Question

A block of mass m lies on wedge of mass m on a smooth fixed wedge of inclination theta. A particle of mass m lies on the wedge of mass m

Answer 1

Given:

• A wedge of mass M lies on a smooth fixed wedge of inclination θ. A particle of mass m lies on the wedge of mass M in the figure. All surfaces are frictionless and the distance of the block from the fixed wedge is l.

To Find:

• Normal reaction between m and M

• The time after which the particle will hit the fixed plane.

Solution:

The given question is incomplete so write the complete question.

The figure is attached below.

In the reference frame attached to wedge of mass M the particle of mass m can only move horizontally.

Free body diagram of wedge of mass M with respect to Free body diagram of particle w.r.t wedge of mass M ground

Applying Newton's second law for the particle  

      $N+ma_{wq} sin \theta = mg ... (i)$

      $ma_{wq} cos \theta =ma_{m}M ...(ii)$

Applying Newton's second law for wedge of mass M

      $N sin \theta +mgsin \theta =Ma_{wq} ...(iii)$

Solving (i), (ii) and (iii) we get

    $a_{wq} = \frac{(M+m)g sin \theta }{M+msin^2 \theta}$

   $N= \frac{Mmg cos^2 \theta}{M+msin^2 \theta}$

    $a_{mM} = \frac{(M+m)gsin \theta cos \theta}{M+msin^2 \theta}$

Answer 2

Explanation:

Here we should take Pseudo force ( it will be toward the opp side of acceleration)

At eqm

macosθ = mgsinθ

           a = g tanθ

Condition for the block to not slip

Normal should be balanced with he net downward forces

N = mg cosθ + m a sinθ

  =mg cosθ + m g tanθ sinθ

 = mgcosθ + mgsin²θ /cosθ

 =  mg cos²θ + mgsin²θ

                 cosθ

                            N  = mg/ cosθ

                Hope you understand

This is Habel sabu .............Isn't it the Brainliest...............