Now, in order for the slope to exert the frictional force specified in Eq. , without any slippage between the slope and cylinder, this force must be less than the maximum allowable static frictional force, , where is the coefficient of static friction.

Rossi stagecoach shotgunThe front braking force based on the static front axle load is F x = (F z)(f site) = (3200)(0.79) = 2528 lb. Consequently, the approximate deceleration based on the static front axle load is: a/g) static = (2528)/6300 = 0.401g Using a deceleration of 0.401g produces a load transfer upon the front axle to further increase the front braking force.

So the static friction force will be the coefficient of static friction multiplied by the normal force and the normal force will be the person’s weight <i>mg</i> so we substitute that in for <i>Fn</i>. So that is 0.016 times 66 kilograms times 9.8 Newtons per kilogram which gives 10 Newtons.