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 x_{1} be the x-position of the center of gravity of the inclined plane.

Let x_{2} 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 m_{1} be the mass of the plane. Let m_{2} 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 F_{p} 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 : m_{2}a_{2}=W_{weight of }_{block}+F_{plan acting on block}+F_{friction from plan to block}+N_{normal from plan to block}

Plane : m_{1}a_{1}=W_{weight of plane}+F_{pushing force}+F_{block acting on plane}+F_{friction from level to plan}+N_{normal from level to plane}+F_{friction from block to plane}+N_{normal from block to plane}

I am quite not sure whether I should include the F_{friction from block to plane} and the N_{normal 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)/(x_{2}-x_{1})

This implies the relation : -a_{2y}=tan()(a_{2x}-a_{1x}) (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) : m_{2}a_{2x}=m_{1} sqrt(a_{1x}^{2}+a_{1y}^{2}) cos() - N_{1} sin() +N_{1 }cos()

(equation 3) : m_{2}a_{2y}= m_{2}g+m_{1} sqrt(a_{1x}^{2}+a_{1y}^{2}) sin() + N_{1} cos() +N_{1 }sin()

(equation 4) : m_{1}a_{1x}=Fp - m_{2} sqrt(a_{2}_{x}^{2}+a_{2}_{y}^{2}) sin() + N_{1} cos() - N_{1 }cos()

(equation 5) : m_{1}a_{1y}=-m_{1}g - m_{2} sqrt(a_{2}_{x}^{2}+a_{2}_{y}^{2}) cos() - N_{1} + N_{1} - N_{1} sin() - N_{1} cos()

Since N_{1}=m_{1}g, equation 5 becomes : m_{1}a_{1y}= - m_{2} sqrt(a_{2}_{x}^{2}+a_{2}_{y}^{2}) cos() - N_{1} sin() - N_{1} cos()

I am confused at this stage because a_{1y}=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 !_{}