What is the Magnitude of Acceleration for Point B in a Moving Piston System?

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SUMMARY

The discussion focuses on calculating the magnitude of acceleration for point B in a moving piston system, where the piston moves at a constant velocity of 2.2 m/s, with distances p = 1 m and q = 0.1 m, and an angle θ of 52 degrees. The participant derived the equations for velocity and acceleration, specifically Vb/a = (V/sin(θ)) and Ab = ([(V/sin(θ))^2]/p)/sin(θ). However, there is a discrepancy between their calculated acceleration of 9.9 m/s² and the expected online answer of approximately 13 m/s², leading to confusion regarding the correct approach.

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Clancy
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Homework Statement



The piston is moving up at constant velocity 2.2m/s, the distance p is 1m, the distance q is 0.1m, the angle ɵ is 52 degrees.
Find the magnitude of the acceleration of B



Homework Equations



Vb = Va + Vb/a
Ab = Aa + Ab/a



The Attempt at a Solution



In general terms I found Vb/a to be (V/sin(ɵ))
The normal component of the acceleration of Ab/a is then [(V/sin(ɵ))^2]/p
I then got a general solution of Ab = ([(V/sin(ɵ))^2]/p)/sin(ɵ)

I don't think this is right however as this is an online homework, my solution gives the acceleration of B as 9.9 m/ss where the online answer is rounded to 13m/ss.

Any help with this would be great, the actual answer won't be 13 as the lecturer rounded it a lot I think. If anyone could just try this and see if they get close to 13 I'd very much appreciate it.
 

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Hi Clancy! :smile:
Clancy said:
In general terms I found Vb/a to be (V/sin(ɵ))
The normal component of the acceleration of Ab/a is then [(V/sin(ɵ))^2]/p
I then got a general solution of Ab = ([(V/sin(ɵ))^2]/p)/sin(ɵ)

Sorry, I don't understand this. :redface:

Start by finding the lengths CA and CB in terms of p and θ. :wink:
 

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