Dynamics; I don't get, nor see the wrong assumption.

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SUMMARY

The discussion focuses on the dynamics of a 2-kg collar C sliding along a smooth shaft AB under various conditions. Specifically, it addresses the acceleration of collar C when collar A moves downward at a constant velocity and when it is subjected to a downward acceleration of 2 m/s². The key takeaway is that the correct expression for the component of acceleration parallel to the shaft must account for both gravitational forces and the motion of collar A. The participant identifies errors in calculating the y component of acceleration and emphasizes the importance of understanding the reference frame for accurate results.

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Pascal1p
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Homework Statement


The 2-kg collar C is free to slide along the smooth shaft AB. Determine the acceleration of collar C if (a) the shaft is fixed from moving. (b) Collar A, which is fixed to shaft AB, moves downward at constant velocity along the vertical rod, and (c) collar A is subjected to a downward acceleration of 2 m/s^2. In all cases, the collar moves in the plane.
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My question is about the case c.

Homework Equations


∑F=m*a


T3. The Attempt at a Solution
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The answer model also says that a(along shaft) does not change, so why doesn't this method work?
 
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You do not have the correct expression for the component of acceleration of the collar that is parallel to the shaft. Your expression would be correct if the collar was stuck to the shaft so that it could not slide along the shaft.

EDIT: Actually, now that I look at it again, I think this part is correct. You correctly found the component of acceleration of the collar that is along 45 degrees below the horizontal. There will also be a component perpendicular to this.

I believe your error is when you tried to find the acceleration component in y direction.
 
Last edited:
TSny said:
View attachment 103081

You do not have the correct expression for the component of acceleration of the collar that is parallel to the shaft. Your expression would be correct if the collar was stuck to the shaft so that it could not slide along the shaft.

But even the solution model says that the value of this m*a equals 2*9.81*sin(45). They use: m*a(along shaft)= 2*9.81*sin45 = 2(2cos45 + a[c relative to ab]).
Even if it not stuck to the shaft, the only force that can make it slide along shaft is the gravitational force, which won't change direction nor magnitude during the motion. The normall force is perpendicular to the motion down the shaft, so it is this 2*9.81*sin45 that is making it accelerate down the shaft.
 
Yes, I was wrong. See the edit to my first post. I don't think you got the y component of acceleration correct.
 
It's important to keep in mind that the acceleration of 6.94 m/s2 "along the shaft" that you obtained is the component of acceleration at 45 degrees below the horizontal as measured in the Earth frame of reference. It is not the acceleration of the collar relative to the shaft frame of reference.
 
upload_2016-7-11_17-8-49.png
 

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