# Ratio of distance before bar slides

1. Apr 17, 2012

### songoku

1. The problem statement, all variables and given/known data
A uniform square bar lies horizontally across two supports, A and B. B is fixed to the floor, and A is moved toward center O of the bar with uniform speed. Initially, A slides along the underside of the bar, while the bar remains at rest with respect to B. When A reaches certain point , however, the bar begins to slide across the pivot B. The coefficient of static friction between the bar and each of A and B is 0.50 and the coefficient of kinetic friction if 0.20

The distances a and b shown in the figure above. What is the value of b/a at the moment the bar begins to slide across the top of B?

a. 0.40
b. 0.50
c. 1.0
d. 2.0
e. 2.5

2. Relevant equations
Newton's Law
Torque

3. The attempt at a solution
There are 4 forces acting on the bar; weight, normal force of A and B and kinetic friction of bar and A.

What does it mean by the bar slides across the pivot B? Is it translational motion or actually rotational motion (the bar topple)?

If it is rotational motion, then when the bar topples, the normal force of B equals zero and the bar will move counter-clockwise. Taking the pivot at A, the only force that produces torque is the weight of the bar and the direction is clockwise, as opposes the direction of the bar topples???

If it is translational motion, there must be force ≥ static friction at point B. What is the force?

Thanks

2. Apr 18, 2012

### I like Serena

The first 3 forces you mention are vertical forces that cancel each other out (as long as the bar is at rest).
The kinetic friction of the bar at A is a horizontal force.
Since the bar remains at rest initialially, there must be a 5th force that cancels this horizontal force.
Which force is that?

Btw, in your problem statement it says: "What is the value of b/a at the moment the bar begins to slide across the top of B?"
In other words, the bar slides, and does not rotate.

Last edited: Apr 18, 2012
3. Apr 18, 2012

### songoku

Let say there is external force acting on A to move it with constant speed. The external force F is directed to the right and has magnitude equals to kinetic friction.

F = μk.N . Is this also the force that will make the bar slides from B?

After A reaches certain point which makes the rod slides from B, the resultant of torque of the bar is still zero.

NA. a = NB. b
b/a = NA / NB ; N = friction / μ
b/a = (fAk) / (fBs)

Is fA = fB ?

Thanks

4. Apr 18, 2012

### I like Serena

Yes.

However, if we only look at the forces acting on the bar, there must be another force to counteract the friction at A.
Which force is that?

Ah, I didn't notice you slipping in a completely new force $f_B$ in here.

You tell me.

In general, if an object (in this case the bar) is or remains at rest, then the following 3 equations must hold:
ƩFH=0
ƩFV=0
ƩM=0

5. Apr 18, 2012

### songoku

I don't think so
The rod will experience kinetic friction directed to the right so there must be force equals to the friction directed to the left. I call this fA It will make the rod tends to move to the left with respect to B.

And there will be second horizontal force directed to the left at point B to counter the static friction. I call this fB

So the force that you are asking me probably this fA and fB

I don't know. There are total 4 horizontal forces acting on the bar; kinetic and static friction, and also forces to balance them. There is possibility like this, kinetic friction = 5 N and static friction = 7 N. The frictions are not the same but the rod will still in equilibrium since there will be 5 N and 7 N forces cancel both frictions.

My intuition says that fA and fB are the same but my mind can't grasp it....

6. Apr 18, 2012

### I like Serena

Support A acts on the bar through a (kinetic) frictional force to the right.
As a result the bar will want to slide to the right.

At the same time, the bar pushes back against support A, also through a frictional force that is equal and opposite (Newton's 3rd law).

Since we only want to look at the forces acting on the bar, we just look at the kinetic frictional force $f_A$ that acts on the bar and that points to the right.

This force must be canceled by a static frictional force at point B that therefore points to the left.
I'm sort of assuming that you named this force $f_B$, which therefore must be equal and opposite to $f_A$ to keep the bar in equilibrium.

7. Apr 18, 2012

### PeterO

Do this experiment yourself.

Get a long ruler [standard wooden metre rule from the lab perhaps] and support in on your two index fingers - one near an end and the other about 1/3 the way from the other end.

Now move you fingers/hands slowly together and notice which finger the ruler slides over and which one "sticks" to the ruler. Keep going until your fingers are together and see if it balances.

(For fun you could load up the end that your one finger is under, with say 100gm mass or a packet of lollies, and repeat the exercise and see when the ruler topples.)

Once you have done that you should be able to answer this question easily.

8. Apr 18, 2012

### songoku

Now I see where my mistake is.

Maybe I'll do it later

Thanks a lot for the help