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

Correct answer: 0 Fcos60 ° = or (F)/(2) = 1 3 3 N (1) & N = 60 ° + 3 g (2) From equation (1) & (2) (F)/(2) = 1 3 3 ( F 3 2 + 3 g) ⇒ F = g = 10 Newton = 3x So x = (10)/(3) = 3 . 33

JEE Main 202126 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 26 February
Shift: Morning Shift
Subject: Physics
Chapter: Laws of Motion

JEE Main 2021 · 26 February · Morning Shift · Physics · Laws of Motion

As shown in the figure, a block of mass 3 kg is kept on a...

As shown in the figure, a block of mass $\sqrt{3}kg$ is kept on a horizontal rough surface of coefficient of friction $\dfrac{1}{3\sqrt{3}}$ . 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

\left[g = 10m{s}^{- 2}; \sin 60 ° = \dfrac{\sqrt{3}}{2}; \cos 60 ° = \dfrac{1}{2}\right]

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 .

Correct answer

Step-by-step solution

Fcos60 ° = \muN $or$ \dfrac{F}{2} = \dfrac{1}{3\sqrt{3}}N \ldots\left(1\right)
& $N = \sin 60 ° + \sqrt{3}g \ldots \left(2\right)$
From equation \left(1\right) $& \left(2\right)$
$\dfrac{F}{2} = \dfrac{1}{3\sqrt{3}}\left(\dfrac{F\sqrt{3}}{2} + \sqrt{3}g\right)$ ⇒ F = g = 10 $Newton$ = 3x $So$ x = $\dfrac{10}{3} =$ 3 . 33 $</blockquote>$

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