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=μmg√1+μ2 has to be applied at an angle θ=tan−1(1μ) with the horizontal.
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B
a minimum force Fmin=μmg has to be applied at an angle θ=tan−1(μ) with the horizontal.
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C
a minimum force Fmin=μmg√1+μ2 has to be applied at an angle θ=tan−1(μ) with the horizontal.
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D
a minimum force Fmin=μmg has to be applied at an angle θ=tan−1(1μ) with the horizontal.
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Solution
The correct option is C a minimum force Fmin=μmg√1+μ2 has to be applied at an angle θ=tan−1(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=mg−Fsinθ...(2)
On substituting (2) in (1), we get Fcosθ=μmg−μFsinθ ⇒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=μ√1+μ2 &cosθ=1√1+μ2
Thus, Fmin=μmg√1+μ2