A Leaning Plank. Find where it looses contact.

[warjunkie]

Homework Statement

A uniform plank of length L leans against a vertical wall, its head a distance h above the horizontal ground. If we let go of the plank from rest, how far above the ground is the head of the plank the moment it loses contact with the wall?

Homework Equations

Fx=0
Fy=mg

The Attempt at a Solution

I'm not sure but wouldn't the head of the plank never lose contact with the wall,
since there is no x-direction forces.

In my knowledge, I only see y- direction force, which is mg (Mass of the plank and gravity), which causes the head of the plank to just fall all the way without losing contact with the wall.

Am I correct?

 

[tiny-tim]

hi warjunkie!

A uniform plank of length L leans against a vertical wall, its head a distance h above the horizontal ground. If we let go of the plank from rest, how far above the ground is the head of the plank the moment it loses contact with the wall?
…
I'm not sure but wouldn't the head of the plank never lose contact with the wall,
since there is no x-direction forces.

In my knowledge, I only see y- direction force, which is mg (Mass of the plank and gravity), which causes the head of the plank to just fall all the way without losing contact with the wall.

(is there friction? )

there are three forces on the plank, not one …

the weight, and the reaction force from the wall and from the ground …

start again​

 

[ashishsinghal]

Do you know what the answer is. If that is so please post it. That we can do our individual calculations and then be sure and help you.

 

[warjunkie]

hi warjunkie! (is there friction? )

there are three forces on the plank, not one …

the weight, and the reaction force from the wall and from the ground …

start again​

Alrighty, thanks for the post. I'll try it again!
And i don't think friction exists in this problem...

--------------------------------------------------------------------

mg -Ng = m*ay
Nw=m*ax

weight=mg= mass x gravity
Ng= normal force from the ground
Nw=normal force from the wall
ay= y-acceleration
ax= x-acceleration

ay = ax? since the plank is one object and if the head slips down, the bottom also slips with same distance/speed.

These are my new equations... but I still don't see why the plank will lose contact with the wall...

wouldn't the y-distance lost by sliding down will always be compensated by the increase in x-distance?

 

[warjunkie]

Do you know what the answer is. If that is so please post it. That we can do our individual calculations and then be sure and help you.

Unfortunately, there's no answer :(

It's an extra credit problem that was posted by my professor.

 

[tiny-tim]

ay = ax? since the plank is one object and if the head slips down, the bottom also slips with same distance/speed.

noooo

(use Pythagoras or trig)

These are my new equations... but I still don't see why the plank will lose contact with the wall...

it will if a_x is zero …

what is the equation for a_x ?

 

[ashishsinghal]

Unfortunately, this thread had stopped to continue before the answer was achieved.Though I am not the OP I would like to resume it. Please help me to get the answer.

 

[tiny-tim]

… Though I am not the OP I would like to resume it. Please help me to get the answer.

hi ashishsinghal!

show us what you've done, and where you're stuck, and then we'll know how to help!

 

[ashishsinghal]

First of all I am not able to understand why the rod will lose contact with the wall.

 

[ashishsinghal]

Let the height of ladder be y and its horizontal distance be x
then
x² + y² = const
x*dx/dt + y*dy/dt=0
x*dx/dt = y*v_y
here v_y is the vertical velocity of the rod and v_y = -dy/dt
hence
v_x=tan$$\theta$$v_y
then,
a_x=tan$$\theta$$a_y

for center of mass : a_com,x = a_x/2
and similarly for y

we know Ma_com,x = N_w(normal from wall)
also Ma_com,y = Mg - N_g

But this cannot give me time. I know that N_w is zero at the required time. I am stuck!

 

[tiny-tim]

conservation of energy?

 

[ashishsinghal]

But it would not give me time, also is my attempt correct?

