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A car is parked on a sloping road with the handbrake on. The handbrake acts on the rear wheels. What is the minimum coefficient of static friction required to prevent the car from slipping?

Solution

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

The coordinate system will be turned by the angle Alpha for the solution.

Balance of forces in x-direction

$\tag{1} 0={F_{\mathit{Bx}}}-G \sin{\left( \alpha \right) }$

Balance of forces in y-direction

$\tag{2} 0=-G \cos{\left( \alpha \right) }+{F_{\mathit{By}}}+{F_A}$

Balance of forces around A (left-turning moments are positive)

$\tag{3} 0=G a \sin{\left( \alpha \right) }-G b \cos{\left( \alpha \right) }+{F_{\mathit{By}}} \left( c+b\right)$

Relationship between tangential and normal force on the rear wheel

$\tag{4} {F_{\mathit{Bx}}}={F_{\mathit{By}}} {µ_0}$

Solved for FBy and used in the balance of moments

$\tag{5} {F_{\mathit{By}}}=\frac{{F_{\mathit{Bx}}}}{{µ_0}}$

$\tag{6} 0=G a \sin{\left( \alpha \right) }-G b \cos{\left( \alpha \right) }+\frac{{F_{\mathit{Bx}}} \left( c+b\right) }{{µ_0}}$

Equation 1 solved for FBx and used in the previous result

$\tag{7} {F_{\mathit{Bx}}}=G \sin{\left( \alpha \right) }$

$\tag{8} 0=\frac{G\, \left( c+b\right) \sin{\left( \alpha \right) }}{{µ_0}}+G a \sin{\left( \alpha \right) }-G b \cos{\left( \alpha \right) }$

The minimum value for the coefficient of static friction is

$\tag{9} {µ_0}=\frac{\left( c+b\right) \sin{\left( \alpha \right) }}{b \cos{\left( \alpha \right) -}a \sin{\left( \alpha \right) }}$

It is obvious that the weight does not play a role in the determination of the required coefficient of static friction, i.e. a discussion of the distribution of the weight force over several wheels per axle is superfluous.