To move a block up to the top of the wedge

[haruspex]

The downward mg does no effect on the motion of the block but the horizontal normal force does. Why?

The "mg" of the block acts on the block, not on the wedge. The vertical force on the wedge is the vertical component of the normal force from the block plus the "mg" of the wedge, and this is exactly balanced by the normal force from the ground.

 

[rudransh verma]

The "mg" of the block acts on the block, not on the wedge. The vertical force on the wedge is the vertical component of the normal force from the block plus the "mg" of the wedge, and this is exactly balanced by the normal force from the ground.

I am asking shouldn’t the force be in the direction or opposite to the direction of the motion in order to work?

 

[haruspex]

I am asking shouldn’t the force be in the direction or opposite to the direction of the motion in order to work?

That's the horizontal component of the normal force, obviously.

 

[rudransh verma]

That's the horizontal component of the normal force, obviously.

I am not able to understand how can only the normal horizontal force can slow the block. The block is not just moving in -x direction but also in the +y direction.I have drawn the fig of the two instants of time 1 and 2. Clearly the block is also moving along y axis. Normal and weight mg ,both should slow down the block.

 

[haruspex]

I am not able to understand how can only the normal horizontal force can slow the block. The block is not just moving in -x direction but also in the +y direction.I have drawn the fig of the two instants of time 1 and 2. Clearly the block is also moving along y axis. Normal and weight mg ,both should slow down the block.

Sorry, I got confused. I thought you were asking about the acceleration of the wedge.
For the block, there is a horizontal component of the normal force slowing the horizontal component of the block's velocity, while in the vertical direction there is the block's weight and the vertical (upward) component of the normal force. The weight has the greater magnitude, so the vertical velocity also reduces.

 

[rudransh verma]

Sorry, I got confused. I thought you were asking about the acceleration of the wedge.
For the block, there is a horizontal component of the normal force slowing the horizontal component of the block's velocity, while in the vertical direction there is the block's weight and the vertical (upward) component of the normal force. The weight has the greater magnitude, so the vertical velocity also reduces.

Yeah! Thank you!

In that case you cannot conserve momentum along the x-direction because the component of the weight in your chosen x-direction is an external force acting on the block.

And there will be no violation of momentum conservation in x direction but in y direction we cannot apply this law because there is mg and by the way we don’t have to.

 

[kuruman]

I am asking if the block is moving up the wedge then it is acted upon by the downward weight and the horizontal normal force. You said the blocks speed will slow down because of the horizontal normal force even though the direction of the force is not opposite to the motion of the block.

Draw a correct free body diagram of the block and you will see that the horizontal component of the normal force on the block wedge is opposite to the motion of the block. While you'e at it, you might as well draw a free body diagram of the wedge.

If you are right then why does mg have no effect on the motion of the block. Why doesn’t mg also slow down the block?

Where did I say that mg has no effect on the motion of the block? In post #29 I wrote

If you isolate the block and call that the system, then its momentum is not conserved because it is acted upon by the normal force exerted by the block wedge and the vertical force of gravity exerted by the Earth[/color]. Its acceleration is dictated by both these forces and I don't see why you exclude the weight.

How does the "vertical force of gravity exerted by the Earth" differ from the weight also known as mg?

 

[kuruman]

And there will be no violation of momentum conservation in x direction but in y direction we cannot apply this law because there is mg and by the way we don’t have to.

That would be true if the x direction is the same as the horizontal direction. If that is what you think, you are contradicting yourself. In post #22 you wrote

By taking the x and y-axis parallel to the slope and perpendicular to the slope of the wedge, I have drawn the fig. Please see. Clearly its the horizontal component of the weight of the block that is slowing down the block.

Unless the angle of the incline is zero, the horizontal direction and the direction parallel to the slope cannot be the same. Furthermore, you mention "the horizontal component of the weigh". You seem to be unaware that the weight does not have a horizontal component and that's serious. The horizontal component of the weght is zero no matter how you choose to orient the x and y axes. That's because the direction of the weight serves as the definition of "vertical". Have you ever wondered how people, since ancient times, were able to build vertical walls on sloped hillsides? How did they figure out which way is vertical?

 

[rudransh verma]

That would be true if the x direction is the same as the horizontal direction. If that is what you think, you are contradicting yourself. In post #22 you wrote

Unless the angle of the incline is zero, the horizontal direction and the direction parallel to the slope cannot be the same. Furthermore, you mention "the horizontal component of the weigh". You seem to be unaware that the weight does not have a horizontal component and that's serious. The horizontal component of the weght is zero no matter how you choose to orient the x and y axes. That's because the direction of the weight serves as the definition of "vertical". Have you ever wondered how people, since ancient times, were able to build vertical walls on sloped hillsides? How did they figure out which way is vertical?

I got it. Thank you to you too.

 

[pbuk]

There has been a big diversion in this thread considering the angle of the block and the dynamics applying while the block is moving up the ramp. These are irrelevant and do not lead to an answer to the initial question so for posterity:

[Edited for clarity]
Homework Statement:: A block of mass $m$ moves with initial velocity $u$ towards a movable wedge of mass $\eta m$ and height $h$ that is initially stationary. All surfaces are smooth.

What is the minimum value of $u$ for which the block will reach the top of the wedge?

The key to answering this question lies in realising that if we are looking for the minimum initial velocity then we are looking for the value which means the block just reaches the top of the wedge i.e. it stops traveling up the wedge when it reaches the top. This also means that the block stops traveling along the wedge i.e. the velocity of the wedge and the block at the final position are equal; let's call this $v$.

We can now write two equations using (i) conservation of momentum and (ii) conservation of energy which we can solve to find the two unknowns $u$, which is the answer to the question, and $v$, which we don't need.

And that's it - no free body diagram, no inventing parameters that are not in the question such as the angle or length of the wedge, no components of forces, no SUVAT equations, none of these will get you any nearer the answer.

 

[nasu]

@pbuk Have you actually tried this method? Is the KE conserved?

 

[kuruman]

@pbuk Have you actually tried this method? Is the KE conserved?

The KE of the block + wedge system is not conserved but the KE + PE (mechanical energy) of the two-mass system is conserved as is the horizontal linear momentum.

 

[haruspex]

There has been a big diversion in this thread considering the angle of the block and the dynamics applying while the block is moving up the ramp. These are irrelevant and do not lead to an answer to the initial question

Quite so, but the OP successfully answered the question, using your approach, in post #17. The digression then occurred because it transpired (post #18) that the OP did not really understand the mechanical process by which the block and wedge arrive at the same speed.

 

[pbuk]

@pbuk Have you actually tried this method?

Yes of course. Have you tried any other method?

Is the KE conserved?

There is no law of conservation of kinetic energy. Total energy is of course conserved (because all surfaces are smooth and there are no external forces), so what other form of energy is involved here?

Quite so, but the OP successfully answered the question, using your approach, in post #17.

Ah yes, I didn't see that - useful then that we refer back to that and the successful method at the end of this thread.

 

[berkeman]

Thread is closed temporarily while the Mentors consider tying it off since the OP's questions have been answered well.

 

[berkeman]

Thread is closed temporarily while the Mentors consider tying it off since the OP's questions have been answered well.

After a Mentor discussion, the thread will remain closed. Thanks all for helping the OP.