How Does Static Friction Maintain Equilibrium in a System of Balls?

In summary, The problem involves two small balls supporting a larger ball at a 45 degree angle with respect to the horizontal. A block is placed on the right to prevent sliding and the vertical surface of the block is frictionless. The static friction coefficient is unknown between all balls and the rectangle block and the bottom floor. The goal is to determine the minimum coefficient of static friction between the bottom floor and the lower right ball for the system to be in equilibrium, and the maximum weight that can be supported by the lower balls before sliding occurs. Equations for equilibrium are set up and solved, taking into account normal forces and frictional forces between the balls and the block. The assumption is made that all frictions are at their maximum to account for impending
  • #1
aerograce
64
1

Homework Statement


Two small balls, each of weight Wo, are supporting a big ball of weight W1, as shown in Figure Q2. The line connecting the center of the upper ball and that of a lower ball forms an angle of 45° with respect to the horizontal, as shown in Figure Q2. A block of weight Wo is placed on the right to prevent sliding of the balls. The vertical surface of the rectangle block is frictionless. The static friction coefficient equals μ between all balls and between the rectangle block and the bottom floor. Answer the following questions and express the answers in terms of Wo and μ. Make assumptions where necessary.
(a) Determine the minimum static friction coefficient between the bottom floor and the lower right ball for the system to be in equilibrium.
(b) Determine the maximum weight W1 that can be supported by the lower balls before sliding occurs.


Homework Equations



f=μN.



The Attempt at a Solution



Honestly, I have no thoughts about how to approach this problem. I am just thinking that, we should analyse the right ball, where it has a static friction μN, and N can be obtained by:
2Nsin(45)=W1;

And we use equilibrium for the right ball, and we will know the normal force it exerts on Wo. And we can then know the static friction coefficient by analyzing the block.

Can anyone give me more thoughts or inspirations to approach this problem?
 

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  • #2
What are all the forces on the right-hand ball? Invent unique variable names for them. How many equations should you be able to write for its equilibrium? What are they?
 
  • #3
There are three places on the ball that come into contact with other objects. Friction is involved at only two of these.
 
  • #4
haruspex said:
What are all the forces on the right-hand ball? Invent unique variable names for them. How many equations should you be able to write for its equilibrium? What are they?

The right-hand ball will experience:
Normal force from the bigger ball;
normal force from the block;
normal force from the ground;
static friction between the right hand ball and the bigger ball;
static friction between the right hand ball and the ground( I am not sure about this,coz in question statement it didnt mention this friction)
 
  • #5
Basic_Physics said:
There are three places on the ball that come into contact with other objects. Friction is involved at only two of these.

Could you see my comment above? I expressed my concerns there
 
  • #6
aerograce said:
The right-hand ball will experience:
Normal force from the bigger ball;
normal force from the block;
normal force from the ground;
static friction between the right hand ball and the bigger ball;
static friction between the right hand ball and the ground( I am not sure about this,coz in question statement it didnt mention this friction)
It doesn't mention it in the set-up, but it is implied in question a). The coefficient there is unknown, it's zero between the ball and the block, and everywhere else that matters it's μ.
So, make up some name for those force and write some equilibrium equations.
 
  • #7
haruspex said:
It doesn't mention it in the set-up, but it is implied in question a). The coefficient there is unknown, it's zero between the ball and the block, and everywhere else that matters it's μ.
So, make up some name for those force and write some equilibrium equations.

I did my calculations as follow. Could you please check for me? This image is rotated wrongly when I upload. Could you please save it to your computer and rotate with windows photo viewer? It will be faster for u to read.
 

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  • #8
aerograce said:
I did my calculations as follow. Could you please check for me? This image is rotated wrongly when I upload. Could you please save it to your computer and rotate with windows photo viewer? It will be faster for u to read.
I can rotate it, but the image posted is far too small, and blowing it up to a readable size makes it all a blur. If you can't post a good clear image of your working, right way up, please take the trouble to type your working in, preferably in LaTex.
 
  • #9
haruspex said:
I can rotate it, but the image posted is far too small, and blowing it up to a readable size makes it all a blur. If you can't post a good clear image of your working, right way up, please take the trouble to type your working in, preferably in LaTex.

You can still refer to my sent image for some notations.

