# Homework Help: Plank leaning against a wall

1. Dec 27, 2013

### Tanya Sharma

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

A plank of length 2L leans against a wall.It starts to slip downward without friction.At what fraction of the initial height does the top of the plank lose contact with the wall .

2. Relevant equations

3. The attempt at a solution

Let the normal force from the vertical wall on top of the plank be N1.
Let the normal force from the floor on bottom of the plank be N2.
Let x and y represent the coordinate of the CM with origin at the bottommost point of the wall.
Let θ be angle which plank makes with the floor at any instant.

The top of the plank loses contact when N1 = 0.

Now,we have

N2-Mg = Md2y/dt2 (1)

N1 = Md2x/dt2 (2)

N1Lsinθ-N2LCosθ = [M(2L)2/12]d2θ/dt2

On simplifying , N1Lsinθ-N2LCosθ = (ML2/3)d2θ/dt2 (3)

Another relation we can obtain is x=Lcosθ

Differentiating,dx/dt = -Lsinθ(dθ/dt)

Again differentiating , d2x/dt2 = -L[cosθ(dθ/dt)2+sinθd2θ/dt2] (4)

How should I proceed further ?

I would be grateful if some member could help me with the problem .

Edit :Fixed errors .Erroneously typed d2x/dt2 in place of d2y/dt2 and vica versa .

#### Attached Files:

• ###### plank.PNG
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Last edited: Dec 27, 2013
2. Dec 27, 2013

### Staff: Mentor

Tanya,

Good start. You need to express both x and y in terms of θ. (Incidentally, in the coordinate system you are using, shouldn't x = Lsinθ)? Once you have the second derivatives of x and y with respect to t expressed in terms of the time derivatives of θ, you can substitute these into eqns 1 and 2, and solve for N1 and N2. You then substitute these results into the moment balance. This gives an equation entirely in terms of θ.

Chet

3. Dec 27, 2013

### TSny

I believe x = Lcosθ is correct for the x coordinate of the cm. [Edit: Oh, I now see the confusion. Equations (1) and (2) imply x is vertical and y is horizontal, while x = Lcosθ implies x is horizontal.]

I think you can avoid the force and torque equations (1) and (3) by just using energy along with your equation (4).

Last edited: Dec 27, 2013
4. Dec 27, 2013

### Tanya Sharma

Sorry for the needless confusion in OP .I have fixed the errors .

5. Dec 27, 2013

### Tanya Sharma

For using energy should I use : Loss in PE = Gain in Kinetic energy

But then , I am not sure what is the initial position of the plank .Is it vertical or is it at an arbitrary angle, say α

6. Dec 27, 2013

### TSny

Yes.

Assume it's an arbitrary angle. It will still work out nicely.

7. Dec 27, 2013

### Tanya Sharma

Let the initial angle be α then,using COE

mgL(sinα-sinθ) = (2/3)ML2(dθ/dt)2

Is it correct ? If yes ,how to proceed further ?

8. Dec 27, 2013

### Staff: Mentor

Did the steps I outlined in my post #2 not work?

9. Dec 27, 2013

### TSny

That looks good.

What do your equations (2) and (4) tell you at the instant the plank loses contact with the vertical wall?

Last edited: Dec 27, 2013
10. Dec 27, 2013

### TSny

I apologize for redirecting Tanya towards energy concepts. I happened to find the energy approach to be a nice way to do it. But, of course, the torque and force equations should work, too.

11. Dec 27, 2013

### Tanya Sharma

Since I had erroneously interchanged d2x/dt2 with d2y/dt2 in OP ,when I first tried the approach outlined by you ,ended up with a complicated DE .But that is mistake on my part .

I will try again and post the working

12. Dec 27, 2013

### Tanya Sharma

This is what i am getting

d2θ/dt2 + (3g/2L)cotθ(sinα-sinθ) = 0

Is it correct ?

13. Dec 27, 2013

### TSny

Yes.

14. Dec 27, 2013

### Tanya Sharma

Now comes the most difficult part ,solving the DE .

This looks to be a second order DE .But the second term is confusing me.

I will have to look at some reference .Could you please help me identify the DE and the method I have to use .

15. Dec 27, 2013

### TSny

You can avoid solving a DE. Can you see a way to use your energy equation to obtain an expression for $\ddot{θ}$?

16. Dec 27, 2013

### Tanya Sharma

This is simply outstanding .

Excellent thinking TSny !!!

Nevertheless ,how can we solve the DE ?

17. Dec 27, 2013

### TSny

I don't know. I don't think $\theta$ as a function of time would be very simple.

#### Attached Files:

• ###### Planck Sliding Down Wall.png
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18. Dec 27, 2013

### Tanya Sharma

:rofl:

terrific sense of humour...

19. Dec 27, 2013

### Tanya Sharma

One thing I need to clarify is that θ is measured from the horizontal with clockwise being the direction of increasing θ .

Does that mean direction of increasing θ has to be considered positive ?
Does this reflect on the direction of torque as well with clockwise torque being considered positive and ccw negative ?

Even though I have used this I am having a little doubt .

20. Dec 27, 2013

### haruspex

I got that time is proportional to ∫cosec(θ/2).dθ, where θ is the angle to the vertical. This is with being vertical initially. That makes sense in that if it really started from vertical with no initial speed then it would take an infinite time.

21. Dec 28, 2013

### TSny

Yes. For general starting points, it looks like you might get elliptic integrals of some sort.

22. Dec 28, 2013

### ehild

Well, it is difficult to follow the thread without a a picture showing what is meant positive and negative (upward and to the right in the picture, counter-clockwise for the angle).

If you use the angle from the vertical, it is zero initially, and increases anticlockwise with time .

You have two equations for the acceleration of the CM

$$m\ddot x=N_x$$
$$m\ddot y=N_y-mg$$

and one for the angular acceleration (with respect to the CM)

$$\frac{mL^2}{3}\ddot\theta=N_y L\sin(\theta)-N_x L \cos(\theta)$$

From the geometry, x=Lsin(θ), y=Lcos(θ). Substitute the second derivatives into the first equations, then substitute the expressions of Nx and Ny into the third one. You get a very simple second order de for theta(t) *(I do not show ). You need theta when Nx=0. You certainly know the following trick, but I write it just for case...

dθ/dt=ω. Your equation * does not have t explicitly. You can consider ω as function of θ.

$$\ddot \theta= \dot \omega = \frac{d \omega}{d \theta}\frac{d \theta}{d t}= \frac{d \omega}{d \theta}\omega =0.5 \frac{d (\omega^2)}{d \theta}$$
Integrating and using that initially θ=0 and ω=0, you get ω2 as function of θ.
Now use the condition for Nx=0.

ehild

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23. Dec 28, 2013

### Tanya Sharma

This doesn't look appropriate .

suits you well .

24. Dec 28, 2013

### ehild

Have you got that equation? I would like to see it, please, to check if my one is correct.:shy:

ehild

25. Dec 28, 2013

### Tanya Sharma

Please give me some time .