As shown in the figure, a block of mass 3kg is kept on a horizontal rough surface of coefficient of friction 133. 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 = 10m/s2 ; sin 60° = 32; cos60° = 12]

0 3 Views | Posted a month ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    a month ago

    N = mg + F sin 60°

    flim=Fcos60°μN=Fcos60°

    μmg=F [cos60°μsin60°]

    F=μmgcos60°μsin60°=133×3×1012133×32=10/31216=10/31/3=10N=3× (103)N

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