Suggestions:
- FBD: Draw FBD of system made up of A, B and both massless links. Determine which, if any, non-conservative forces do work.
- Work-energy equation: You will find from you FBD that there are no non-conservative forces that do work. Therefore, the work-energy equation becomes T1 + V1 = T2 + V2 where: T1 = 0 and T2 = (1/2)*mA*vA^2 + (1/2)*mB*vB^2. Write down the potential energy of the system at states 1 and 2. This will depend on your choice of datum position (I recommend a datum position that passes through point O.)
- Kinematics: You need to determine the relationship between vA and vB at position 2. