Questions

In the arrangement shown, mass of the object W is 3 kg, angle of friction between the object W and the inclined surface is $15^{o}$ . Then the minimum value of the horizontal force F for which the object W will remain in equilibrium is $(Take g = 10 m / s^{2})$

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303N

103N

30 N

60 N

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detailed solution

Correct option is C

N=mgcosα+Fminsinα∴Fmincosα−mgsinα=μN⇒Fmincosα−mgsinα=μmgcosα+μFminsinα⇒Fmin=mgtan(α−λ)Where λ=angle of friction=30o∴ Fmin=3×10×tan(60o−15o)N=30N

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