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प्रश्न
An inclined plane is bent in such a way that the vertical cross-section is given by Y = `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.
विकल्प
22
23
25
26
उत्तर
An inclined plane is bent in such a way that the vertical cross-section is given by Y = `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 25 cm.
Explanation:
Given, equation of vertical cross-section
Y= `x^2/4`
Coefficient of friction (µ) = 0.5
Now, applying the force balance equation,
mg sinθ = µ N = fs
⇒ mg sinθ = µ mg cosθ
⇒ tanθ = µ ... (i)
Now, y = `x^2/4`
tanθ = `"dy"/("d"x)`
= `(2x)/4`
= `x/2`
From equation (i)
⇒ tanθ = µ `x/2`
⇒ x = 2µ = 2 × 0.5
⇒ x = 1
Hence, y = `x^2/4`
y = `1/4`
y = 0.25 cm
or Y = 25 cm.