# Question:[Physics] a moving inclined plane with block

## Question:[Physics] a moving inclined plane with block

Hello everybody,

I'm trying to solve for a challenging problem : a moving inclined plane with a block

I want to solve for the acceleration components for the block and the plane and the normal force acting on the block.

Let O=(0,0) be an external origin.

Let h be the upper left height of the inclined plane.

Let x1 be the x-position of the center of gravity of the inclined plane.

Let x2 be the x-postion of the center of gravity of the block.

Let y be the y-position of the center of gravity of the block.

Let m1 be the mass of the plane. Let m2 be the mass of the block.

Let   be the coeffiction of kinetic friction between the bottom of the inclined plane and the level surface.

Let  be the coeffiction of kinetic friction between the block and the upper surface of the inclined plane.

Let  be the angle of the plane with the horizontal.

Let Fp a force applied to the inclined plane.

With those defined variables, I make two separable free body diagrams for the block and for the inclined plane, indicating all of the external forces acting on each. It then comes those two vectorial equations :

Block : m2a2=Wweight of block+Fplan acting on block+Ffriction from plan to block+Nnormal from plan to block

Plane : m1a1=Wweight of plane+Fpushing force+Fblock acting on plane+Ffriction from level to plan+Nnormal from level to plane+Ffriction from block to plane+Nnormal from block to plane

I am quite not sure whether I should include the Ffriction from block to plane and the Nnormal from block to plane into the plane's acceleration calculation. Am I right ?

I notice that from the geometry of the figure, I can write down the relation : tan()=(h-y)/(x2-x1)

This implies the relation : -a2y=tan()(a2x-a1x) (equation 1)

Writing down the equations for the x- and y- components of the accelerations of the block and of the plane , this yields :

( equation 2) : m2a2x=m1 sqrt(a1x2+a1y2) cos() -     N1 sin() +Ncos(

(equation 3) : m2a2y= m2g+m1 sqrt(a1x2+a1y2) sin() +   N1 cos() +Nsin(

(equation 4) : m1a1x=Fp - m2 sqrt(a2x2+a2y2)  sin() +   N1 cos() - Ncos(

(equation 5) : m1a1y=-m1g  - m2 sqrt(a2x2+a2y2)  cos() -  N1 + N1 -   N1 sin() -   N1 cos()

Since N1=m1g,  equation 5 becomes : m1a1y= - m2 sqrt(a2x2+a2y2)  cos()   -   N1 sin() -   N1 cos()

I am confused at this stage because a1y=0, that is to say, the plane remains at the ground level surface.

Where am I wrong ? Does this comes from my previous question ?

I want to solve this problem with Maple and plot the solutions. Thank you for any answer !

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