Question

A cubical block rests on an inclined plane of μ = 1/√3, determine the angle of inclination when the block just slides down the inclined plane?

Answer 1

Answer:

Explanation: draw the diagram first , Normal force = mgcosФ

     just slides means n mg cosФ = mg sinФ

                means tanФ = 1/3

         Ф = tan inverse of 1/3

        hope u get that if not let me know i will help

Answer 2

Answer:

The angle of inclination is 30°

Explanation:

Given a cubical block rests on an inclined plane

Coefficient of friction, μ = 1/√3

The resisting force of motion acting between two surfaces is called frictional force, and is given by, $F=\mu N$

From the diagram, it can be seen that the normal reaction N is balanced by mgcosθ

Therefore, $F=\mu cos\theta$            ...(1)

And the force F is balanced by mgsinθ, therefore equation (1) can be written as

$sin\theta=\mu cos\theta$

$\implies \mu=\dfrac{sin\theta}{cos\theta} =tan\theta$

Substituting the value of $\mu$,

$\implies tan\theta=\frac{1}{\sqrt{3} }$

$\implies \theta =tan^{-1}\frac{1}{\sqrt{3} }=30^{o}$

Therefore, the angle of inclination when the block just slides down the inclined plane is 30°.