Why the forces in the arch are carried to the ground

  • #26
205
101
Hello paisiello2. Is the ground force well located?. I've considered it as opposite to [itex]\vec{F}_{yAC}[/itex] and [itex]m_{C}\vec{g}[/itex], as shown in diagram at message #19. And I've considered [itex]\vec{F}_{yAC}[/itex] and [itex]m_{C}\vec{g}[/itex] as parallel forces. So the resultant module is the sum of modules, and the point of application is given by this ecuation: [itex]F_{yAC}\;d_1=m_{C}g\;d_2[/itex], where [itex]d_1[/itex] is the distance from [itex]\vec{F}_{yAC}[/itex] to the point of application of the sum of the two forces, and [itex]d_2[/itex] is also the distance from [itex]m_{C}\vec{g}[/itex] to the sum of the two forces. [itex]\vec{F}_{NC}[/itex] is opposite and equal in module. I've moved [itex]\vec{F}_{yAC}[/itex] and [itex]m_{C}\vec{g}[/itex] along their line of action to the ground, where the ground acts.

The center of mass is at the center of simmetry. It can be calculated by polar integration: consider a circular sector with angle [itex]\pi/3[/itex] wich axis of simmetry is [itex]X[/itex] axis. The coordenate [itex]Y[/itex] is zero, and [itex]X[/itex] coordenate is:

[itex]\displaystyle\frac{1}{\frac{\pi}{6}(r_1^2-r_0^2)}\displaystyle\int_{r_0}^{r_1}\int_{-\pi/6}^{\pi/6}\rho^2\cos\theta\;d\theta\;d\rho[/itex]

[itex]r_1=2[/itex], and [itex]r_0=1[/itex], so the result is approximately 1.49.

Regarding the total torque, the piece named [itex]C[/itex] might rotate, given the enough weight to the piece [itex]A[/itex], around the point [itex]O[/itex], but it is not the case: The arch with which we are working is a three pieces arch, same in shape and weight. When I say, for example, that piece [itex]A[/itex] weights 0.5, it could mean newtons or pounds; it could be expressed to scale, or not. When I draw the arch from 1 to 2, it could be any measure: feet, meters...the arch works. I will do the calculations:


[itex]\vec{M}_O(\vec{F}_{AC})=\vec{M}_O(\overrightarrow{F_x}_{AC})+\vec{M}_O(\overrightarrow{F_y}_{AC})[/itex];

[itex]\vec{F}_{AC}=(1'18-0'75,1'05-1'3)=(0'43,-0'25)[/itex];

[itex]\overrightarrow{OM}=(0'75-2,1'3-0)=(-1'25,1'3)[/itex];

[itex]\vec{M}_{O}
(\vec{F}_{AC})=\overrightarrow{OM}\land{\vec{F}_{AC}}=\begin{vmatrix} \vec{i}&\vec{j}&\vec{k}\\-1'25&1'3&0\\0'43&-0'25&0\end{vmatrix}=-0'25\vec{k}[/itex].

[itex]\overrightarrow{ON}=(1'3-2,0'75-0)=(-0'7,0'75)[/itex];

[itex]m_{C}\vec{g}=(1'3-1'3,0'25-0'75)=(0,-0'5)[/itex];

[itex]\vec{M}_{O}(m_{C}\vec{g})=\overrightarrow{ON}\land{m_{C}\vec{g}}=\begin{vmatrix}\vec i&\vec{j}&\vec{k}\\-0'7&0'75&0\\0&-0'5&0\end{vmatrix}=0'35\vec{k}[/itex].

The ground force restrains the counterclockwise rotation, so the total torque is [itex]\vec{0}[/itex]
 
Last edited:
  • #27
205
101
Hello, if I place [itex]\vec{F}_{NC}[/itex] as shown in diagram, everything seems right:

1- Sum of forces:

[itex]\sum{\vec{F}_{yC}}=m_{C}\vec{g}+\vec{F}_{yAC}+\vec{F}_{NC}=\vec{0}[/itex]

[itex]\sum{\vec{F}_{xC}}=\vec{F}_{xAC}+\vec{f}_{eC}=\vec{0}[/itex]

2-Total torque about O and Q (because it might be clockwise or counterclockwise rotation).[itex]\vec{f}_{eC}[/itex] does not generate torque:

About O:

[itex]\vec{M}_{O}(\vec{F}_{AC})=\vec{M}_{O}(\overrightarrow{F_x}_{AC})+\vec{M}_{O}(\overrightarrow{F_y}_{AC})[/itex];


[itex]\vec{F}_{AC}=(1'18-0'75,1'05-1'3)=(0'43,-0'25)[/itex];


[itex]\overrightarrow{OM}=(0'75-2,1'3-0)=(-1'25,1'3)[/itex];


[itex]\vec{M}_{O}(\vec{F}_{AC})=\overrightarrow{OM}\land{\vec{F}_{AC}}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-1'25&1'3&0\\0'43&-0'25&0\end{vmatrix}=-0'25\vec{k}[/itex].


[itex]\overrightarrow{ON}=(1'3-2,0'75-0)=(-0'7,0'75)[/itex];


[itex]m_{C}\vec{g}=(1'3-1'3,0'25-0'75)=(0,-0'5)[/itex];


[itex]\vec{M}_{O}(m_{C}\vec{g})=\overrightarrow{ON}\land{m_{C}\vec{g}}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-0'7&0'75&0\\0&-0'5&0\end{vmatrix}=0'35\vec{k}[/itex].

[itex]\overrightarrow{OP}=(-0'13,0)[/itex];

[itex]\vec{F}_{NC}=(0,0'75)[/itex];

[itex]\vec{M}_{O}(\vec{F}_{NC})=\overrightarrow{OP}\land{\vec{F}_{NC}}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-0'13&0&0\\0&0'75&0\end{vmatrix}=-0'10\vec{k}[/itex].

Torque about Q:

[itex]\overrightarrow{QM}=(-0'25,1'30)[/itex];

[itex]\vec{F}_{AC}=(0'43,-0'25)[/itex];

[itex]\vec{M}_{Q}(\vec{F}_{AC})=\overrightarrow{QM}\land{\vec{F}_{AC}}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-0'25&1'30&0\\0'43&-0'25&0\end{vmatrix}=-0'50\vec{k}[/itex].

[itex]\overrightarrow{QN}=(0'3,0'75)[/itex];

[itex]m_{C}\vec{g}=(0,-0'5)[/itex];

[itex]\vec{M}_{Q}(m_{C}\vec{g})=\overrightarrow{QN}\land{m_{C}\vec{g}}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\0'3&0'75&0\\0&-0'5&0\end{vmatrix}=-0'15\vec{k}[/itex].

[itex]\overrightarrow{QP}=(0'87,0)[/itex];

[itex]\vec{F}_{NC}=(0,0'75)[/itex]

[itex]\vec{M}_{Q}(\vec{F}_{NC})=\overrightarrow{QP}\land{\vec{F}_{NC}}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\0'87&0&0\\0&0'75&0\end{vmatrix}=0'65\vec{k}[/itex].


I don't really know if it is right. The only thing I am sure of is the place of the center of mass of the pieces of my arch, because I've got the proof. Just wanted to publish my last effort. I've already got back to the subject I left almost three months ago.

Thanks a lot!
 

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