# Find acceleration of Moving incline with a block on it

iceninja3
Homework Statement:
A block (mass m) slides on an inclined plane(mass M, angle = 60˚ to the horizontal). Ignoring friction between the surfaces, find the acceleration of the inclined plane when the block is sliding.
Relevant Equations:
F=ma
Components of force of gravity on inclined plane: mgsinθ, mgcosθ My attempt:
As I need to find acceleration I believe that I need to use F=ma(and thus draw a free body diagram).
I drew the block's weight components(mgsinθ, mgcosθ) and concluded that the only force acting on the plane in the horizontal direction is the horizontal component of mgcosθ--->mgcosθsinθ . I then took the mass of the system since the block and plane move horizontally together(?) to be m+M. This led me to:
a = (mgcosθsinθ)/(m+M)
a = (mg√3/4)/(M+m) which is obviously wrong.

Possible places I went wrong:
1. Is the small mass m maybe not "effectively" m anymore?
2. Perhaps I'm missing some other force components (or misrepresenting the ones I already have)?

Any small hints would be welcome.
Note: The answer is E but I am not sure how to get there.

Homework Helper
The normal force, in this case, is not ##mg cos( \theta)##. In the static case, you get ##N=mg cos( \theta)## from the condition that there is no acceleration along the direction normal to the plane. This is not the case here.

• iceninja3 and topsquark
iceninja3
The normal force, in this case, is not ##mg cos( \theta)##. In the static case, you get ##N=mg cos( \theta)## from the condition that there is no acceleration along the direction normal to the plane. This is not the case here.
How would I actually figure out the force exerted on the inclined plane?

Homework Helper
Gold Member
2022 Award
How would I actually figure out the force exerted on the inclined plane?
First thing is to choose your reference frame. You could choose the inertial frame or the frame of the accelerating ramp.
Next, assign variables to the relevant forces and accelerations.
Thirdly, draw a free body diagram for each object.
Now you can write some equations.
Somewhere in there you need an equation expressing the fact that the block stays in contact with the ramp.

• iceninja3 and nasu
Gold Member
Are you allowed to choose any path forward, or are you supposed to be using a certain analysis?

I got there by fixing a frame to the ground, applying conservation of momentum, conservation of energy, and a block velocity component constraint (down the slope ##v## has a fixed direction), and some light calculus involving the time derivative of the ramp velocity squared ##u^2##.

• iceninja3
Gold Member
Are you allowed to choose any path forward, or are you supposed to be using a certain analysis?

I got there by fixing a frame to the ground, applying conservation of momentum, conservation of energy, and a block velocity component constraint (down the slope ##v## has a fixed direction), and some light calculus involving the time derivative of the ramp velocity squared ##u^2##.
You know what. I got ##a## as the answer...I guess I goofed somewhere.

• iceninja3
iceninja3
Are you allowed to choose any path forward, or are you supposed to be using a certain analysis?

I got there by fixing a frame to the ground, applying conservation of momentum, conservation of energy, and a block velocity component constraint (down the slope ##v## has a fixed direction), and some light calculus involving the time derivative of the ramp velocity squared ##u^2##.
Any approach (forces, energy, etc.) is valid.

Gold Member
Any approach (forces, energy, etc.) is valid.
Well, what I said should work but what I got was ##a##. Not ##e##. So now I'm concerned.

• iceninja3
Homework Helper
Gold Member
The part of the solution where one needs to be careful is setting the correct constraint equation "expressing the fact that the block stays in contact with the ramp" as @haruspex noted.

• iceninja3 and erobz
iceninja3
You know what. I got ##a## as the answer...I guess I goofed somewhere.
Wouldn't conservation of momentum not work because gravity (on the block) is an external force? Unless of course, the impulse from gravity was accounted for(?)

Homework Helper
Gold Member
Momentum is conserved in the horizontal direction only because there are no horizontal forces acting on the block + incline system.

• Lnewqban, iceninja3 and erobz
vcsharp2003
Homework Statement:: A block (mass m) slides on an inclined plane(mass M, angle = 60˚ to the horizontal). Ignoring friction between the surfaces, find the acceleration of the inclined plane when the block is sliding.
Relevant Equations:: F=ma
Components of force of gravity on inclined plane: mgsinθ, mgcosθ

View attachment 319049
My attempt:
As I need to find acceleration I believe that I need to use F=ma(and thus draw a free body diagram).
I drew the block's weight components(mgsinθ, mgcosθ) and concluded that the only force acting on the plane in the horizontal direction is the horizontal component of mgcosθ--->mgcosθsinθ . I then took the mass of the system since the block and plane move horizontally together(?) to be m+M. This led me to:
a = (mgcosθsinθ)/(m+M)
a = (mg√3/4)/(M+m) which is obviously wrong.

Possible places I went wrong:
1. Is the small mass m maybe not "effectively" m anymore?
2. Perhaps I'm missing some other force components (or misrepresenting the ones I already have)?

Any small hints would be welcome.
Note: The answer is E but I am not sure how to get there.
I have always used the pseudo-force concept successfully in such type of problems, where the motion of object is complex (the object in this problem is the block ##m##) and there is relative motion between parts of the system. It makes the problem solving much more simpler than other solutions.

Normally, a block on a fixed incline plane would follow a clear path of motion down the incline, but that's not the case here as the incline plane is allowed to move as the block the slides and the result is that if an observer at rest on ground observes the path of the block then he or she will find that its not a simple straight line path or any other simple path.

But if we take an observer who is fixed to the accelerating incline plane then relative to such an observer, the path of the block will be a simple path as if the incline plane was at rest and the block simply slides down the fixed plane. Then in such a frame of reference, the block will have an acceleration down the incline just like in a typical incline plan-block problem where the incline plane does not move. When we use such an accelerating frame of reference having an acceleration of ##a## then we must include a pseudo-force ##-ma## on the block being observed in addition to other forces acting on the block (the negative sign means the pseudo-force must act in a direction opposite to the acceleration of the accelerating frame of reference).

• topsquark and iceninja3