In this problem, Points C and B are related by a cable. Doesnt this mean that aBx=aCx?
And this way there s no such thing as alpha(BC).
Thank you.
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6 comments:
That is correct.
I would think so too. But the thing is, proff. Krousgrill insisted on that alpha(BC) would disappear from the equations without being zero.
--to nour--
Having B and C connected by a rigid member does NOT guarantee that the x-components of acceleration of B and C are equal. They might be equal but that would be only for special situations.
I am not sure what you are saying with "there is no such thing as alpha_BC". Any rigid body moving in a plane has an angular acceleration. This angular acceleration could be equal to zero; however the angular acceleration does exist!
In general, you can solve the kinematics equations given in the solution to relate alpha_w and alpha_AB by algebraically eliminating alpha_BC.
For this particular problem, it turns out that alpha_BC is zero. However, if the system were not at rest, then you would find that alpha_BC is NOT zero. Your kinematics equations will tell you this; you do not need to know this a priori.
I am not sure that what I am saying is clear. Please let me know.
Yes it is clear, thank you.
But if having two points connected by a rigid member does not guarantee them having the same acceleration, then how come we did use that to solve sample problem 28, where we said that aBx=aO ?
Recall that the kinematics equation that relates the acceleration of two points A and B on the same rigid body is given by:
aA = aB + alpha x rA/B - omega^2*rA/B
If aA = aB, this says that the difference of the last two terms in the RHS of this equation must be zero, or: alpha x rA/B = omega^2*rA/B
When does the above relationship hold? One case is when the body is in pure TRANSLATION; that is if: omega = 0 and alpha = 0. There are many other cases of when this could be true. Note also if, say, the x-component of this equation is true, then aAx = aBx.
But, the thing to remember is that aA is NOT equal to aB in general. For Problem 28, the cables are in pure translation (no rotation). Therefore, the components of acceleration along the cable must be the same.
This is an excellent question. Asking it on the blog is good so that others can see the answer and have the opportunity to ask followup questions on their own. Thanks for asking!
waw. That s very clear.
Thank you.
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