# Kinematics (maybe)

1. Jan 25, 2010

### songoku

1. The problem statement, all variables and given/known data
A 10 metre ladder rests against a wall at $$\frac{\pi}{4}$$.If the ladder slips so that when its bottom moving at 0.02 m/s, how fast would the ladder be moving down the wall?

a. 0.02 m/s
b. 0.0025 m/s
c. 0.015 m/s
d. 0.12 m/s
e. 0.15 m/s

2. Relevant equations
not sure

3. The attempt at a solution
My answer is 0.02 m/s because the speed when the ladder falls down will be the same as when its bottom moves. Actually, I think the speed will be independent of the angle and at all condition, the speed will be the same. Am I right?

Thanks

Last edited: Jan 25, 2010
2. Jan 25, 2010

### ZeroNowhere

I assume you mean pi/4 radians, that is, 45 degrees ((pi/4)*(180/pi)), given the lack of units. In which case, it shouldn't be too complicated to draw up a diagram, containing the ladder both before and after motion (after 1 second, so with the bottom having moved 0.02m in the second ladder), in order to confirm the answer. Given the length of the ladder, you can calculate the height of it on the wall as well as the distance of the bottom from the wall, and then it should be fairly easy to derive one of the lengths on the second ladder (you already have the ladder's length), and then figure out the other through Pythagoras' Theorem.

As for whether it would be independent of the angle, if it happens to be with 45 degrees, repeat the calculations with a sufficiently different angle, like 70 degrees, between the ladder and wall, and find out if they remain the same.

3. Jan 25, 2010

### songoku

Hi ZeroNowhere

I've tried it and the answer for this question is 0.02 m/s. I tried it for 30 degrees and it came out different (0.0348 m/s). The question is I don't get the logic behind it because I think when the ladder moves 0.02 m to, say, left, then it also should move 0.02 m downward.

I also tried to find the velocity using trigonometry (tan) when the angle 30 degrees. Vx = 0.02 m/s, then Vy = Vx tan (30) = 0.0115 m/s, differs from when I used your method (find the vertical distance traveled by the ladder, then find the velocity). Why can't I use tan ?

Basically, I confuse why the speed depends to the angle and why tan doesn't work.

Thanks

4. Jan 26, 2010

### tiny-tim

Welcome to PF!

Hi ZeroNowhere! Welcome to PF!
songoku, why are you using angles?

Just follow ZeroNowhere's suggestion … use lengths and Pythagoras.
… you'll get y as a function of x, and then you can find how dy/dt depends on dx/dt.

5. Jan 27, 2010

### songoku

Re: Welcome to PF!

Hi tiny-tim
It just came across my mind. In parabolic motion, we can find the angle of projection using tan (angle) = (Vy)/(Vx). So I tried it here. Why doesn't it work?

This is what I've tried :
Let :
θ = angle between ladder and floor
a = horizontal distance between the bottom of the ladder and the wall = c cos θ
b = height of top of the ladder from wall = c sin θ

After the bottom moves x meters (further from the wall ) and the top falls y metres, we have:

c2=(a+x)2+(b-y)2

c2=(c cos θ+x)2+(c sin θ-y)2

c2=c2cos2θ+2cx cos θ+x2+c2sin2θ-2cy sin θ+y2

c2=c2+2cx cos θ+x2-2cy sin θ+y2

2cx cos θ+x2=2cy sin θ-y2

differentiate with respect to t

2c cos θ dx/dt + 2x dx/dt = 2c sin θ dy/dt - 2y dy/dt

Is this what you mean?

Thanks

6. Jan 30, 2010

### songoku

Sorry to bump up but can anyone help please?

Thanks

7. Jan 30, 2010

### tiny-tim

No, θ depends on t also, so you've left out all the terms with dθ/dt.

Can't you see that by introducing θ, you've only made it more complicated? … you're only interested in dx/dt and dy/dt, so what is the point of dragging θ into it … you'll only have to eliminate θ later.

