What is the Minimum Coefficient of Static Friction to Keep the Wedge Stationary?

In summary, the minimum coefficient of static friction between a wedge and a horizontal surface for the wedge to be stationary when struck by air of density ρ and velocity v is calculated by breaking down the force into components and taking into account the change of momentum in 1 second. The resulting equation is μ2ρ*A*v^2sinθ/(mg + ρ*A*v^2cosθ).
  • #1
UchihaClan13
145
12

Homework Statement


Air of density ρ, moving with velocity v strikes normally on an inclined surface (having area A) of a wedge of mass m kept on a horizontal surface. Collisions are perfectly elastic (No loss of kinetic energy). Minimum coefficient of static friction between wedge and the horizontal surface, for the wedge to be stationary, is[/B]
upload_2016-1-25_23-38-47.png


Homework Equations


Force imparted to the wedge=ρ*A*v^2
Forces exerted are broken down into their respective components[/B]

The Attempt at a Solution


Breaking down the force (exerted by the air collisions)on the wedge into components we get,
ρ*A*v^2costheta + mg= N(the normal force exerted by the ground surface on the wedge)
Here i assumed that the wedge will be in equilibrium in the y-direction
and in the x-direction(after drawing the f.b.d of the wedge),one can see that
ρ*A*v^2sintheta - ffriction = m*a(here a is zero since in the question it says so)
now comes the hardest part
we know that
0c02725f116ef3cad99527413171d4bb.png

static friction is self adjustingt
so for the minimum coefficient of static friction,i first assumed that the friction force wouldn't be equal to the limiting friction and got stuck with an inequality
Ffriction≤ [PLAIN]https://upload.wikimedia.org/math/9/3/9/939974a71dda1b83cce5ab82a2d2cec1.png(mg + ρ*A*v^2costheta) hence i left this case
and now took the frictional force as limiting friction yielding [PLAIN]https://upload.wikimedia.org/math/9/3/9/939974a71dda1b83cce5ab82a2d2cec1.png=[B][B][B][B][B]ρ*A*v^2sintheta/([B][B][B]ρ*A*v^2costheta + mg)
However my answer doesn't match with the actual answer
I think i am going wrong somewhere or i might have overlooked the "collisions are elastic" or i might have made a wrong assumption
[/B][/B][/B][/B][/B][/B][/B][/B]
Anyways,as usual
Help is much appreciated!:)[/B]
 
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  • #2
UchihaClan13 said:

Homework Statement


Air of density ρ, moving with velocity v strikes normally on an inclined surface (having area A) of a wedge of mass m kept on a horizontal surface. Collisions are perfectly elastic (No loss of kinetic energy). Minimum coefficient of static friction between wedge and the horizontal surface, for the wedge to be stationary, is[/B]
View attachment 94806

Homework Equations


Force imparted to the wedge=ρ*A*v^2[/B]
Is it right?
imagine a particle of mass m and velocity v collides into the wedge elastically. What is the velocity of the particle after the the collision? How much is the change of its momentum?
 
  • #3
I don't know,for sure
And as for your second question,the particle which strikes the wedge elastically with a velocity v would rebound back with the same velocity v.
hence the change in momentum would be Pfinal - Pinitial = -mv-mv=-2mv or 2mv in the opposite direction
 
  • #4
Wait
are you trying to imply that the change in momentum w.r.t time equals the impulsive force imparted by the wedge to the "supposed" block.
Therefore,by Newton's third law ,the force exerted by the block on the wedge equals the impulsive force but in the opposite direction
But if you take a certain time of collision "t" into consideration then the mg force of the wedge gets a factor of t,doesn't it?
How do you proceed further?
 
  • #5
UchihaClan13 said:
I don't know,for sure
And as for your second question,the particle which strikes the wedge elastically with a velocity v would rebound back with the same velocity v.
hence the change in momentum would be Pfinal - Pinitial = -mv-mv=-2mv or 2mv in the opposite direction
Yes, you have to take the total change of momentum of the air flow which is twice the original momentum. During the collision, which is very short time for an individual air molecules, gravity plays no role.
The momentum imparted to the wedge in unit tine is equal the force exerted on it by the flow of air molecules. You need to take the vertical component of this force in addition to the weight of the wedge to calculate the normal force, as you did it in the OP.
 
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  • #6
Thanks a lot ehild for that particular insight
I completely overlooked it
Now i am getting the correct answer
Let me write it down again(This time correctly:smile:)
Mass of air falling per second on the wedge=ρ*A*v (in 1 second,this much mass falls)
Hence momentum of the mass of air =ρ*A*v^2
final mometum of the mass of air after an elastic collision =- ρ*A*v^2 (as it's in the opposite direction)
therefore change in momentum in 1 second=-2ρ*A*v^2
therefore force imparted by the air on the wedgeρ*A*v^2
elementary vector resolutions and algebraic manipulations yield μ2ρ*A*v^2sinθ/(mg + ρ*A*v^2cosθ)
which is the correct answer
thanks againo_O:biggrin:
 
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  • #7
sorry for that odd message Something went wrong
 
  • #8
Here i considered the mass flow rate as mass per second
and calculated the change of momentum in 1 second
hence my time factor "t" is 1 second
I neglected the time of an individual collision since it's negligible
Is my reasoning correct?
 
Last edited by a moderator:
  • #9
It was correct, only the force exerted by the air flow was wrong.
 
  • #10
Thanks once again :)
 

Related to What is the Minimum Coefficient of Static Friction to Keep the Wedge Stationary?

What is the Wedge-Air Collision Problem?

The Wedge-Air Collision Problem is a mathematical physics problem that involves a wedge-shaped object moving on a frictionless surface and colliding with an air-filled cavity. It is often used as a case study in introductory mechanics courses.

What are the main factors that influence the outcome of the Wedge-Air Collision Problem?

The main factors that influence the outcome of the Wedge-Air Collision Problem include the initial velocity and angle of the wedge, the geometry of the wedge and cavity, and the air pressure and density within the cavity.

How is the Wedge-Air Collision Problem solved?

The Wedge-Air Collision Problem can be solved using principles of conservation of momentum and energy. It involves setting up and solving equations for the velocity and trajectory of the wedge and air particles before and after the collision.

What are some real-world applications of the Wedge-Air Collision Problem?

The Wedge-Air Collision Problem has applications in fields such as fluid dynamics, impact mechanics, and vehicle safety. It can also be used to study the motion of objects in low-gravity environments, such as in space.

Are there any limitations to the Wedge-Air Collision Problem model?

Yes, the Wedge-Air Collision Problem model has some limitations. It assumes an idealized scenario with a perfectly smooth wedge and cavity, and does not take into account factors such as air turbulence or the deformation of the wedge and cavity upon impact. In real-world situations, these factors may affect the outcome of the collision.

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