A block of mass 10 kg is kept on an inclined plane of inclination 30° and friction coefficient between the block and inclined plane is μ=13 Find the minimum value of force (in newton) required to move the block up the inclined plane( Take 3=1.732, g=10.m/s2)

N=mgcosα-FsinθandFcosθ=mgsinα+μNAfter solving we get,F=mgsinα+μmgcosαcosθ+μsinθTo find minimum value of F,dFdθ=0⇒μcosθ=sinθsinθ=μ1+μ2=12 ; cosθ=11+μ2=32Fmin=10012+13.3232+13.12=503=86.6N

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A block of mass 10 kg is kept on an inclined plane of inclination 30° and friction coefficient between the block and inclined plane is $μ = \frac{1}{\sqrt{3}}$ Find the minimum value of force (in newton) required to move the block up the inclined plane( Take $\sqrt{3} = 1 .732, g = 10 . m / s^{2}$ )

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answer is 86.6.

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Detailed Solution

$N = mgcos α - Fsinθ$

and

$Fcosθ = mgsin α + μN$

After solving we get,

$F = \frac{mgsinα + μmgcosα}{cosθ + μsinθ}$

To find minimum value of F,

$\frac{dF}{dθ} = 0$

$\Rightarrow μ cosθ = sinθ$

$sinθ = \frac{μ}{\sqrt{1 + μ^{2}}} = \frac{1}{2}; cosθ = \frac{1}{\sqrt{1 + μ^{2}}} = \frac{\sqrt{3}}{2}$

$F_{\min} = \frac{100 (\frac{1}{2} + \frac{1}{\sqrt{3}} . \frac{\sqrt{3}}{2})}{(\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{3}} . \frac{1}{2})} = 50 \sqrt{3} = 86.6 N$