 

[ehild]

Theta changes during the motion, you can not ignore its time derivatives. Express x and y with theta and write the equations for acceleration of CM and the rotation around the CM in terms of theta and its time derivatives. The time is not needed to find the value of theta where the normal force from the wall is zero.

ehild

 

[ashishsinghal]

Oh, my mistake I thought the question is "Find the time when it loses contact"

 

[ashishsinghal]

Can anyone please first tell me that why will the plank lose contact? If it loses contact a_x =0 then either tan$$\theta$$=0 or a_y=0 (using equations from my earlier post). If tan$$\theta$$ is zero then it lies horizontal. If a_y is zero then N_g = Mg. Also for rotational equilibrium $$\tau$$_Mg = $$\tau$$_N. Hence plank needs to be horizontal.

 

[ehild]

Revise your equations.

x=Lsinθ,

v_x= dx/dt=Lcosθ dθ/dt,

a_x=dv_x/dt=

= -Lsinθ(dθ/dt)²+Lcosθ d²θ/dθ²

and so on...

You can also write up the Lagrangian in terms of theta and get the relationship between theta and derivatives from the Euler equation.

ehild

 

[ashishsinghal]

You can write up the Lagrangian in terms of theta.


ehild

What is Lagrangian? Please first tell me why will the plank lose contact.

 

[ehild]

For Euler-Lagrange equations, see http://www.physics.thetangentbundle.net/wiki/Classical_mechanics/Euler-Lagrange_equations

If you are not familiar with it, write up the equation of motions for the CM and rotation about the CM.
I do not know when the plank loses contact. That is this problem for.

ehild

 

[ehild]

Conservation of energy provides also a very easy solution. Write up the total mechanical energy in therms of theta, find the angular speed and acceleration, from that you can get ax, and solve for ax=0.

ehild

 

[ashishsinghal]

v_x=tan$$\theta$$v_y

for center of mass : v_com,x = v_x/2
and similarly for y

So in the frame of com the plank is doing rotational motion with angular velocity = sqrt(v_x²/4 + v_y²/4)/(L/2)
=sqrt(v_y²tan²$$\theta$$/4 + v_y²/4)/(L/2)
=v_ysec$$\theta$$/L

From this I can find energy. Is all this correct?

Now Rotational KE = 1/2*M[STRIKE]L[/STRIKE]²/12 * (v_ysec$$\theta$$/[STRIKE]L[/STRIKE])²

Translational KE = 1/2*M(v_ysec$$\theta$$)²

Potential energy = Mg*Lsin$$\theta$$/2

I think I have calculated everything required. But I have not received any confirmation if the calculations are correct or not. Please do so.
Also I am mainly having a conceptual problem over here. I am not able to understand why it will lose contact. I am not able to visualize it. Please help me do the same. I would be highly thankful.

 

[ehild]

You made quite a few mistakes :)

v_x/v_y = - tan(θ).

The translational KE is

1/2 *m ( (v_x/2)²+(v_y/2)²)

Potentional energy = mgL/2 cos(θ)

The question if the plank loses contact is answered by the solution of the problem. I can not explain it in words. The answer is a derivation which results in a distance x or angle theta when the force from the wall is zero, and would be negative after that instant if the same equations hold. But the force can not be negative, you have a different problem from that instant. It is just a plank, touching the ground in one point , traveling away from the wall and falling at the same time so the edge on the ground has higher horizontal speed than the other one.

Anyway, a little experiment can help. I tried it with a ruler and a book as wall, and it lost contact , in spite of friction.

ehild

 

[tiny-tim]

Also I am mainly having a conceptual problem over here. I am not able to understand why it will lose contact. I am not able to visualize it.

when it gets verrrrry close to the ground, if it was still in contact with the wall, its velocity would be very nearly vertical …

ie its horizontal velocity would be very small …

so at some point the horizontal velocity has to start decreasing, which is not possible since there is no force towards the wall

 

[ehild]

Tiny-Tim, I like your explanation so verrrrrrrrrrry much :)ehild

 

[SammyS]

Tiny-Tim, I like your explanation so verrrrrrrrrrry much :)
ehild

I agree with echild.    The fish RULES !

 

[ashishsinghal]

So should I use conservation of energy to find θ for which v_x is max

 

[ehild]

Yes. I would say ax=0, but it is at the same time when the velocity of the bottom point of the plank reaches its maximum value.

ehild

 

[zorro]

I agree with echild.    The fish RULES !

