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Fall 2008 -- Purdue University -- West Lafayette, IN
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9 comments:
Also, it seems from the picture that the w term should be in the negative K direction, but it doesn't work out that way with the math. Am I missing something?
this happened with the previous problem as well, with the math, omega turns out positive, but it should be negative to match up with the clockwise direction
For tim:
The angular acceleration of links AB and CD are definitely not zero for the position of interest. However, your answers for the loads carried by link DC will depend on the x-component of the force equation for the plate, as well as the Euler equation for the plate. Neither of these two equations involve the angular acceleration of the links. Therefore, setting alpha = 0 or alpha = 1000 or ... will not affect your answers for the force in the links. Leave alpha as an unknown and solve for it if you need to know it.
For eric/kyle:
The speed term in the work/energy equation appears as a "square". When you take the square root in solving for v, you can use either the "+" or "-" sign. It looks like you are choosing the "+" sign for this and other problems. Take the "-" sign when you feel that it is correct. It is just as correct as the "+" sign.
I have used the fact that CD and AB are two force members to come up with my three initial equations.
Sum Forces X
Sum Forces Y
Moment About G
These give me the force of CD in the X and Y direction in terms of the acceleration of the plate in the given direction.
I also used the T1+V1+U1-2=T2+V2, and said that
T1=0 at rest
V1=mgh where h is .4
V2=0 since datum is there
T2=.5mV^2+.5Iw^2 where I solved for I using 1/12 m(b^2+h^2), and V I solved for using Vc=Vd+w x r.
This leaves me with an unknown acceleration in the x
acceleration in the y
force of CD
W of CD
and only the 3 equations.
Does anyone know what other equation would give me a value that I need, or did I use a wrong equation some place. Thanks for any help.
Mark
Your work energy equation is just for the plate (since AB and CD have negligible mass, they contribute to neither the KE nor to the PE).
Note that the plate is in PURE TRANSLATION (no rotation). Therefore, the KE for your system (the plate) is simply:
T2 = (1/2)*m*v^2
From the work/energy equation, you can now solve for v (the speed of G, as well as of ANY point on the plate).
Since the plate is in pure translation, all points on the plate also have the same acceleration. In particular, you know that a_G = a_C. Since C is on a circular path with its center at D, you can write (path description):
a_C = vdot_C*j - (v^2/rho)*i
where rho is the radius of curvature of the path of C ( = 0.8 meters). Therefore,
a_Gx = a_Cx = -v^2/rho
Also note that in your Euler equation for the plate you have alpha = 0 (since plate is in pure translation).
Let us know if this does not help.
I am confused as to why links AB and CD don’t do any work. The rectangular plate moves a distance in the x-direction due to the forces of AB and CD. Why is it that these forces don’t do any work yet they caused a displacement?
Good question.
As mbricher points out above, AB and CD are two-force members. Therefore, the forces of these links on the plate are aligned with the links. Importantly, these forces are PERPENDICULAR to the paths of points B and C where they act on the plate. A force that acts in a direction that is perpendicular to the path of the point of application can NOT do work (recall the definition of work).
In summary, although these links do dictate the motion of the plate, they do not do work on the plate.
Let us know if this is not clear.
Ok, that makes sense. I guess I did not notice that the forces act perpendicular. The fact that the forces are perpendicular clears things up.
Thanks,
Tyler
Yeah, thanks Professor, I knew I didn't know it, so I decided to try it as zero... I got lucky..
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