Do what ZeroNowhere and I suggested … forget the angle, and just use Pythagoras to find y as a function of x, and then differentiate.

8. Jan 30, 2010

### songoku

Hi tiny-tim

Ok, I tried a new one :

c2 = (a+x)2 + (b-y)2

c2 = a2 + 2ax + x2 + b2 - 2by + y2

Differentiate with respect to t (a, b, and c are constants) :

0 = 0 + 2a dx/dt + 2x dx/dt + 0 - 2b dy/dt + 2y dy/dt

Is this what you mean?

9. Jan 30, 2010

### rl.bhat

In the problem it stated that the ladder is at π/4 in the beginning. Now the question is whether the ladder starts moving from that position or already in motion? To find dy/dt, you must know where dx/dt = 0.2 m/s. If the floor is frictionless, then only in the vertical position the ladder will not slip. Otherwise it will start slipping and KE of the center of mass of the ladder will increase. Consequently dx/dt will increase.

10. Jan 30, 2010

### songoku

Hi rl.bhat
I think it starts moving from that position. The ladder is at rest, then slips.

I don't get what "where dx/dt = 0.02 m/s" means. If dx/dt = 0.02 m/s, then the the bottom will move 0.02 m in 1 second. Is that your point?

I also don't get this. The question doesn't say about friction so I think it's frictionless and if the floor is frictionless, why the vertical position doesn't slip?

Thanks

11. Jan 31, 2010

### rl.bhat

If the ladder stars slipping from rest, it must accelerate to reach 0.02 m/s. At what distance from the starting point, the velocity of the foot of the ladder will be 0.02 m/s?
Now let us analyze the problem. Let us search for the origin of the acceleration.
Find the component of g along the ladder. It is g*sinθ.
Find its component along floor. It is g*sinθ*cosθ. Or acceleration a = g/2*sin2θ.
Now you can the purpose of setting the ladder at π/4. At that angle acceleration of the foot of the ladder is maximum and that is equal to g/2. After this point acceleration decrease with increase in x.
Now v^2 = 2*a*x. = 2*g*sinθ*cosθ*x = 2*g*(1 - cos^2θ)^1/2*cosθ*x
= 2*g*[1-{(x+a)/r}^2]^1/2*(x+a)/r*x
v is given and r is known. Find x. Distance of the foot of the ladder from the wall is
X = xo + x, where xo = r*sin(π/4). Then find Y and find dy/dt using the equation given by you in #8.

Last edited: Jan 31, 2010
12. Jan 31, 2010

### tiny-tim

Hi songoku!

(just got up :zzz: …)
Yup!

And so dy/dt = [(b - y)/(a + x)]dx/dt.

13. Jan 31, 2010

### songoku

Hi tiny-tim and rl.bhat

Now I want to ask the logic behind it. From the equation derived by rl.bhat or by myself (with tiny-tim and zeroNowhere advice, of course ) it can be seen that dy/dt is not always the same as dx/dt. I can't understand why because if I think about it, it seems that the top will fall down with the same speed as the bottom. I imagine a real ladder, if the bottom moves 0.5 m, I think the top will also move 0.5 m in the same time. What will cause the different in the distance traveled by the top and the bottom?

Thanks

14. Jan 31, 2010

### rl.bhat

OK.
Now initially x^2 + y^2 = r^2.
According to your suggestion, after the bottom moves through 0.5 m and top also move through 0.5 m, then
(x+0.5)^2 + (y-0.5)^2 must be equal to r^2.
Is it true? Check with any numerical value.

15. Feb 1, 2010

### songoku

Hi rl.bhat

I get it now. Thanks !! :

16. Feb 1, 2010

### rl.bhat

The problem is not phrased properly. It should be
A 10 metre ladder rests against a wall. If the ladder slips so that its bottom is moving at 0.02 m/s when it makes an angle π/4 with the floor, how fast would the ladder be moving down the wall?

17. Feb 1, 2010

### songoku

Hi rl.bat
Yes you're right, so we don't consider the acceleration. Thanks :)