# Problem with Dynamic blocks on a fixed inclined plane

• physicsissohard
physicsissohard
Homework Statement
Block a of mass m is placed over a wedge of the same mass m. Both the block and wedge are placed on a fixed inclined plane. assuming all surfaces to be smooth calculate the displacement of block in ground frame in 1 s
Relevant Equations
the homework question.
This is what I thought. since the y component of both their accelerations will be same. we can do this. Mg-N=ma where N is the normal force. and for the wedge it is N+Mgsin^2(theta)=ma using second law of motion where N is the normal force. sin^2(theta) because i resolved gravity twice. now if you elminate N from the equation you get g(1+sin^2(theta))/2 and if you put it in the law of kinematics with t=1 it becomes g(1+sin^2(theta))/4. But apparently thats not the correct answer and I have no idea why.

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physicsissohard said:
Homework Statement: Block a of mass m is placed over a wedge of the same mass m. Both the block and wedge are placed on a fixed inclined plane. assuming all surfaces to be smooth calculate the displacement of block in ground frame in 1 s
Relevant Equations: the homework question.

This is what I thought. since the y component of both their accelerations will be same. we can do this. Mg-N=ma where N is the normal force. and for the wedge it is N+Mgsin^2(theta)=ma using second law of motion where N is the normal force. sin^2(theta) because i resolved gravity twice. now if you elminate N from the equation you get g(1+sin^2(theta))/2 and if you put it in the law of kinematics with t=1 it becomes g(1+sin^2(theta))/4. But apparently thats not the correct answer and I have no idea why.
Do you have an accompanying diagram by any chance?

erobz
Write Newtons 2nd for the wedged shaped block in the vertical direction. A FBD would be good practice.

Another tip would be to designate the normal forces by double subscripts. i.e. the normal force acting on block B from the fixed incline might be ##N_{IB}##, and the normal force from block A acting on B might be designated as ##N_{AB}##.

Try to take a minute to learn LaTeX Guide. We use it to make the math easily readable and quotable.

Last edited:
berkeman and Lnewqban
physicsissohard said:
for the wedge it is N+Mgsin^2(theta)=ma
You have overlooked that N, the normal force between block and wedge, increases the normal force between wedge and ramp, and that this last has a vertical component.

Lnewqban
erobz said:
Write Newtons 2nd for the wedged shaped block in the vertical direction. A FBD would be good practice.

Another tip would be to designate the normal forces by double subscripts. i.e. the normal force acting on block B from the fixed incline might be ##N_{IB}##, and the normal force from block A acting on B might be designated as ##N_{AB}##.

Try to take a minute to learn LaTeX Guide. We use it to make the math easily readable and quotable.

haruspex said:
You have overlooked that N, the normal force between block and wedge, increases the normal force between wedge and ramp, and that this last has a vertical component.
can u complete the solution

physicsissohard said:
can u complete the solution
We aren't permitted to hand out solutions right away. You need to make an effort into applying the advice given to you, and we will help you along the way if you get stuck. The objective is for you to learn. If you have specific questions about what was asked of you please let us know.

physicsissohard said:
can u complete the solution
@erobz told you how to proceed in post #4. Draw separate FBDs for the two bodies and write their F=ma equations. I merely pointed out where your attempt went wrong.

erobz
erobz said:
We aren't permitted to hand out solutions right away. You need to make an effort into applying the advice given to you, and we will help you along the way if you get stuck. The objective is for you to learn. If you have specific questions about what was asked of you please let us know.

Thanks, I re-solved it. I am getting the solution. yeah the normal force exerted on the wedge by the block also needs to be resolved into components. im so dumb

Last edited by a moderator:
physicsissohard said:
Thanks, I re-solved it. I am getting the solution. yeah the normal force exerted on the wedge by the block also needs to be resolved into components. im so dumb
Inexperienced is not dumb. Don't be hard on yourself, it takes time and sometimes struggle.

Lnewqban and berkeman

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