JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 29)
As shown in the figure, a block of mass $$\sqrt 3 $$ kg is kept on a horizontal rough surface of coefficient of friction $${1 \over {3\sqrt 3 }}$$. 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$$^\circ$$ = $${{\sqrt 3 } \over 2}$$; cos60$$^\circ$$ = $${1 \over 2}$$]
_26th_February_Morning_Shift_en_29_1.png)
[g = 10 m/s2; sin60$$^\circ$$ = $${{\sqrt 3 } \over 2}$$; cos60$$^\circ$$ = $${1 \over 2}$$]
Answer
3.33
Explanation
_26th_February_Morning_Shift_en_29_2.png)
N = Mg + Fsin60$$^\circ$$
N = $$\sqrt 3 g + {{F\sqrt 3 } \over 2}$$
For No slipping
Fcos60$$^\circ$$ = Friction
$${F \over 2} = \mu N = {1 \over {3\sqrt 3 }}\left( {\sqrt 3 g + {{F\sqrt 3 } \over 2}} \right)$$
$${F \over 2} = {g \over 3} + {F \over 6}$$
$${F \over 2} - {F \over 6} = {g \over 3}$$
$${{6F - 2F} \over {12}} = {g \over 3}$$
4F = 4g
F = 10
F = 3x
x = $${F \over 3}$$ = $${10 \over 3}$$ = 3.33
x = 3.33
Comments (0)
