Question

A block of mass m is placed on a surface with a vertical cross-section given by $y = \frac{x^3}{6}$ 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. $\frac{1}{6}m$
• B. $\frac{2}{3}m$
• C. $\frac{1}{3}m$
• D. $\frac{1}{2}m$

Answer 1

The Correct Option is A

\(\tan \theta=\frac{dy}{dx}=\frac{x^{2}}{2}\)
At limiting equilibrium,
\(\mu=\tan \theta\)
\(0.5=\frac{x^{2}}{2}\)
\(\Rightarrow x=\pm 1\)
Now, \(y=\frac{1}{6}\)

Answer 2

After placing the block at the maximum height so that it is not slipping. So, now the block is under limiting friction (i.e. if we move the block further by an inch it will slip). Hence,  the limiting friction is equal to the slope of the surface at that point of time.
So, now the coefficient of the friction is denoted by (μ)
And the slope is nothing but \(\frac{dy}{dx}\) = tanθ
⇒ μ= \(\frac{dy}{dx}\)= tanθ..................(1)
Hence, the equation of the surface is given as:
y= \(\frac{x^{3}}{6}\)
So now differentiate it w.r.t x we have,
⇒ \(\frac{dy}{dx}\) = \(\frac{d}{dx}\) (\(\frac{x^{3}}{6}\))
So now as we know, d/dx xⁿ= nx^n-1 
After using this property in above equation we get,
⇒ \(\frac{dy}{dx}\)= \(\frac{1}{6.3x^{2}}\)
Now from the equation (1) we have,
⇒ μ= \(\frac{3x^{2}}{6}\)= x²2
Now it is given that the coefficient of friction is 0.5
⇒ 0.5= \(\frac{x^{2}}{2}\)
⇒ x²= 1
So now we take square root on both sides,
⇒ x= \(\sqrt{1}\)= 1
So now after substituting this value in the equation of surface we have,
⇒ y= \(\frac{x^{3}}{6}\)= \(\frac{1}{6}\)
Hence the maximum height above the ground at which the block can be placed without slipping is (\(\frac{1}{6}\)) m.
Therefore, the correct  option is ‘A’

Answer 3

Here, a block of a certain mass (m) is placed on a surface with a vertical cross-section.

So firstly, 

µ_s= 0.5 

tan θ= 0.5

µ_s= tan θ

\(\frac{dy}{dx}\)= tan θ . \(\frac{dy}{dx}\)= \(\frac{3x^{2}}{6}\)

tan θ= \(\frac{x^{2}}{2}\) = 0.5 = \(\frac{1}{2}\)

x²= \(\frac{2}{2}\)= 1

Hence, x= ±1

So now, 

y= \(\frac{x^{3}}{6}\) | x=1

y= \(\frac{(1)^{3}}{6}\) = \(\frac{1}{6}\)

Therefore, the maximum height at which the block can be placed without slipping above the ground is \(\frac{1}{6}\)m.

Therefore, the correct option is ‘A’.