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Question

A particle of mass m rests on a horizontal floor which has coefficient of static friction μ. For the block to just move

A
a minimum force Fmin=μmg1+μ2 has to be applied at an angle θ=tan1(1μ) with the horizontal.
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B
a minimum force Fmin=μmg has to be applied at an angle θ=tan1(μ) with the horizontal.
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C
a minimum force Fmin=μmg1+μ2 has to be applied at an angle θ=tan1(μ) with the horizontal.
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D
a minimum force Fmin=μmg has to be applied at an angle θ=tan1(1μ) with the horizontal.
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Solution

The correct option is C a minimum force Fmin=μmg1+μ2 has to be applied at an angle θ=tan1(μ) with the horizontal.
Let the FBD of the block be as shown in the figure, when it's just about to move.

From the FBD, we have
Fcosθ=f=μN...(1) and
N+Fsinθ=mg
N=mgFsinθ...(2)

On substituting (2) in (1), we get
Fcosθ=μmgμFsinθ
F=μmgcosθ+μsinθ...(3)

Let Z=cosθ+μsinθ
For F to be minimum, Z must be maximum
dZdθ=0
sinθ+μcosθ=0
tanθ=μ
sinθ=μ1+μ2
& cosθ=11+μ2
Thus,
Fmin=μ mg1+μ2

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