Question

The maximum value of the force Fis nx4N, such that the block shown in Fig, does not move. Find n
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Answer 1

Few things that are clear from the above given figure are

★ Angle made by an object is 60° when the force is applied to the object.

★m= √3 kg

Solⁿ:-

And also we know that if the surface is rough the friction with for sure oppose relative motion between the any particular object and surface.

Now,

Fx = Fcos 60° , Fy = Fsin60°

→At vertical equilibrium we have,

N = Fsin60° + mg

so friction = μN

→μ(Fsin60° + mg) ....... [ by substituting the value of N]

→ At horizontal equilibrium we have,

Fcos 60°= μ(Fsin60° + mg)

$\implies F = \frac{μmg}{(cos \: 60 °\ - μsin \: 60°) }$

Now plug in the values of all the known terms in above equation.

$\implies \frac{1/2√3 \times √3 \times 10}{(1/2-1/2√3×√3/2)}$

$\implies \frac{5}{1/2-1/4}$

$\implies \:F = 20N$