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I have been drawn the following figure, showing that the sphere with radius R and cone with base radius 2R and height 2R, away from the pivot with distance 2R, is rotational equilibrium with cylinder that has base radius and height of 2R.

enter image description here

The lengths are indicated in the figure above, and the red dots represent the centre of masses of the objects.

This is the MWE:

\documentclass[parskip]{scrartcl}
\usepackage[margin=15mm]{geometry}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usetikzlibrary{3d,calc}
\usetikzlibrary{shapes.geometric}

\begin{document}

\begin{tikzpicture}
    %Wall
    \draw [fill,pattern=north east lines,draw=none] (-3,3) rectangle (3,3.25);
    \draw (-3,3)--(3,3);
    
    %Segment
    \draw[|<->|]  (-2,2.45) -- (0.67,2.45) node[midway,fill=white] {$2R$};
    \draw[|<->|]  (0.67,2.45) -- (2,2.45) node[midway,fill=white] {$R$};
    \draw[|<->|]  (4,1) -- (4,-0.5) node[midway,fill=white] {$2R$};
    \draw[|<->|]  (-0.5,1) -- (-0.5,-1) node[midway,fill=white] {$2R$};
    \draw[|<->|]  (-0.5,-2) -- (-0.5,-4) node[midway,fill=white] {$2R$};
    \draw[|<->|]  (4,1) -- (4,-0.5) node[midway,fill=white] {$2R$};
    \draw[|<->|]  (1,1.25) -- (2,1.25) node[midway,fill=white] {$R$};
    \draw[|<->|]  (3,1.25) -- (2,1.25) node[midway,fill=white] {$R$};
    
    %Fulcrum
    \draw[thick, fill=yellow, yellow] (-2.01,2) rectangle (2.01,2.25);
    
    %Lines hanging objects
    \draw[thick] (-2,2)--(-2,1)  (0.67,2.25)--(0.67,3) (-2,-1)--(-2,-2) (2,1)--(2,2);
    
    %Sphere
    \draw (-3,0) arc (180:360:1cm and 0.5cm);
    \draw[dashed] (-3,0) arc (180:0:1cm and 0.5cm);
    \draw (-2,1) arc (90:270:0.5cm and 1cm);
    \draw[dashed] (-2,1) arc (90:-90:0.5cm and 1cm);
    \draw (-2,0) circle (1cm);
    \shade[ball color=blue!10!white,opacity=0.20] (-2,0) circle (1cm);
    \tkzDefPoint(-2,0){A} 
    \tkzDrawPoints[color=red, fill=red](A)
    
    %Cone
    \draw (-3,-4) arc (180:360:1cm and 0.5cm) -- (-2,-2) -- cycle;
    \draw[dashed] (-3,-4) arc (180:0:1cm and 0.5cm);
    \shade[left color=blue!5!white,right color=blue!40!white,opacity=0.3] (-3,-4) arc (180:360:1cm and 0.5cm) -- (-2,-2) -- cycle;
    \draw (-2,-4)--(-1,-4);
    \node at (-1.5,-3.7) {$2R$};
     \tkzDefPoint(-2,-3){B} 
    \tkzDrawPoints[color=red, fill=red](B)
    
    %Cylinder
    \draw (1,1) arc (90:270:0.75cm and 1.5cm);
    \draw[dashed] (1,1) arc (90:-90:0.75cm and 1.5cm);
    \draw (3,1) arc (90:270:0.75cm and 1.5cm);
    \draw (3,1) arc (90:-90:0.75cm and 1.5cm);
    \draw (1,1)--(3,1) (1,-2)--(3,-2);
    \shade[left color=green!5!white,right color=green!40!white,opacity=0.3] (1,1) arc (90:270:0.75cm and 1.5cm)--(1,-2)--(3,-2)--(3,-2) arc (-90:90:0.75cm and 1.5cm)--cycle;
    \tkzDefPoint(2,-0.5){C} 
    \tkzDrawPoints[color=red, fill=red](C)
    
    
\end{tikzpicture}

\end{document}

On the other hand, I would like to take infinitesimal thickness of Δx, away with distance x (in blue) as shown in below:

enter image description here

It seems there would be hard to take the infinitesimal thickness of Δx for sphere since it is hard to calculate the height of a small portion explicitly. Is there any ways to draw it nicely?

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  • 2
    Hi, maybe you can try to start with tex.stackexchange.com/questions/42812/3d-bodies-in-tikz
    – Colo
    Oct 11, 2020 at 13:02
  • @Colo I have edited the post after refer to your guidance. But the figure is not that smooth to me.
    – weilam06
    Oct 11, 2020 at 16:24
  • Could you develop your question? My uncertainties: 1) Is there a connection between your blue curves and the centers of mass? 2) Is the final drawing close to reality? Are the two $\Delta x$ on the left and $\Delta x$ on the right different such that the blue volumes verify the formula $2(V(sphere)+V(cone))=V(cylinder)$? By the way, the cone 's center of mass is not correctly placed.
    – Daniel N
    Oct 17, 2020 at 5:12
  • Note that the drawing of your sphere is not in equilibrium position, because the string is not fixed at the pole. In your drawing you shall only see the north pole, but not the south pole which shall be behind the contour. See <upload.wikimedia.org/wikipedia/commons/b/b3/Ello-u12.svg>. Oct 20, 2020 at 12:29

1 Answer 1

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You can try this code

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc,3dtools}% https://github.com/marmotghost/tikz-3dtools
\begin{document}
\begin{tikzpicture}[3d/install view={phi=110,theta=70},line join = round, line cap = round,declare function={R=3;r=R;h=R;}]
        \path
        (0,0,0) coordinate (O)
        (0,0,R)  coordinate (N)
        (0,0,-R)  coordinate (S)
(0,0,3) coordinate (T);
    \begin{scope}[3d/install view={phi=120,psi=0,
        theta=70}]
    \draw[3d/screen coords] (O) circle[radius=R]; 
    \path pic{3d/circle on sphere={R=R,C={(0,0,0)},n={(0,0,1)}}}
    ; 
    \path pic{3d/circle on sphere={R=R,C={(O)}}};
\path  pic{3d/circle on sphere={R=R,C={(O)},P={(O)}, n={(1,-5,0)}}};

\end{scope}

\begin{scope}
\path (0,0,-4*R)
pic{3d/cone={r=R,h/.evaluated=2*R}};
\draw[3d/hidden] (N) -- (0,0,-4*R); 
\end{scope}

\begin{scope}[xshift=5cm,3d/install view={phi=10,theta=70,psi=0}]
    \path[3d/record physical components] 
    (1,0,0) coordinate (ez') 
    (0,1,0) coordinate (ex') 
    (0,0,1) coordinate (ey');
    \begin{scope}[x={(ex')},y={(ey')},z={(ez')}]
    \pic{3d/frustum={r=r,R=r,h=2*R}};
    \end{scope}
\end{scope}     
\end{tikzpicture}
\end{document}

enter image description here

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