## 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.

## Solution

The following video is in German language, but English subtitles are available.

\( \DeclareMathOperator{\abs}{abs} \newcommand{\ensuremath}[1]{\mbox{$#1$}} \)

\[\tag{1} 0={F_S} \sin{\left( \alpha \right) }-{F_x}\]

\[\tag{2} 0={F_y}-{F_S} \cos{\left( \alpha \right) }\]

\[\tag{3} {F_x}={F_y} {µ_0}\]

\[\tag{4} {F_S} \cos{\left( \alpha \right) }=\frac{G}{2}\]

\[\tag{5} {F_S}=\frac{G}{2 \cos{\left( \alpha \right) }}\]

\[\tag{6} 0=\frac{G \sin{\left( \alpha \right) }}{2 \cos{\left( \alpha \right) }}-{F_x}\]

_{x}

\[\tag{7} {F_x}=\frac{G \sin{\left( \alpha \right) }}{2 \cos{\left( \alpha \right) }}\]

_{y}and substitute F

_{S}

\[\tag{8} {F_y}={F_S} \cos{\left( \alpha \right) }\]

\[\tag{9} {F_y}=\frac{G}{2}\]

\[\tag{10} {µ_0}=\frac{{F_x}}{{F_y}}\]

_{y}from equation 9

\[\tag{11} {µ_0}=\frac{2 {F_x}}{G}\]

_{x}from equation 7 we get finally the minimum value for µ

_{0}

\[\tag{12} {µ_0}=\frac{\sin{\left( \alpha \right) }}{\cos{\left( \alpha \right) }}\]

\[\tag{13} {µ_0}=\tan{\left( \alpha \right) }\]

the height of the sliding blocks is neglected