Question: [Physics] a moving inclined plane with block

August 30 2014 CompuTux 5


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  mu[1] be the coeffiction of kinetic friction between the bottom of the inclined plane and the level surface.

Let mu[2]  be the coeffiction of kinetic friction between the block and the upper surface of the inclined plane.

Let theta 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(theta)=(h-y)/(x2-x1)

This implies the relation : -a2y=tan(theta)(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(theta) -    mu[2]  N1 sin(theta) +Ncos(theta


(equation 3) : m2a2y= m2g+m1 sqrt(a1x2+a1y2) sin(theta) +  mu[2]  N1 cos(theta) +Nsin(theta

(equation 4) : m1a1x=Fp - m2 sqrt(a2x2+a2y2)  sin(theta) +  mu[2]  N1 cos(theta) - Ncos(theta

(equation 5) : m1a1y=-m1g  - m2 sqrt(a2x2+a2y2)  cos(theta) - mu[1] N1 + N1 -  mu[2]  N1 sin(theta) -  mu[2]  N1 cos(theta)

Since N1=m1g,  equation 5 becomes : m1a1y= - m2 sqrt(a2x2+a2y2)  cos(theta)   -  mu[2]  N1 sin(theta) -  mu[2]  N1 cos(theta)

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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