Q117PP - 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 0.90.

If the person steps onto a smooth rock surface that’s inclined at an angle large enough that these shoes begin to slip, what will happen? (a) She will slide a short distance and stop; (b) she will accelerate down the surface; (c) she will slide down the surface at constant speed; (d) we can’t tell what will happen without knowing her mass.

Step-by-Step Solution

Answer

Since the acceleration is negative, so the person is accelerating down the surface. Therefore, the correct answer is (b).

Step 1: identification of given data:

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

Step 2: Concept/Significance of kinetic friction force

When the body is in motion, then the kinetic friction force will be the opposing force, which is then lesser than the static frictional force.

Step 3: Find the condition when a person steps onto a smooth rock surface that’s inclined at an angle large enough that these shoes begin to slip

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

 

Draw the free-body diagram for the forces acting on the shoes 

In the figure above, n is the normal force exerted by the plane on shoes,   is the gravitation weight of the person, and $f_k$  is the kinetic frictional force between the shoes and inclined plane.

 

The net force along the y-direction is given by,

$$\begin{gathered} \underset{ }{ \sum }{F}_y=0 \\ n-mg\cos \theta =0 \\ n=mg\cos \theta \end{gathered}$$  

 

The person is accelerating down the incline and can be modelled as a particle under a net force in the x-direction.

$$\begin{gathered} \sum _{ }{\mathrm{F}}_{\mathrm{x}}={\mathrm{ma}}_{\mathrm{x}} \\ {\mathrm{\mu }}_{\mathrm{k}}\mathrm{n}-\mathrm{mgsin\theta }={\mathrm{ma}}_{\mathrm{x}} \\ {\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgcos\theta }-\mathrm{mgsin\theta }={\mathrm{ma}}_{\mathrm{x}} \\ {\mathrm{a}}_{\mathrm{x}}=\mathrm{g}({\mathrm{\mu }}_{\mathrm{k}}\cos \mathrm{\theta }-\sin \mathrm{\theta }) \end{gathered}$$ 

 

Here, $a_x$ is acceleration in x-direct, $\theta$ is inclined angle, and g is acceleration due to gravity.

 

Substitute all the values in the above equation.

$$\begin{gathered} {\mathrm{a}}_{\mathrm{x}}=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left[0.9\cos \left(50.19°\right)-\sin \left(50.19°\right)\right] \\ =-1.88\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$  

 

Since the acceleration is negative, so the person is accelerating down the surface. Therefore, the correct answer is (b).

 

There is a net acceleration in the sliding direction. Hence, the person will not stop at the end. Therefore, option (a) is incorrect.

 

Since the acceleration of the person is not zero. So, option (c) is incorrect.

 

And the motion is independent of mass. So, option (d) is incorrect.