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Newton Laws of motion question

by Pranav-Arora
Tags: laws, motion, newton
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Pranav-Arora
#1
Jun19-12, 01:18 PM
P: 3,807
1. The problem statement, all variables and given/known data
A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity √3 m/s towards right. The velocity of end B when rod makes an angle of 60o with the ground.




2. Relevant equations



3. The attempt at a solution
I assumed that at any instant the distance of A from block is x and distance of B from ground is y. The length of rod is l.
l^2=x^2+y^2
Differentiating with respect to time.
0=2x(dx/dt)+2y(dy/dt)
y=xtan(60o)
dx/dt=√3 m/s
Solving, i get

(dy/dt)=-3 m/s
But that's not correct, the answer says its 2 m/s. I don't seem to get where i am wrong.
Any help is appreciated.

Thanks!
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Infinitum
#2
Jun19-12, 01:37 PM
P: 854
Hi Pranav!

Quote Quote by Pranav-Arora View Post
y=xtan(60o)
dx/dt=√3 m/s
Solving, i get

(dy/dt)=-3 m/s
Are you sure about that step?

Recheck your solving,

[tex]0 = 2x\sqrt{3} + 2\sqrt{3} x \frac{dy}{dt}[/tex]
Pranav-Arora
#3
Jun19-12, 01:44 PM
P: 3,807
Quote Quote by Infinitum View Post
Hi Pranav!

Are you sure about that step?

Recheck your solving,

[tex]0 = 2x\sqrt{3} + 2\sqrt 3 x \frac{dy}{dt}[/tex]
Oh yes, sorry, my mistake, i wrote tan60=1/√3.

But now i get (dy/dt)=-1.

Infinitum
#4
Jun19-12, 01:47 PM
P: 854
Newton Laws of motion question

Quote Quote by Pranav-Arora View Post
Oh yes, sorry, my mistake, i wrote tan60=1/√3.

But now i get (dy/dt)=-1.
Yes. That's correct. So, whats the velocity of the end B of the rod now?
Pranav-Arora
#5
Jun19-12, 01:49 PM
P: 3,807
Quote Quote by Infinitum View Post
Yes. That's correct. So, whats the velocity of the end B of the rod now?
Isn't (dy/dt) the velocity of the end B?
Infinitum
#6
Jun19-12, 01:51 PM
P: 854
Quote Quote by Pranav-Arora View Post
Isn't (dy/dt) the velocity of the end B?
That is the vertical component of B's velocity
Pranav-Arora
#7
Jun19-12, 01:54 PM
P: 3,807
Quote Quote by Infinitum View Post
That is the vertical component of B's velocity
Ah i get it, thanks for the help Infinitum!
I shouldn't be studying Physics at midnight.
Infinitum
#8
Jun19-12, 01:58 PM
P: 854
Quote Quote by Pranav-Arora View Post
Ah i get it, thanks for the help Infinitum!

I shouldn't be studying Physics at midnight.
Good night, then
Pranav-Arora
#9
Jun19-12, 01:59 PM
P: 3,807
Good night, then
Hehe, no, i am posting one more question and then i will go to sleep. :D


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