Q:A block of mass m is placed on a surface with a vertical cross-section given by y = x3/6. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is
(a) 1/6m
(b)2/3m
(c)1/3m
(d) 1/2m
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Ans: (a)
Sol:$\large y = \frac{x^3}{6}$
$\large \frac{dy}{dx}=\frac{x^2}{2}$
$\large tan\theta = \frac{x^2}{2}$
$\large 0.5 = \frac{x^2}{2}$
x2 = 1
$x = \pm 1$
y = 1/6