First slide

Angle of friction

Question

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 $(Take g = 10 m / s^{2})$

Easy

A

$30 \sqrt{3} N$
B

$10 \sqrt{3} N$
C

30 N
D

60 N

Solution

$\begin{array}{l} N = mgcosα + F_{\min} sinα \\ ∴ F_{\min} cosα - mgsinα = μN \\ \Rightarrow F_{\min} cosα - mgsinα = μmgcosα + {μF}_{\min} sinα \\ \Rightarrow F_{\min} = mgtan (α - λ) \\ Where λ = angle of friction = 30^{o} \\ ∴ F_{\min} = 3 \times 10 \times \tan (60^{o} - 15^{o}) N = 30 N \end{array}$

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