:rofl:
When I first joined PF, I read ehild as 'echild' everytime.

 

[ehild]

:rofl:
When I first joined PF, I read ehild as 'echild' everytime.

Never mind. Don't you want to show a solution? If I do I will get a serious warning. But you can present yours and ask if it is correct. It would be a pity to left this very nice problem unsolved.

ehild

 

[zorro]

I already solved the question b4 posting
I used energy-conservation, Lagrange's equations and a bit of math to get the answer as h/3.

 

[zorro]

I tried using the simple torque approach in place of Langrangian's method but I got a sign change.

Let the length of the plank be 2l (for simplicity) and let it make an angle θ with the horizontal when normal force by the vertical wall is 0.

1. Energy Conservation-

ω² = 3g/2l (sinΦ - sinθ), where Φ is the initial angle of the base with horizontal.

2.

Angular acceleration of the rod when N=0 = α = -ω²cotθ

a_{COM along vertical direction} = -lω²sinθ + lαcosθ

3. Torque

N₂cosθ = mlα/3, where N₂ is the normal force by the ground
(Takin the extreme case when N=0)

Also,
N₂ = m(g-a_{COM along vertical direction})

Substituting and simplifying, I got ω² = -3gsinθ/4l

whereas ω² = +3gsinθ/4l using Lagrange's method.

I have an intuition that there is some problem with the torque method.

 

[ehild]

Let the length of the plank be 2l (for simplicity) and let it make an angle θ with the horizontal when normal force by the vertical wall is 0.

1. Energy Conservation-

ω² = 3g/2l (sinΦ - sinθ), where Φ is the initial angle of the base with horizontal.

Correct so far, but I but I can not follow you after. I write the details to get the angular acceleration

The angular acceleration is obtained by derivation of

ω²=3g/(2L) (sinΦ - sinθ). *

2ω dω/dt=-3g/(2L)(cosθ)ω --->dω/dt=-3g/(4L)cosθ. **

N=0 means that the horizontal acceleration of the centre of mass is zero. The x coordinate of the CM is x_cm=L(cosθ), v_cm=-Lω(sinθ) and

a_cm=-L*(ω²cosθ +dω/dt sinθ ).***

a_cm=0 means that dω/dt =-ω²cotθ, the same you got.

Substituting (*) for ω² and (**) for dω/dt in (***) for the horizontal acceleration, you get the angle when the contact is lost.

I do not understand the signs in your torque method. Show what you consider positive directions. Also, dω/dt is missing from your equation for N2. And you got an expression (*) for ω² already, what are these different new ones?

ehild

 

[zorro]

Substituting (*) for ω² and (**) for dω/dt in (***) for the horizontal acceleration, you get the angle when the contact is lost.

No need of any Langrangian or torque equations?

I do not understand the signs in your torque method. Show what you consider positive directions. Also, dω/dt is missing from your equation for N2. And you got an expression (*) for ω² already, what are these different new ones?

ehild

I took upward direction positive. dw/dt = α in my equations.
I solved for α from the torque equation and Newton's law to find one more expression for α and hence ω.

btw I got the height as 2h/3 not h/3

 

[ehild]

No need of any Langrangian or torque equations?

No need any of them, just get the accelerations dω/dt and a_x by derivation with respect to time.

I took upward direction positive. dw/dt = α in my equations.
I solved for α from the torque equation and Newton's law to find one more expression for α and hence ω.

Check the signs in your equations. But the torque method is quite complicated.

btw I got the height as 2h/3 not h/3

2h/3 is correct. Great work!

First I wrote up the equations of motion for the translation of the CM and for the rotation around it. It was quite complicated. Then I used the Lagrangian to find the equation of motion in terms of the angle. But I had to solve a differential equation to get the angular speed. The easiest way was using conservation of energy.

ehild

 

[ashishsinghal]

Thanks to all for this wonderful effort. I am really thankful.Never could have possibly got to that without your help.
Special thanks to ehild.