Coefficient Of Friction Between A Body And Wedge MCQ

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The figure shows a horizontal force $\vec{F}$ acting on a block of mass ' $M$ ' on an inclined plane. What is the normal reaction of the block?

$m g \sin θ + F \cos θ$

$m g \sin θ - F \cos θ$

$m g \cos θ - F \sin θ$

$m g \cos θ + F \sin θ$

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A block rests on a rough inclined plane making an angle of 30^o with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is

(take g = 10 m/s² )

2.0

4.0

1.6

2.5

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Concepts Covered - 1

Coefficient of Friction Between a Body and Wedge

If the same wedge is made rough then the time taken by it to come down becomes n times more (nt)

Then find the Coefficient of Friction between body and wedge in term of n?

For this make 2 cases

Case 1- A body slides on a smooth wedge of angle θ and its time of descent is t.

Case 2- If the same wedge made rough then the time taken by it to come down becomes n times more (i.e., nt)

(The length of the path in both cases are the same)

For smooth wedge

$\begin{aligned} S = u \cdot t + \frac{1}{2} a t^{2} \\ S = \frac{1}{2} (g \sin θ) t^{2} \\ u = 0 \\ a = g \sin θ \end{aligned}$

For Rough wedge

$S = \frac{1}{2} g [\sin θ - μ \cos θ] (n t)^{2}$

(i) $=$ (ii)

$μ = \tan θ [1 - \frac{1}{n^{2}}]$

$μ = coefficient of friction$
$θ =$ Angle of inclination
$n =$ an integer

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