First slide

Friction on inclined plane of angle more than angle of repose

Question

A given object takes n times as much time to slide down a 45° rough incline as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline is given by

Moderate

A

$(1 - \frac{1}{n^{2}})$
B

$\frac{1}{(1 - n^{2})}$
C

$\sqrt{(1 - \frac{1}{n^{2}})}$
D

$\sqrt{(\frac{1}{1 - n^{2}})}$

Solution

On smooth inclined plane, the acceleration of the body,
$a = gsin θ$
If s be the distance traveled, then
$s = \frac{1}{2} gsin {θt}_{1}^{2}$ …(1)
On rough inclined plane, the acceleration is given by
$\begin{array}{ll} a & = gsin θ - μgcos θ \\ ∴ s & = \frac{1}{2} (gsin θ - μgcos θ) t_{2}^{2} . . . (2) \end{array}$
From eqs. (1) and (2),
$\frac{t_{2}^{2}}{t_{1}^{2}} = \frac{\sin θ}{\sin θ - μcos θ}$
$But t_{2} = {nt}_{1} or t_{2}^{2} / t_{1}^{2} = n^{2}$
$∴ n^{2} = \frac{\sin θ}{\sin θ - μcos θ}$
Solving, we get
$\begin{aligned} μ & = \frac{n^{2} - 1}{n^{2}} \times \frac{\sin θ}{\cos θ} = \frac{n^{2} - 1}{n^{2}} \times \frac{\sin 45^{\circ}}{\cos 45^{\circ}} \\ = \frac{(n^{2} - 1)}{n^{2}} \times \frac{(1 / \sqrt{2})}{(1 / \sqrt{2})} = \frac{n^{2} - 1}{n^{2}} = (1 - \frac{1}{n^{2}}) . \end{aligned}$

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