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

1. Sep 26, 2014

### Satvik Pandey

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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?

2. Sep 26, 2014

### ehild

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

3. Sep 26, 2014

### Satvik Pandey

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?

4. Sep 26, 2014

### ehild

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 ...

5. Sep 26, 2014

### Satvik Pandey

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

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

6. Sep 26, 2014

### Satvik Pandey

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

7. Sep 26, 2014

### ehild

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.

8. Sep 26, 2014

### Satvik Pandey

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

9. Sep 26, 2014

### ehild

Solve for mu and find the smallest value.

10. Sep 26, 2014

### Satvik Pandey

Should $\theta=45$?

11. Sep 26, 2014

### ehild

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

12. Sep 26, 2014

### Satvik Pandey

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

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

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 $\theta$?

13. Sep 26, 2014

### ehild

$\mu$ need to be minimum.

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

14. Sep 27, 2014

### Satvik Pandey

That was a typo.

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

15. Sep 27, 2014

### Satvik Pandey

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.

16. Sep 27, 2014

### ehild

Splendid !

ehild

17. Sep 27, 2014

### Satvik Pandey

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