Finding Velocity of Connected Objects on Perpendicular Guide Rails

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The problem involves two connected objects A and B sliding along perpendicular guide rails, with A moving at a constant speed of 35.2379 m/s along the x-axis. To find the velocity of B along the y-axis when the rod makes a 27-degree angle with the x-axis, the relationship x^2 + y^2 = 35^2 must be used. The correct approach involves differentiating this equation with respect to time and applying the known velocity of A. The confusion arises from the initial attempt to use sine without considering the full geometric relationship between the variables. Understanding the correct application of trigonometric relationships and differentiation is crucial for solving the problem.
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Homework Statement



Two objects A and B are connected by a rigid rod that has a length 35 m. The objects slide along perpendicular guide rails. If A slides to the left with a constant speed 35.2379 m/s along the x-axis, find the velocity of B along the y-axis when the rod makes an angle 27* with the x-axis. Answer in units of m/s

Homework Equations



I'm not even sure.

The Attempt at a Solution



I attempted to find the velocity using the equation:

\vec{Ay}=\vec{A}sin\Theta

so,

\vec{Ay}=35sin(27*)

\vec{Ay}= 15.890 m/s

This, however, was not correct and I really am confused about what even would be correct or how I should go about solving the problem.

Thank you in advance for any help or information (or direction...).
 
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Since x^2+y^2=35, then \frac{y}{x}=tan(\psi). Then you can differentiate both sides of that by the time t. You know the required angle, and you know \frac{dx}{dt} as well.
 
bel said:
Since x^2+y^2=35, then \frac{y}{x}=tan(\psi). Then you can differentiate both sides of that by the time t. You know the required angle, and you know \frac{dx}{dt} as well.


so,

y=xtan\Theta

\frac{dy}{dt}=\frac{dx}{dt}tan27

\frac{dy}{dt}=35.2379*tan27 ?
 
No, you have to make use of the relation x^2+y^2=35 as well and treat \psi as a function of time t.
 
I think one wants - x^2+y^2=35^2
 
Oh right, it is the square of 35, silly me, how could I have forgotten? I apologise.
 
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