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Question

A block of mass m is placed on a surface with a vertical cross-section represented by the equation y=x36. If the coefficient of static friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is

A
13 m
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B
12 m
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C
16 m
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D
23 m
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Solution

The correct option is C 16 m
The equation of the surface is given by y=x36.
Normal reaction on the block (N) at a given point, will act to the tangent to the curve at that point.

Static friction (fs) will act in a tangential direction as shown in figure, to prevent the slipping of the block triggered by mgsinθ component.


For maximum height of block on the surface without slipping, friction fs will act at its limiting value.
fs=μsN ...(i)

Applying equilibrium condition, along tangential and normal direction,
fs=mgsinθ ....(ii)
N=mgcosθ ...(iii)

Hence, from the above equations,
μsmgcosθ=mgsinθ
tanθ=μs ...(iv)

Slope of the surface represented by the equation y=x36 is given by,
tanθ=dydx
Or, μs=16×3x2
0.5×63=x2
x=1=1 m

Now, the maximum height for the block on the surface will correspond to a value of x=1 m
y=x36
ymax=(1)36=16 m

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