Block on wedge which is not fixed.Find minimum μ to keep the wedge at rest.

[Satvik Pandey]

Homework Statement

Consider a wedge of mass m whose upper surface is a quarter circle and is smooth. The lower part is rough. A small block of same mass is kept at the topmost position and the initial position is shown in figure. The block slides down the wedge.

Find the minimum value of coefficient of friction between floor and lower surface of wedge so that the wedge remains stationery during the block's motion on it.

Homework Equations

The Attempt at a Solution

I tried to find the angle 'theta' at which horizontal force on the wedge is maximum.

By conservation of energy
$mgr=mgr(1-sin\theta )+\frac { m{ v }^{ 2 } }{ 2 }$

So ${ v }^{ 2 }=2grsin\theta$

Horizontal force on the wedge(Fx) is $\frac { m{ v }^{ 2 } }{ r } cos\theta$So $Fx=\frac { m }{ r } 2grsin\theta cos\theta$

$Fx=mgsin2\theta$

For Fx to be maximum $d(Fx)/d\theta=0$

so
$mg\frac { d }{ d\theta } sin2\theta =0$

$\theta$=45

So ${ v }^{ 2 }=\sqrt { 2 } \ gr$

Normal force on the wedge = $mg+mg+\frac { m{ v }^{ 2 } }{ 2 } sin\theta$=3mg

Horizontal force on wedge= $mgsin2\theta$=mg

As wedge should remain at rest so
$3\mu mg=mg$
So $\mu =1/3$

But this is not correct.Where did I go wrong?

 

[ehild]

Homework Statement

Consider a wedge of mass m whose upper surface is a quarter circle and is smooth. The lower part is rough. A small block of same mass is kept at the topmost position and the initial position is shown in figure. The block slides down the wedge.

Find the minimum value of coefficient of friction between floor and lower surface of wedge so that the wedge remains stationery during the block's motion on it.

Homework Equations

The Attempt at a Solution

I tried to find the angle 'theta' at which horizontal force on the wedge is maximum.

Are you sure? The force of friction depends on the normal force between the ground and the wedge, and it is not equal to the weight of the wedge.

View attachment 73667
By conservation of energy
$mgr=mgr(1-sin\theta )+\frac { m{ v }^{ 2 } }{ 2 }$

So ${ v }^{ 2 }=2grsin\theta$

Horizontal force on the wedge(Fx) is $\frac { m{ v }^{ 2 } }{ r } cos\theta$

The horizontal force of the wedge is the horizontal component of the normal force between the small block and the wedge. The normal force is not equal to the centripetal force. Do not forget agravity.

ehild

 

[Satvik Pandey]

Are you sure? The force of friction depends on the normal force between the ground and the wedge, and it is not equal to the weight of the wedge.
The horizontal force of the wedge is the horizontal component of the normal force between the small block and the wedge. The normal force is not equal to the centripetal force. Do not forget agravity.

ehild

From figure 2

$N=\frac { m{ v }^{ 2 } }{ R } +mgcos(90-\theta )$

or $N=\frac { m{ v }^{ 2 } }{ R } +mgsin\theta$

Horizontal force in wedge $Fx=Ncos\theta =\left( \frac { m{ v }^{ 2 } }{ R } +mgsin\theta \right) cos\theta$

From figure 3

$R=mg+Nsin\theta$

$R=mg+\left( \frac { m{ v }^{ 2 } }{ R } +mgsin\theta \right) sin\theta$

Are these equations correct?
Is my expression of v² in #post1 correct?

 

[ehild]

View attachment 73671

From figure 2

$N=\frac { m{ v }^{ 2 } }{ R } +mgcos(90-\theta )$

or $N=\frac { m{ v }^{ 2 } }{ R } +mgsin\theta$

And you got already that $v^2=2gR\sin(\theta)$, so N= ??

Horizontal force in wedge $Fx=Ncos\theta =\left( \frac { m{ v }^{ 2 } }{ R } +mgsin\theta \right) cos\theta$

[
$R=mg+Nsin\theta$

$R=mg+\left( \frac { m{ v }^{ 2 } }{ R } +mgsin\theta \right) sin\theta$

Are these equations correct?

Certainly not, R is the radius, a length, it can not be force ...

 

[Satvik Pandey]

And you got already that $v^2=2gR\sin(\theta)$, so N= ??

Horizontal force in wedge $Fx=Ncos\theta =\left( \frac { m{ v }^{ 2 } }{ R } +mgsin\theta \right) cos\theta$Certainly not, R is the radius, a length, it can not be force ...

$N=\frac { m2gR\sin (\theta ) }{ R } +mgsin\theta =3 mg\sin (\theta )$

$Fx=3 mg\sin (\theta )cos\theta$...(1)

The 'R' in the RHS is not radius.It is reactionary force on wedge.Sorry I should have choose other variable for it, let it be A.
So
$A=mg+Nsin\theta$

$A=mg+\left( \frac { m{ v }^{ 2 } }{ R } +mgsin\theta \right) sin\theta$
Please consider reactionary force on wedge in figure(3) in #post 3 be A.

 

[Satvik Pandey]

On differentiating eq(1) wrt to $\theta$ and equating it to 0 I got $\theta=45$ maximum value of $Fx=3mg/2$

 

[ehild]

On differentiating eq(1) wrt to $\theta$ and equating it to 0 I got $\theta=45$ maximum value of $Fx=3mg/2$

You need maximum coefficient of friction, not the maximum horizontal force. The force of friction also depends on the normal force and on theta.

