A block is placed on a surface with vertical cross-section given by the equation y = 20 x 2 . If the coefficient of friction is 0.5 , the maximum height above the ground at which a block can be placed without slipping is

For equilibrium mg sin θ = μ mg cos θ ⇒ tan θ = μ But tan θ = dx dy = 10 x So 10 x = 0.5 ⇒ x = 5 ⇒ y = 20 5 2 = 1.25 m

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A block is placed on a surface with vertical cross-section given by the equation $y = \frac{x ^{2}}{20}$ . If the coefficient of friction is 0.5 , the maximum height above the ground at which a block can be placed without slipping is

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Solution:

For equilibrium $mg sin θ = μ mg cos θ \Rightarrow tan θ = μ$

But $tan θ = \frac{dy}{dx} = \frac{x}{10}$

So $\frac{x}{10} = 0.5 \Rightarrow x = 5 \Rightarrow y = \frac{5 ^{2}}{20} = 1.25 m$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Friction

⏱ 2mℹ️ Source: QB

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