Finding Friction in Equilibrium Rod on Ball

[crom1]

Hi guys. I have one problem from competition that I can't solve. I'll try to translate it.

"rod of mass m and length l is lying on a ball of radius R in state of equilibrium. The rod makes angle θ with the horizontal. There is friction in every point of contact, and it's assumed that it is large enough to hold the system in equilibrium. Find the friction between surface and a ball.

I tried to sketch it in sketch toy, it looks like this: http://sketchtoy.com/55802563

 

[Simon Bridge]

Your diagram appears to have the rod touching the ground - is this what the problem describes?

Please show us your best attempt with your reasoning at each stage.
The approach to these is almost always the same - draw each component in isolation and draw the linear forces at their points of action on each in turn.

 

[crom1]

Yes, the rod is touching the ground. So, my attempt is :
Just to say that the length of a rod is not given but it's easy to find it over R and θ. l=Rcot(θ/2)
And the mass is not given but linear density ρ( I think) so m=ρl

Okay I first started with the fact that if the system is in equilibrium that the sum of forces and moments is 0.
First for rod : ƩF=0
Forces acting on rod are gravitational force from middle of a rod mg,friction between rod and surface, and force that the ball is acting on the rod, let call it F_r(Newtons 3rd law, its normal on the rod)
Now in x and y directions I get

ƩF_x=F_rsinθ-fr (fr- friction)
ƩF_y=mg-F_rcosθ (1)

For the ball

ƩF_x=F_rsinθ-fr-frcosθ
ƩF_y=Mg-N-F_rcosθ ( N is the force that surface is acting on the ball)

ƩM (sum of moments should be 0)
for rod
mg cosθ *l/2=Nl (2)

for ball fr*R=fr*R so frictions between rod and ball and ball and surface are the same.

But from (1) I get mg=F_rcosθ, and from (2) mg=2F_r/cosθ, which leads to contradiction, so obviously I am doing something wrong. Hope you can me show me what.

 

[Simon Bridge]

Yes, the rod is touching the ground. So, my attempt is :
Just to say that the length of a rod is not given but it's easy to find it over R and θ. l=Rcot(θ/2)
And the mass is not given but linear density ρ( I think) so m=ρl

Okay I first started with the fact that if the system is in equilibrium that the sum of forces and moments is 0.
First for rod : ƩF=0
Forces acting on rod are gravitational force from middle of a rod mg,friction between rod and surface, and force that the ball is acting on the rod, let call it F_r(Newtons 3rd law, its normal on the rod)

Did you remember: reaction on rod from the floor, friction from ball and friction from the floor.

This may help talk about things:
Rod goes from A to B.
The center of the ball is at point C.
The Rod touches the ground at point A and the ball at point D.
The ball touches the ground at point E.

So you can define some lengths, eg.|AB|=l, |CD|=|CE|=R and describe relationships, eg.|AC|=|AE|, and so on.
The ball does not slide or roll ... the friction at point D $f_D$ opposes the rolling, and the friction at point E $f_E$ stops the sliding.

The triangle ADE may be useful to you here.

 

[crom1]

Ok. Rod goes from A to B. It touches the ball in point B ( so it ends where it touches the ball).
Center of the ball is in C and it touches the floor in point D.

Forces acting on the rod : gravitational force(mg), reaction from floor(N₂, reaction from the ball(N₁, friction in point B(f_b), and friction in point A(f_a.)

ƩF=0

ƩF_x=0

f_bcosθ+N₁sinθ-f_a+N₂sinθ=0 ( we can't get anything from here)

ƩF_y=mg-N₁cosθ+f_bsinθ-N₂cosθ=0 (same here)

ƩM=0

mg cosθ *l/2=N₁ ----> N₁=(mgcosθ)/2


ƩM=0 on the ball

f_b*R=f_d*R ---> f_b=f_d

ƩF=0

ƩF_y= Mg-N₃+f_bsinθ-N₁cosθ (can't get anything from here)

ƩF_x=N₁sinθ-f_b-f_bcosθ --->

f_b=Nsinθ/(1+cosθ)=mgsinθcosθ/(2(1+cosθ)

(using m=ρl=Rρcot(θ/2) and fact that tan(θ/2)=sinθ/(1+cosθ) )

f_b=1/2*(ρgRcosθ)

 

[crom1]

And by the way, did you had something particular in mind when you said "The triangle ADE may be useful to you here." ?

 

[Tanya Sharma]

Ok. Rod goes from A to B. It touches the ball in point B ( so it ends where it touches the ball).
Center of the ball is in C and it touches the floor in point D.

Forces acting on the rod : gravitational force(mg), reaction from floor(N₂, reaction from the ball(N₁, friction in point B(f_b), and friction in point A(f_a.)

ƩF=0

ƩF_x=0

f_bcosθ+N₁sinθ-f_a+N₂sinθ=0 ( we can't get anything from here)

What is the direction of N₂ ? Does it have a component in x-direction?

 

[crom1]

What is the direction of N₂ ? Does it have a component in x-direction?

Yes, it does N₂ sinθ, it says there. I could be wrong but I made the direction of N₁ the same as N₂, that is normal on the rod.

 

[Tanya Sharma]

Yes, it does N₂ sinθ, it says there. I could be wrong but I made the direction of N₁ the same as N₂, that is normal on the rod.

That is incorrect .The direction of N₂ i.e normal force on the rod from the floor is vertically upwards .

 

[crom1]

That is incorrect .The direction of N₂ i.e normal force on the rod from the floor is vertically upwards .

Thanks for your correction, I'll keep that in mind next time but I don't think that affects the final result.

 

[Tanya Sharma]

Thanks for your correction, I'll keep that in mind next time but I don't think that affects the final result.

The signs of the components of forces are also wrong.

1) What is the direction of friction on the bottom tip of the rod from the floor?
2) What is the direction of friction on the upper tip of the rod from the ball?

Assume rightwards direction to be +x, and vertically downwards to be +y.What is the equation of forces acting in the x-direction on the rod ?What is the equation of forces acting in the y-direction on the rod ?

 

[crom1]

1.)to the right,+x direction

2.)In the line of a rod, so it has x and y components.

O

ƩF=0

ƩF_x=0

f_bcosθ+N₁sinθ-f_a=0 ( we can't get anything from here)

ƩF_y=mg-N₁cosθ+f_bsinθ-N₂=0 (same here)
)

 

[Tanya Sharma]

You are messing up with the signs.

Please upload a picture clearly marking all the forces.

 

[crom1]

It would be something like this. http://sketchtoy.com/55942979

If it's still unclear how I marked the forces acting on the rod, I'll write it on a paper and then post it.
Thanks for you time, Tanya.

 

[crom1]

Oops it seems that I marked friction f_b in wrong direction..

 

[Tanya Sharma]

Yes...the direction of f_b is opposite to what you have marked in the figure.

What about the sign of component of normal force from the ball ,on the rod in x-direction ? From the figure it looks in the -x direction .But in your equation it is positive.

Please reconsider your two equations and write them again (force equations on the rod in x and y direction)