And substitute for v² also in the expression of A.

 

[Satvik Pandey]

You need maximum coefficient of friction, not the maximum horizontal force. The force of friction also depends on the normal force and on theta.

And substitute for v² also in the expression of A.

You need maximum coefficient of friction, not the maximum horizontal force. The force of friction also depends on the normal force and on theta.

And substitute for v² also in the expression of A.

$A=mg+3mg{ sin }^{ 2 }\theta$
I thought by equating $\mu A=Fx$ I can solve for $\mu$?

 

[ehild]

$A=mg+3mg{ sin }^{ 2 }\theta$
I thought by equating $\mu A=Fx$ I can solve for $\mu$?

Solve for mu and find the smallest value.

 

[Satvik Pandey]

Solve for mu and find the smallest value.

Should $\theta=45$?

 

[ehild]

Should $\theta=45$?

No, why? Solve for mu in terms of theta, and find the minimum value of mu.

 

[Satvik Pandey]

I got $\mu=3/5$

No, why? Solve for mu in terms of theta, and find the minimum value of mu.

Oh! Sorry.
Putting value in $\mu A=Fx$

I got $μ(mg+3mg{ sin }^{ 2 }\theta )=3mgsin\theta cos\theta$

or $μ=\frac { 3sin\theta cos\theta }{ 1+3{ sin }^{ 2 }\theta }$

In order to have $\mu$ maximum

$\frac { d }{ d\theta } \left( \frac { 3sin\theta cos\theta }{ 1+3{ sin }^{ 2 }\theta } \right) =0$

or $\frac { d }{ d\theta } \left( \frac { sin\theta cos\theta }{ 1+3{ sin }^{ 2 }\theta } \right) =0$or $\frac { (1+3{ sin }^{ 2 }\theta )\frac { d }{ d\theta } (sin\theta cos\theta )-(sin\theta cos\theta )\frac { d }{ d\theta } (1+3{ sin }^{ 2 }\theta ) }{ { (1+3{ sin }^{ 2 }\theta ) }^{ 2 } } =0$or $(1+3{ sin }^{ 2 }\theta )\frac { d }{ d\theta } (sin\theta cos\theta )-(sin\theta cos\theta )\frac { d }{ d\theta } (1+3{ sin }^{ 2 }\theta )=0$

or $(1+3{ sin }^{ 2 }\theta )cos2\theta -(sin\theta cos\theta )(3sin2\theta )=0$

or $2(1+3{ sin }^{ 2 }\theta )cos2\theta =2(sin\theta cos\theta )(3sin2\theta )$

or $2(1+3{ sin }^{ 2 }\theta )cos2\theta =3{ sin }^{ 2 }2\theta$

or $2(1+3{ sin }^{ 2 }\theta )(1-2{ sin }^{ 2 }\theta )=3{ sin }^{ 2 }2\theta$

How to solve for $\theta$?

 

[ehild]

I got $\mu=3/5$

Oh! Sorry.
Putting value in $\mu A=Fx$

I got $μ(mg+3mg{ sin }^{ 2 }\theta )=3mgsin\theta cos\theta$

or $μ=\frac { 3sin\theta cos\theta }{ 1+3{ sin }^{ 2 }\theta }$

In order to have $\mu$ maximum

$\mu$ need to be minimum.

$\frac { d }{ d\theta } \left( \frac { 3sin\theta cos\theta }{ 1+3{ sin }^{ 2 }\theta } \right) =0$

Change to double angles. You have turned $\sin(\theta)\cos(\theta)$ to $0.5 \sin(2\theta)$ already, but remember the very important formulas

$\cos^2(\theta)= \frac{1+\cos(2\theta )}{2}$

and

$\sin^2(\theta)= \frac{1-\cos(2\theta) }{2}$

Store them in your memory!

ehild

 

[Satvik Pandey]

$\mu$ need to be minimum.

That was a typo.

Change to double angles. You have turned $\sin(\theta)\cos(\theta)$ to $0.5 \sin(2\theta)$ already, but remember the very important formulas

$\cos^2(\theta)= \frac{1+\cos(2\theta )}{2}$

and

$\sin^2(\theta)= \frac{1-\cos(2\theta) }{2}$

Store them in your memory!

ehild

Is $\theta ={ cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \times \frac { 1 }{ 2 }$??

 

[Satvik Pandey]

μ=3sinθcosθ1+3sin2θμ=\frac { 3sin\theta cos\theta }{ 1+3{ sin }^{ 2 }\theta }

That was a typo.

Is $\theta ={ cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \times \frac { 1 }{ 2 }$??

So $\theta =26.565$

As $μ=\frac { 3sin\theta cos\theta }{ 1+3{ sin }^{ 2 }\theta }$

or $μ=\frac { 3sin2\theta }{ 2(1+3{ sin }^{ 2 }\theta ) } =\frac { 3\times 0.8 }{ 3.199 } =0.75$

$\mu=0.75$
Got correct answer :D.
Thank you ehild.

 

[ehild]

So $\theta =26.565$

As $μ=\frac { 3sin\theta cos\theta }{ 1+3{ sin }^{ 2 }\theta }$

or $μ=\frac { 3sin2\theta }{ 2(1+3{ sin }^{ 2 }\theta ) } =\frac { 3\times 0.8 }{ 3.199 } =0.75$

$\mu=0.75$
Got correct answer :D.
Thank you ehild.

Splendid !

ehild

 

[Satvik Pandey]

Splendid !

ehild

Thank you ehild. I could't have solved it without your help.:)