D and the force in cable AE.





The structure AD has negligible weight and has a ball and socket joint at A. The structure AD supports a uniform rectangular sign. Determine the magnitude of the force in cables EB and DC and all the reaction forces at A.



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