# Newton Laws of motion question

by Pranav-Arora
Tags: laws, motion, newton
 P: 3,616 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!
P: 836
Hi Pranav!

 Quote by Pranav-Arora y=xtan(60o) dx/dt=√3 m/s Solving, i get (dy/dt)=-3 m/s
Are you sure about that step?

$$0 = 2x\sqrt{3} + 2\sqrt{3} x \frac{dy}{dt}$$
P: 3,616
 Quote by Infinitum Hi Pranav! Are you sure about that step? Recheck your solving, $$0 = 2x\sqrt{3} + 2\sqrt 3 x \frac{dy}{dt}$$
Oh yes, sorry, my mistake, i wrote tan60=1/√3.

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

P: 836

## Newton Laws of motion question

 Quote by Pranav-Arora 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?
P: 3,616
 Quote by Infinitum 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?
P: 836
 Quote by Pranav-Arora Isn't (dy/dt) the velocity of the end B?
That is the vertical component of B's velocity
P: 3,616
 Quote by Infinitum 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.
P: 836
 Quote by Pranav-Arora Ah i get it, thanks for the help Infinitum!

 I shouldn't be studying Physics at midnight.
Good night, then
P: 3,616
 Good night, then
Hehe, no, i am posting one more question and then i will go to sleep. :D

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