Problem with Dynamic blocks on a fixed inclined plane

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physicsissohard
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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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  • #2
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?
 
  • #3
1693445210093.png
 
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  • #4
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.
 
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  • #5
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.
 
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  • #6
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.
 
  • #7
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
 
  • #8
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.
 
  • #9
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.
 
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  • #10
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
 
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  • #11
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.
 
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Related to Problem with Dynamic blocks on a fixed inclined plane

What is the effect of the angle of inclination on the motion of a dynamic block on a fixed inclined plane?

The angle of inclination affects the component of gravitational force acting along the plane, which in turn influences the acceleration of the block. As the angle increases, the parallel component of gravitational force increases, resulting in greater acceleration of the block down the plane.

How do frictional forces impact the movement of a block on an inclined plane?

Frictional forces oppose the motion of the block. The magnitude of the frictional force depends on the coefficient of friction between the block and the plane and the normal force. Higher frictional forces can significantly reduce the acceleration of the block or even prevent it from moving if the force is sufficiently large.

How can we calculate the acceleration of a block on an inclined plane?

The acceleration of the block can be calculated using the equation: \( a = g \sin(\theta) - \mu g \cos(\theta) \), where \( g \) is the acceleration due to gravity, \( \theta \) is the angle of inclination, and \( \mu \) is the coefficient of friction. This equation accounts for both the gravitational component along the plane and the frictional force opposing the motion.

What role does the mass of the block play in its motion on an inclined plane?

The mass of the block affects the normal force, which in turn influences the frictional force. However, in the absence of friction, the mass does not affect the acceleration of the block down the plane, as both gravitational force and inertial resistance increase proportionally with mass, canceling each other out.

How do we determine whether a block will start moving on an inclined plane?

To determine if a block will start moving, we compare the component of gravitational force parallel to the plane (\( mg \sin(\theta) \)) with the maximum static friction force (\( \mu_s mg \cos(\theta) \)). If the parallel component exceeds the maximum static friction, the block will start moving. Otherwise, it will remain stationary.

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