A child weighing $\mathbf{140}\mathbf{N}$sits at rest at the top of a playground slide that makes an angle of$\mathbf{25}\mathbf{°}$with the horizontal. The child keeps from sliding by holding onto the sides of the slide. After letting go of the sides, the child has a constant acceleration of$\mathbf{0}\mathbf{.}\mathbf{86}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$(down the slide, of course).

(a) What is the coefficient of kinetic friction between the child and the slide?

(b) What maximum and minimum values for the coefficient of static friction between the child and the slide are consistent with the information given here?

Weight of child:$140\text{N}$

Angle:$\mathrm{\theta }={25}^{\text{o}}$

Acceleration:$\mathrm{a}=0.86{\text{m/s}}^{\text{2}}$

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. Use ${\overrightarrow{\mathbf{w}}}_{\mathbf{x}}$and${\overrightarrow{\mathbf{w}}}_{\mathbf{y}}$as the components of the gravitational pull of Earth on the block; their magnitudes are${\mathbf{w}}_{\mathbf{x}}\mathbf{=}\mathbf{mgsin\theta }$and${\mathbf{w}}_{\mathbf{y}}\mathbf{=}\mathbf{mgcos\theta }$. Apply Newton's second law to the x and y-axis. If the downhill component of the weight force were insufficient to overcome static friction, the child would not slide at all.

Formula:

The frictional force acting on a body, ${\mathbf{f}}_{\mathbf{k}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{k}}{\mathbf{F}}_{\mathbf{N}}$ (i)

(a)

The mass of the child can be given as:

$\begin{array}{c}\mathrm{m}=\frac{\mathrm{F}}{\mathrm{g}}\\ =\frac{140}{9.8}\\ =14.3\text{g}\end{array}$

With the $\mathrm{x}$axis directed up along the incline (so that $\mathrm{a}=-0.86{\text{m/s}}^{\text{2}}$), applying Newton's second law along the inclination direction we get that

role="math" localid="1661144564949" $\begin{array}{c}{\mathrm{f}}_{\mathrm{k}}-\mathrm{mg}\text{sin}{25}^{\text{o}}=\mathrm{ma}\mathrm{..............................}\left(\mathrm{a}\right)\\ {\mathrm{f}}_{\mathrm{k}}-140\text{sin}{25}^{\text{o}}=14.3(-0.86)\\ {\mathrm{f}}_{\mathrm{k}}=59.17\text{N}-12.29\text{N}\\ =46.87\text{N}\end{array}$

Also, applying Newton's second law to the$y$axis (perpendicular to the inclined surface), where the acceleration-component is zero, we can get the normal force value applied on the body as follows:

$\begin{array}{c}{\mathrm{F}}_{\mathrm{N}}-140\cos {25}^{\text{o}}=0\\ {\mathrm{F}}_{\mathrm{N}}=140\cos {25}^{\text{o}}\\ =126.88\text{N}\\ \approx 127\text{N}\end{array}$

Thus, the coefficient of friction can be calculated using equation (i) as follows:

$\begin{array}{c}{\mathrm{\mu }}_{\mathrm{k}}={\mathrm{f}}_{\mathrm{k}}/{\mathrm{F}}_{\mathrm{N}}\\ =47\text{N}/127\text{N}\\ =0.37\end{array}$

Hence, the value of the coefficient of friction is $0.37$.

(b)

From the first equation (a) in step (4) calculations, we get that the downhill component of the weight force were insufficient to overcome static friction, the child would not slide at all.

Thus, it requires a greater value that is given by:

$\begin{array}{c}140\text{sin}{25}^{\text{o}}>{\mathrm{f}}_{\mathrm{s},\max }\\ 140\text{sin}{25}^{\text{o}}>{\mathrm{\mu }}_{\mathrm{s}}{\mathrm{F}}_{\mathrm{N}}({\mathrm{f}}_{\mathrm{s},\max }={\mathrm{\mu }}_{\mathrm{s}}{\mathrm{F}}_{\mathrm{N}})\\ {\mathrm{\mu }}_{\mathrm{s}}<\frac{140\text{sin}{25}^{\text{o}}}{127\text{N}}\\ <0.47\end{array}$

When, the minimum value of ${\mathrm{\mu }}_{\mathrm{s}}$equals ${\mathrm{\mu }}_{\mathrm{k}}$and the body in frictional motion is more subtle. But if ${\mathrm{\mu }}_{\mathrm{k}}$exceeded ${\mathrm{\mu }}_{\mathrm{s}}$then kinetic friction will be greater than the static friction (as the incline is raised). So, the body will start to move - which is impossible if ${\mathrm{f}}_{\mathrm{k}}$is large enough to cause deceleration.

Thus, the bounds on ${\mathrm{\mu }}_{\mathrm{s}}$are therefore given by $0.47>{\mathrm{\mu }}_{\mathrm{s}}>0.37$.