Q116P - University Physics with Modern Physics

Question

Friction and climbing SHOES. Shoes made for the sports of bouldering and rock climbing are designed to provide a great deal of friction between the foot and the surface of the ground. Such shoes on the smooth rock might have a coefficient of static friction of 1.2 and a coefficient of kinetic friction of   .

For a person wearing these shoes, what’s the maximum angle (with respect to the horizontal) of a smooth rock that can be walked on without slipping? (a) $\mathbf{42}\mathbf{°}$ ; (b) $\mathbf{50}\mathbf{°}$; (c) $\mathbf{64}\mathbf{°}$; (d) larger than $\mathbf{90}\mathbf{°}$ .

Step-by-Step Solution

Answer

The correct answer is (b).

Step 1: identification of given data:

• The coefficient of static friction is ${\mu }_s=1.2$.
• The coefficient of kinetic friction is ${\mu }_k=0.90$.

Step 1: Concept/Significance of Angle of repose

The angle formed by an inclined surface with the horizontal surface when the object placed on the inclined plane is on the verge of sliding down is known as the angle of repose.

Step 2: Find the maximum angle (with respect to the horizontal) of a smooth rock that can be walked on without slipping

The limiting condition for slipping is at $50.19°$.

 

Draw the free-body diagram for the forces acting on the block.

Apply the equilibrium of the forces along y-direction.

$n-mg \cos \theta =0$  

$n=mg \cos \theta$                                                                                                          ….. (1)

 

Apply the equilibrium of the forces along the x-direction.

$$\begin{gathered} \mathrm{mgsin} \mathrm{\theta }-{\mathrm{f}}_{\mathrm{s}}=0 \\ \mathrm{mgsin} \mathrm{\theta }-{\mathrm{\mu }}_{\mathrm{s}}\mathrm{n}=0 \\ \mathrm{mgsin} \mathrm{\theta }={\mathrm{\mu }}_{\mathrm{s}}\mathrm{mgcos} \mathrm{\theta } \\ \sin \mathrm{\theta }={\mathrm{\mu }}_{\mathrm{s}}\cos \mathrm{\theta } \end{gathered}$$  

 

Simplify further and you get,

$$\begin{gathered} \frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}={\mathrm{\mu }}_{\mathrm{s}} \\ \mathrm{tan\theta }={\mathrm{\mu }}_{\mathrm{s}} \\ \mathrm{\theta }={\tan }^{-1}\left(1.2\right) \\ =50.19° \end{gathered}$$ 

  

The maximum angle is $50.19°$. So, the options (a), (c), and (d) are incorrect.

 

Therefore, the correct answer is (b).