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

In summary, when P is equal to 0, the wedge being massless, there is no external horizontal force on the system and the center of mass does not move in the horizontal direction. The block will move down the slope, and the wedge will move to the left, cancelling out the horizontal component of the block's motion. This results in no horizontal acceleration of the block and no force in that direction. The normal forces on the wedge will also be equal to 0 in this case.
  • #1
Shivam
32
2
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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  • #2
Shivam said:
the wedge being massless, there shoul not be any force acting on
No net force.
 
  • #3
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
 
  • #4
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 ?
 
  • #5
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
 
  • #6
What about the force mass m exerts on the wedge ?
 
  • #7
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
 
  • #8
Shivam said:
the normal due to ground on wedge is also zero
Why do you think that?
 
  • #9
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.
 

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

1. What is a massless wedge?

A massless wedge is a theoretical object that has no mass and is often used in physics problems to simplify calculations. It is usually represented as a triangle with a flat surface.

2. How is a smooth massless wedge different from a regular wedge?

A smooth massless wedge has no friction, whereas a regular wedge has friction that can affect its movement and interactions with other objects. This means that a smooth massless wedge will not experience any resistance when being pushed or pulled.

3. What does it mean for a wedge to be pushed by a horizontal force?

When a wedge is pushed by a horizontal force, it means that the force is acting in a direction parallel to the surface of the wedge. This force can cause the wedge to move or change direction, depending on its placement and the magnitude of the force.

4. How does the force P affect the movement of the wedge?

The force P will cause the wedge to accelerate in the direction of the force. This acceleration will depend on the magnitude of the force and the mass of the wedge. If the force is greater than the weight of the wedge, it will cause the wedge to move in the direction of the force.

5. Can a smooth massless wedge be used in real-life scenarios?

No, a smooth massless wedge is a theoretical concept and cannot exist in the physical world. In real-life scenarios, wedges will always have some amount of mass and friction, which will affect their movement and interactions with other objects.

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