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Standard VIII

Physics

Free Body Diagram

The minimum v...

Question

The minimum value of F such that the block is at rest is $(tan θ > μ .)$

m g (sin θ - μ cos θ)

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m g (sin θ + μ cos θ)

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m g (tan θ - μ sin θ)

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m g (cos θ - μ sin θ)

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Solution

The correct option is A $m g (sin θ - μ cos θ)$
For minimum value of F friction will act upward to the plane
So
$F + f = m g s i n θ$
$F = m g s i n θ - f$
For minimum value of F friction should be limiting
So
$F = m g s i n θ - μ m g c o s θ$
$F = m g (s i n θ - μ g c o s θ)$

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Similar questions

Q. A block of mass

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θ

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tan θ > μ

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is varied from

P_{1} = m g (sin θ - μ cos θ)

P_{2} = m g (sin θ + μ cos θ)

, the frictional force

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The minimum value of F such that the block is at rest is (tanθ>μ.)

The correct option is A mg(sinθ−μ cosθ)For minimum value of F friction will act upward to the plane SoF+f=mgsinθF=mgsinθ−fFor minimum value of F friction should be limiting SoF=mgsinθ−μmgcosθF=mg(sinθ−μgcosθ)

The minimum value of F such that the block is at rest is $(tan θ > μ .)$

The correct option is A $m g (sin θ - μ cos θ)$
For minimum value of F friction will act upward to the plane
So
$F + f = m g s i n θ$
$F = m g s i n θ - f$
For minimum value of F friction should be limiting
So
$F = m g s i n θ - μ m g c o s θ$
$F = m g (s i n θ - μ g c o s θ)$