
#1
Jun1912, 01:18 PM

P: 3,628

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 60^{o} 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(60^{o}) 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! 



#2
Jun1912, 01:37 PM

P: 836

Hi Pranav!
Recheck your solving, [tex]0 = 2x\sqrt{3} + 2\sqrt{3} x \frac{dy}{dt}[/tex] 



#3
Jun1912, 01:44 PM

P: 3,628

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



#4
Jun1912, 01:47 PM

P: 836

Newton Laws of motion question 



#6
Jun1912, 01:51 PM

P: 836





#7
Jun1912, 01:54 PM

P: 3,628

I shouldn't be studying Physics at midnight. 



#8
Jun1912, 01:58 PM

P: 836




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