A smooth massless wedge is pushed by a horizontal Force P....

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SUMMARY

The discussion centers on the dynamics of a massless wedge subjected to a horizontal force P and the implications of this scenario on the forces acting on a block resting on the wedge. Participants agree that if P equals zero, the only forces acting on the block are the gravitational force and the normal reaction from the wedge. The normal force between the block and the wedge is not simply mg cos(θ) due to the absence of horizontal acceleration, leading to a balance of forces where the wedge moves leftward to maintain equilibrium. The analysis emphasizes the importance of understanding the components of forces acting on both the block and the wedge.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of free body diagrams (FBD)
  • Familiarity with the concepts of normal force and gravitational force components
  • Basic principles of dynamics involving massless objects
NEXT STEPS
  • Study the effects of different values of P on the motion of the wedge and block
  • Learn about the implications of massless systems in classical mechanics
  • Explore advanced topics in dynamics, such as non-inertial reference frames
  • Investigate the role of friction in similar wedge-block systems
USEFUL FOR

Students of physics, particularly those studying mechanics, educators explaining dynamics concepts, and anyone interested in the behavior of massless systems in classical mechanics.

Shivam
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Homework Statement
A smooth massless wedge is pushed by a horizontal Force P, then
(1) If P=0, than N<mgcos(theta); N=Normal reaction received by block
(2) If P>0, than N=mgcos(theta)
(3) If a=0, than N=mgcos(theta);a=acceleration of the wedge
(4)If P=mgtan(theta), than N=mg/cos(theta)
Relevant Equations
F=ma, Cpmponents of weight
245648

Answers- 1,3,4

My attempt, the wedge being massless, there shoul not be any force acting on as it will then have infinite acceleration, so by that i really can't think of how force is applied on pully.
 
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Shivam said:
the wedge being massless, there shoul not be any force acting on
No net force.
 
haruspex said:
No net force.
Can you help, in first option N<mgcos(theta) but if there is even a little bit of Normal force on wedge that how can it be true(the option is correct) ?... If you can explain the 1,3,4 option then please help me
 
As Haruspex said, no net force. P is not the only force working on the block.
Can you describe what you think will happen when P = 0 ?
 
BvU said:
As Haruspex said, no net force. P is not the only force working on the block.
Can you describe what you think will happen when P = 0 ?
There is only two force acting on the block .. one is p which is zero here and the other is normal reaction between wedge and block also the the normal due to ground on wedge is also zero, so only force acting is normal reaction but how will it be balanced by and by whom. That's all is could I think of
 
What about the force mass m exerts on the wedge ?
 
BvU said:
What about the force mass m exerts on the wedge ?
the force mass m exerts is the Normal reaction between them. The mg will have two component mgcos(theta) and mgSin(theta), the mgSin(theta) is just parallel to inclined surface so it will not affect the wedge, the mgCos(theta) is pushing on the wedge, so what's next
 
Shivam said:
the normal due to ground on wedge is also zero
Why do you think that?
 
haruspex said:
Why do you think that?
ok, it won't be, let me try drawing FBD.
 
  • #10
245688
What to do next I can't figure it out...
 
  • #11
Shivam said:
View attachment 245688What to do next I can't figure it out...
Since you have now omitted P I assume you are thinking of (1).
The normal force between block and wedge might not be mg cos θ in this case.

For the net force on the wedge to be zero, what must the two normal forces on it be?
 
  • #12
haruspex said:
Since you have now omitted P I assume you are thinking of (1).
The normal force between block and wedge might not be mg cos θ in this case.

For the net force on the wedge to be zero, what must the two normal forces on it be?
Let's think of this, even if N1 <mgCos(theta) then let's make it's components one will be in horizontal and one will be in vertical and the vertical component of N1 will be canceled by the Ng given by the ground, then what will cancel horizontal component. ?
 
  • #13
Shivam said:
(1.)Let's think of this, even if N1 <mgCos(theta) then let's make it's components one will be in horizontal and one will be in vertical and the vertical component of N1 will be canceled by the Ng given by the ground, then what will cancel horizontal component. ?

(2.) Another thought of mine is, wedge is massless, so it doesn't even exist then the block should only move in downward direction with mg force then N1 should be equal to zero
 
  • #14
Shivam said:
ok, it won't be, let me try drawing FBD.
It won't be in general. It could be in some cases. I'm not sure which of 1 to 4 you have in mind.
 
  • #15
haruspex said:
It won't be in general. It could be in some cases.
Can you please explain the answer, I have been stuck with this for 3 days now, 2 days I thought about it myself and then yesterday I posted this question
 
  • #16
Shivam said:
then what will cancel horizontal component. ?
Clearly, nothing.. so what do you deduce about the normal forces on the wedge when P=0?
 
  • #17
Shivam said:
(2.) Another thought of mine is, wedge is massless, so it doesn't even exist then the block should only move in downward direction with mg force then N1 should be equal to zero
That is correct... If P=0, there is no horizontal external force on the system, the centre of mass does not move in horizontal direction: but the wedge is massless, the centre of mass is in the block. The block moves down the slope, and the wedge moves to the left and that motion cancels the horizontal component of motion of the block along the wedge. No horizontal acceleration of the block, no force in that direction...
 
  • #18
ehild said:
That is correct... If P=0, there is no horizontal external force on the system, the centre of mass does not move in horizontal direction: but the wedge is massless, the centre of mass is in the block. The block moves down the slope, and the wedge moves to the left and that motion cancels the horizontal component of motion of the block along the wedge. No horizontal acceleration of the block, no force in that direction...
Yes, if P=0, but I took the prefix (2) in @Shivam 's post #13 as referring to case 2 in post #1.
 

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