Question

a block is place on a ramp of parabolic shape given by the equation y = x²/20. If us =0.5, what is the maximum height above the ground at which the block can be placed without slipping?( tan theta= us = dy/dx)​

Answer 1

Equation of the ramp is :

$\sf \: y = \dfrac{ {x}^{2} }{20}$

• Coefficient of Static Friction = 0.5

Given Relation,

$\sf \: tan( \alpha ) = \dfrac{dy}{dx} = \mu$

Now,

$\sf \: \dfrac{dy}{dx} = \dfrac{2x}{20} \\ \\ \implies \sf \: \dfrac{dy}{dx} = \dfrac{x}{10} \\ \\ \implies \sf \: 0.5 = \dfrac{x}{10} \\ \\ \implies \boxed{ \boxed{ \sf \: x = 5 \: m}}$

Thus,

$\longrightarrow \sf \: y = \dfrac{5 {}^{2} }{20} \\ \\ \longrightarrow \boxed{ \boxed{ \sf y = 1.25 \: m}}$

The block should be placed at an height of 1.25 m from the ground so that it doesn't slip.

Answer 2

Given $y = \dfrac{x^2}{20}$, Also $\mu _{s} = 0.5 = \dfrac{1}{2}$

differentiating with respect to x, we get

$\dfrac{dy}{dx}=\dfrac{1}{20}2x\:\:\:\Rightarrow\:\:\:\dfrac{dy}{dx}=\dfrac{x}{10}$

The value of $\dfrac{dy}{dx}$ gives the slope of the tangent $tan\theta$ at any point on the curve i.e $tan\theta = \dfrac{x}{10}.….(1)$

We know that, $tan\theta = \mu_{s} \Rightarrow tan\theta = \dfrac{1}{2}.....(2)$

From equations (1) & (2)

$\dfrac{x}{10} = \dfrac{1}{2} \Rightarrow x = 5m$

$\therefore$ The maximum height above the ground at which the block can be placed without slipping is

$\implies y = \dfrac{x^2}{20}$

$\implies y = \dfrac{(5)^2}{20}$

$\implies y = \dfrac{25}{20}$

$\implies y = \dfrac{5}{4}$

$\implies y = 1.25m$