91Ó°ÊÓ

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

Angle:$\mathrm{θ}={25}^{\text{o}}$

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 ${\stackrel{\mathbf{→}}{\mathbf{w}}}_{\mathbf{x}}$and${\stackrel{\mathbf{→}}{\mathbf{w}}}_{\mathbf{y}}$as the components of the gravitational pull of Earth on the block; their magnitudes are${\mathbf{w}}_{\mathbf{x}}\mathbf{=}\mathbf{³¾²\mu ²õ¾\pm ²Ôθ}$and${\mathbf{w}}_{\mathbf{y}}\mathbf{=}\mathbf{³¾²\mu ³¦´Ç²õθ}$. 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, ${\mathbf{f}}_{\mathbf{k}}\mathbf{=}{\mathbf{μ}}_{\mathbf{k}}{\mathbf{F}}_{\mathbf{N}}$ (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:

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

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

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

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:

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

When, the minimum value of ${\mathrm{μ}}_{\mathrm{s}}$equals ${\mathrm{μ}}_{\mathrm{k}}$and the body in frictional motion is more subtle. But if ${\mathrm{μ}}_{\mathrm{k}}$exceeded ${\mathrm{μ}}_{\mathrm{s}}$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 ${\mathrm{μ}}_{\mathrm{s}}$are therefore given by $0.47>{\mathrm{μ}}_{\mathrm{s}}>0.37$.