Task
A weight G pulls symmetrically on two sliding carriages at the angle alpha. What is the minimum static friction required for equilibrium to prevail? The dimensions of the slide can be neglected here.
![Slide carriage at rest](https://pickedshares.com/wp-content/uploads/2020/11/Screenshot_20201114-190442_Sketch.jpg)
Solution
The following video is in German language, but English subtitles are available.
![Equilibrium of forces for the whole system](https://pickedshares.com/wp-content/uploads/2020/11/Screenshot_20201116-174445_Sketch-1024x867.jpg)
\( \DeclareMathOperator{\abs}{abs} \newcommand{\ensuremath}[1]{\mbox{$#1$}} \)
\[\tag{1} 0={F_S} \sin{\left( \alpha \right) }-{F_x}\]
\[\tag{2} 0={F_y}-{F_S} \cos{\left( \alpha \right) }\]
\[\tag{3} {F_x}={F_y} {µ_0}\]
\[\tag{4} {F_S} \cos{\left( \alpha \right) }=\frac{G}{2}\]
\[\tag{5} {F_S}=\frac{G}{2 \cos{\left( \alpha \right) }}\]
\[\tag{6} 0=\frac{G \sin{\left( \alpha \right) }}{2 \cos{\left( \alpha \right) }}-{F_x}\]
\[\tag{7} {F_x}=\frac{G \sin{\left( \alpha \right) }}{2 \cos{\left( \alpha \right) }}\]
\[\tag{8} {F_y}={F_S} \cos{\left( \alpha \right) }\]
\[\tag{9} {F_y}=\frac{G}{2}\]
\[\tag{10} {µ_0}=\frac{{F_x}}{{F_y}}\]
\[\tag{11} {µ_0}=\frac{2 {F_x}}{G}\]
\[\tag{12} {µ_0}=\frac{\sin{\left( \alpha \right) }}{\cos{\left( \alpha \right) }}\]
\[\tag{13} {µ_0}=\tan{\left( \alpha \right) }\]
the height of the sliding blocks is neglected