Question

In the early afternoon, a car is parked on a street that runs down a steep hill, at an angle of$\mathbf{35}\mathbf{.}\mathbf{0}\mathbf{°}$relative to the horizontal. Just then the coefficient of static friction between the tires and the street surface is 0.725. Later, after nightfall, a sleet storm hits the area, and the coefficient decreases due to both the ice and a chemical change in the road surface because of the temperature decrease. By what percentage must the coefficient decrease if the car is to be in danger of sliding down the street?

Short answer

At 3.4% the coefficient decreases from the given 0.725 value of coefficient of static friction.

Step-by-step solution

Given

Original coefficient of static friction:${\mu }_s=0.725$

Understanding the concept

The problem deals with the Newton’s second law of motion which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object. Draw the free body diagram and then use Newton’s second law.

Draw the free body diagram

Free body diagram:

Calculate the percentage decrease in coefficient if the car is to be in danger of sliding down the street

The car is in “danger of sliding” down when,

$$\begin{gathered} mg\sin \left(35°\right)=f_s \\ =f_{s.max} \\ mg\sin \left(35°\right)={\mu }_s\times mg\cos \left(35°\right) \\ \sin \left(35°\right)={\mu }_s\times \cos \left(35°\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow \tan \left(35°\right)={\mu }_s \\ \Rightarrow {\mu }_s=0.700 \end{gathered}$$

Comparison with original value:$\Rightarrow \left(1-\frac{0.700}{0.725}\right)=0.034\Rightarrow 3.4\%less$

This value represents a 3.45% decrease from the given 0.725 value of coefficient of static friction.