Question 139: Laws of Motion

Question

A block of mass $$m$$ is placed on a surface with a vertical cross section given by $$y = {{{x^3}} \over 6}.$$ If the coefficient of friction is $$0.5,$$  the maximum height above the ground at which the block can be placed without slipping is:

Accepted Answer

At limiting equilibrium, $$\mu  = 	an 	heta $$
Equation of the surface,
$$y = {{{x^3}} \over 6}$$
Slope, $$	an 	heta  = \mu  = {{dy} \over {dx}} = {{{x^2}} \over 2}$$
Given that, Coefficient of friction $$\mu  = 0.5$$
$$	herefore$$ $$\,\,\,\,0.5 = {{{x^2}} \over 2}$$
$$ \Rightarrow \,\,\,x =  \pm \,1$$
Now, $$y = {{{x^3}} \over 6} = {1 \over 6}m$$

Answer

$${1 \over 6}m$$

Answer

$${2 \over 3}m$$

Answer

$${1 \over 3}m$$

Answer

$${1 \over 2}m$$

Laws of Motion

A block of mass $m is placed on a surface with a vertical cross section given by y = {{{x^3}} \over 6}. If the coefficient of friction is 0.5,$ the maximum height above the ground at which the block can be placed without slipping is:

Choose the correct answer: