# An object, previously at rest, slides 6 m down a ramp, with an incline of (3pi)/8 , and then slides horizontally on the floor for another 3 m. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Mar 23, 2016

$\mu = \frac{2 \sin \left(\frac{3 \pi}{8}\right)}{1 + 2 \cos \left(\frac{3 \pi}{8}\right)}$

#### Explanation:

The problem may split in two parts:

1. The slipping on the ramp.
2. The slipping on the horizontal floor.

The reason for that is because the friction force depends on the force which stick the object on the floor and it is different in each section. The linkage between both stretches is the speed of the object just at the end of the ramp.

The force of gravity has two component, the first which parallel to the ramp

${F}_{x} = m \cdot g \cdot \sin \left(\frac{3 \pi}{8}\right)$

and the second one, which is perpendicular to the ramp

${F}_{y} = m \cdot g \cdot \cos \left(\frac{3 \pi}{8}\right)$

The force which moves the ball is the first component of force of gravity minus the friction force, ${F}_{f} = \mu \cdot {F}_{x}$

$F = {F}_{x} - {F}_{f} = m \cdot g \cdot \sin \left(\frac{3 \pi}{8}\right) - \mu \cdot m \cdot g \cdot \cos \left(\frac{3 \pi}{8}\right)$

This will be equivalent to

F = mg * [sin((3pi)/8 - mu cos((3pi)/8)]

Thus, the speed at the end of ramp is

${v}_{1}^{2} = {0}^{2} + 2 \cdot a \cdot \text{6 m}$

${v}_{1}^{2} = 2 \cdot \frac{F}{m} \cdot \text{6 m}$

${v}_{1}^{2} = 2 \cdot \frac{\cancel{m} g \cdot \left[\sin \left(\frac{3 \pi}{8} - \mu \cos \left(\frac{3 \pi}{8}\right)\right]\right)}{\cancel{m}} \cdot \text{6 m}$

${v}_{1}^{2} = 2 \cdot 6 \cdot g \left[\sin \left(\frac{3 \pi}{8}\right) - \mu \cos \left(\frac{3 \pi}{8}\right)\right] \text{ " " } \textcolor{red}{\left(1\right)}$

In the 2nd stretch, the only force in action is the friction force, ${F}_{f} = \mu \cdot {F}_{N}$

but now ${F}_{N} = m \cdot g \implies {F}_{h} = \mu \cdot m \cdot g$

If you take ${v}_{2} = 0$ to be the speed of the object when it stops, you can say that

${v}_{2}^{2} = {v}_{1}^{2} - 2 \cdot {a}_{h} \cdot \text{3 m}$

${v}_{1}^{2} = 2 \cdot {a}_{h} \cdot \text{3 m}$

Here you have

${a}_{h} = {F}_{h} / m = \frac{\mu \cdot \cancel{m} \cdot g}{\cancel{m}} = \mu \cdot g$

which will get you

${v}_{1}^{2} = 2 \cdot \mu \cdot g \cdot \text{3 m" = 6 * mu * g" " " } \textcolor{red}{\left(2\right)}$

Use equations $\textcolor{red}{\left(1\right)}$ and $\textcolor{red}{\left(2\right)}$ to get

2 * cancel(6 * g) [sin((3pi)/8) - mu cos((3[i)/8)] = mu * cancel(6 * g)

Rearrange to isolate $\mu$ on one side of the equation

$2 \sin \left(\frac{3 \pi}{8}\right) = \mu + 2 \mu \cos \left(\frac{3 \pi}{8}\right)$

$\mu = \frac{2 \sin \left(\frac{3 \pi}{8}\right)}{1 + 2 \cos \left(\frac{3 \pi}{8}\right)}$

and since

$\sin \left(\frac{3 \pi}{8}\right) = \frac{\sqrt{\sqrt{2} + 2}}{2} \text{ }$ and $\text{ } \cos \left(\frac{3 \pi}{8}\right) = \frac{\sqrt{\left(2 - \sqrt{2}\right)}}{2}$

you can say that

$\mu = \frac{\cancel{2} \cdot \frac{\sqrt{\sqrt{2} + 2}}{\cancel{2}}}{1 + \cancel{2} \cdot \frac{\sqrt{2 - \sqrt{2}}}{\cancel{2}}} = \frac{\sqrt{\sqrt{2} + 2}}{1 + \sqrt{2 - \sqrt{2}}} \approx 1.05$

Since you have $\mu > 1$, you can say that the values given to you by the problem are incorrect, either for the length of the platform, the distance traveled horizontally, or angle of the inclined plane, are incorrect.

Apr 3, 2016

Here length of the ramp (l)= 6m

Angle of inclination of the ramp $\theta = \frac{3 \pi}{8}$
Height of the object from horizontal floor,$h = l \sin \theta$
If mass ofthe body is $m k g$
The initial gravitational potential energy of the body $P E = m g h = m g l \sin \theta = 6 m g \sin \theta J$
The normal reaction acting on the body when it is sliding down the ramp is ${N}_{r} = m g \cos \theta$
and the corresponding frictional force ${F}_{r} = \mu {N}_{r} = \mu m g \cos \theta$
where $\mu =$coefficient of kinetc friction
work done against frictional force when sliding down the ramp ${W}_{1} = {F}_{r} \times l = \mu m g \cos \theta \times 6 J$

when the body slides on horizontal force,then normal reaction ${N}_{f} = m g$ and corresponding frictional force ${F}_{f} = \mu {N}_{f} = \mu m g$
work done against frictional force when sliding along floor
${W}_{2} = {F}_{f} \times 3 = 3 \mu m g J$
Now applying conservation of mechanical energy we can write

The initial KE being zero
Initial PE = total work done against frictional force $= {W}_{1} + {W}_{2}$
$\therefore 6 m g \sin \theta = {W}_{1} + {W}_{2}$
$\implies 6 m g \sin \theta = 6 \mu m g \cos \theta + 3 \mu m g$
$\implies \mu = \frac{2 \sin \theta}{1 + 2 \cos \theta} = \frac{2 \sin \left(3 \frac{\pi}{8}\right)}{1 + 2 \cos \left(3 \frac{\pi}{8}\right)} > 1$ It is not feasible as $\mu$ is always <1

If we consider that the ramp subtends angle $\theta$ with the vertical then $\theta = \frac{\pi}{2} - 3 \frac{\pi}{8}$

and the Eq for $\mu$becomes
$\mu = \frac{2 \sin \left(\frac{\pi}{2} - 3 \frac{\pi}{8}\right)}{1 + 2 \cos \left(\frac{\pi}{2} - 3 \frac{\pi}{8}\right)} = \frac{2 \cos \left(3 \frac{\pi}{8}\right)}{1 + 2 \sin \left(3 \frac{\pi}{8}\right)} = 0.26$ and It is possible

Is it OK?