On the right handball, I denote:

normal force from bigger ball: N2'
normal force from ground: N
normal force from block: No'
weight: Wo
frictional force between the right hand ball and the bigger ball: f1=μN2'
frictional force between the right hand ball and the block: f2=μ'N

For the block:
No=Woμ=No'

For the bigger ball:

N2=(sqr(2)/2)W1

For the right hand ball:

Vertical equilibrium:

N=N2'cos45°+f1cos45°+Wo
==>
N=((1+μ)/2)W1+Wo

Horizontal equilibrium:

μN2'Sin45°+No'+f2=N2'Sin45°
==>
1/2 W1μ + Woμ+μ'N=1/2 W1
==>
μ'= (1/2 W1 (1-μ) - Woμ) / (1/2 W1 (1+μ) + Wo)
 
  • #10
You are assuming that all frictions are at their maxima. Can you justify that?
aerograce said:
frictional force between the right hand ball and the block: f2=μ'N
You're told that's frictionless.
For the bigger ball:
N2=(sqr(2)/2)W1
f1 acts also from the r-hand ball on the large ball, and it has a vertical component.
For the right hand ball:
Horizontal equilibrium:

μN2'Sin45°+No'+f2=N2'Sin45°
There is friction between the r-hand ball and the ground, μ'. That is what you're trying to find.
 
  • #11
haruspex said:
You are assuming that all frictions are at their maxima. Can you justify that?

You're told that's frictionless.

f1 acts also from the r-hand ball on the large ball, and it has a vertical component.

There is friction between the r-hand ball and the ground, μ'. That is what you're trying to find.

For the first problem, I typed wrongly, it is friction between the ball and the floor.
And yes, I forgot to account frictional force on the bigger ball
For why I assume all frictions to be at its maxima, it is becoz I think once a part of this system is impending motion, any other parts of this system will be at impending motion since they are geometrically connected with each other.
 
  • #12
aerograce said:
For why I assume all frictions to be at its maxima, it is becoz I think once a part of this system is impending motion, any other parts of this system will be at impending motion since they are geometrically connected with each other.
I'm sure I could construct a system where everything is geometrically connected but the frictional forces don't all have to reach a maximum before something gives.
Suppose μ' is at its minimum value. Reducing it a tiny bit means something will move.
Can something move if friction between block and ground is still below max?
Can something move if friction between top ball and r-hand ball is still below max?
 
  • #13
haruspex said:
I'm sure I could construct a system where everything is geometrically connected but the frictional forces don't all have to reach a maximum before something gives.
Suppose μ' is at its minimum value. Reducing it a tiny bit means something will move.
Can something move if friction between block and ground is still below max?
Can something move if friction between top ball and r-hand ball is still below max?

Um YEAH you are right. So I need to assume one friction is at maxima, and consider two cases where the frictional force between the bigger ball and the right hand ball is maxima and the frictional force between the block and the ground is at maxima?
 
  • #14
You can "get rid" of the problem of not knowing the coefficient of friction between the floor and the ball if you take moments about this contact point.
 
  • #15
aerograce said:
Um YEAH you are right. So I need to assume one friction is at maxima, and consider two cases where the frictional force between the bigger ball and the right hand ball is maxima and the frictional force between the block and the ground is at maxima?
The answers to my two questions are different. If the block doesn't move, nothing else can, so we can assume friction there is at max. When the block slips, there are two possibilities for what else does. Note, however, that it does say "Make assumptions where necessary." Since it is asking for a μ' that prevents slipping, you might need to make an assumption there. E.g. if μ is too low, could it be that no matter how large μ' is slipping will occur?
 
  • #16
Basic_Physics said:
You can "get rid" of the problem of not knowing the coefficient of friction between the floor and the ball if you take moments about this contact point.
Since it's the friction there needs to be assessed, that would seem to be an unhelpful move.
 
  • #17
True. Then maybe about the contact point between the two balls?
 

FAQ: How Does Static Friction Maintain Equilibrium in a System of Balls?

What is static friction and how does it differ from kinetic friction?

Static friction is the force that prevents two stationary surfaces from sliding past each other when a force is applied. Kinetic friction, on the other hand, is the force that opposes the motion of two surfaces that are already in motion.

What factors affect the magnitude of static friction?

The magnitude of static friction depends on the normal force between the two surfaces, the coefficient of static friction, and the roughness of the surfaces.

How is the coefficient of static friction determined?

The coefficient of static friction is determined by conducting experiments to measure the force required to initiate motion between two surfaces, divided by the normal force applied.

What is the maximum possible static friction force?

The maximum possible static friction force is equal to the coefficient of static friction multiplied by the normal force between two surfaces.

How can static friction be reduced?

Static friction can be reduced by using lubricants between two surfaces, increasing the surface area in contact, or decreasing the coefficient of static friction by using different materials.

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