As shown in the figure, a block of mass √(3) kg is kept on a horizontal rough surface of coefficient of friction (1)/(3√(3)). The critical force to be applied on the vertical surface as shown at an angle 60 ° with horizontal such that it does not move, will be 3x. The value of x will be _________. [g = 10 m/s2; sin60 ° = √(3)/2; cos60 ° = 1/2 ] Numerical answer type. Work it out, then open the answer.

Answer: 3.33. F60°Nmgf The push acts downward at 60°: horizontal part Fcos 60° = F/2, vertical (downward) part Fsin 60° = (√(3)F)/(2). Normal force N = mg + (√(3)F)/(2) = 10√(3) + (√(3)F)/(2). No motion: F/2 ≤ μ N = (1)/(3√(3))(10√(3) + (√(3)F)/(2)) = 10/3 + F/6 F/3 ≤ 10/3 ⇒ Fcritical = 10 N 3x = 10 ⇒ x = 3.33

As shown in the figure, a block of mass √(3) kg is kept on a… | JEE Main 2021 (26 Feb, Shift 1)

Q18JEE Main 2021 · 26 FEB · SHIFT 1Medium

Friction Basics

As shown in the figure, a block of mass $\sqrt{3}$ kg is kept on a horizontal rough surface of coefficient of friction $\frac{1}{3\sqrt{3}}$ . The critical force to be applied on the vertical surface as shown at an angle 60 ${}^\circ$ with horizontal such that it does not move, will be 3x. The value of x will be _________. [g = 10 m/s2; sin60 ${}^\circ = \frac{\sqrt{3}}{2}$ ; cos60 ${}^\circ = \frac{1}{2}$ ]

Numerical answer type. Work it out, then open the answer.

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Answer: 3.33

The push acts downward at $60^\circ$ : horizontal part $F\cos 60^\circ = \dfrac{F}{2}$ , vertical (downward) part $F\sin 60^\circ = \dfrac{\sqrt{3}F}{2}$ .
Normal force $N = mg + \dfrac{\sqrt{3}F}{2} = 10\sqrt{3} + \dfrac{\sqrt{3}F}{2}$ .
No motion: $\dfrac{F}{2} \le \mu N = \dfrac{1}{3\sqrt{3}}\left(10\sqrt{3} + \dfrac{\sqrt{3}F}{2}\right) = \dfrac{10}{3} + \dfrac{F}{6}$
$\dfrac{F}{3} \le \dfrac{10}{3} \Rightarrow F_{critical} = 10\text{ N}$
$3x = 10 \Rightarrow x = 3.33$

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