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Q.

In the arrangement shown, mass of the object W is 3 kg, angle of friction between the object W and the inclined surface is15o . Then the minimum value of the horizontal force F for which the object W will remain in equilibrium is (Take g=10m/s2)

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a

303N

b

103N

c

30 N

d

60 N

answer is C.

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Detailed Solution

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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