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A log of weight W is pulled at a constant velocity and with a force F by means of a rope of length l. The distance between the free end of the rope and ground is h. Neglecting the thickness of the log, the coefficient of friction between log and ground is

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$\frac{Wl - Fh}{F \sqrt{l^{2} - h^{2}}}$

$\frac{\sqrt{l^{2} - h^{2}}}{Wl - Fh}$

$\frac{F \sqrt{l^{2} - h^{2}}}{Wl - Fh}$

$\frac{Wl - Fh}{\sqrt{l^{2} - h^{2}}}$

answer is A.

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Detailed Solution

See figure

$\cos α = \frac{\sqrt{l^{2} - h^{2}}}{l} and \sin α = \frac{h}{l}$

Now $f = Fcos α = F \frac{\sqrt{l^{2} - h^{2}}}{l}$ …(1)

and $Fsin α + R = W or R = W - Fsin α$

Substituting the values of f and R from equations. (t) and (2) in eq. (3), we get

$\begin{array}{ll} F [\frac{\sqrt{l^{2} - h^{2}}}{l}] = μ [W - F (\frac{h}{l})] \\ ∴ & μ = \frac{F \sqrt{(l^{2} - h^{2})}}{l} \times \frac{1}{[W - F (h / l)]} \\ or & μ = \frac{F \sqrt{(l^{2} - h^{2})}}{Wl - Fh} \end{array}$

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