An inclined plane is bent in such a way that the vertical cross-section is given by \(y = \frac{{{x^2}}}{4}\) where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction μ = 0.5, the maximum height in cm at which a stationary block will not slip downward is ____________ cm.

Explanation:

riction force concept will be used in this then we can solve using slope of tangent line.

condition for block remaining stationary is that downward force is = to the friction force (in upward direction ) 

mg sin(θ) = µ mg cos(θ)  => tanθ = µ ....................(1)

as we know slope of the equation can be written as: 

tanθ = dy/dx where \(y = \frac{{{x^2}}}{4}\) ........................(2)

Calculation:

Given: 

vertical cross- section \(y = \frac{{{x^2}}}{4}\) .....................(3)

coefficient of friction μ = 0.5

here we can differentiate equation 3 w.r.to. x we get:  dy/dx = x/2

then, from 1 and 2 equation we know  tanθ = µ = dy/dx

so x/2 = 0.5  => x = 1m ...........(4)

put the value from equation 4 in equation 3 we get:

y = 0.25m = 25cm   hence 25cm is height at which a stationary will not slip.

Hence answer is 25 cm.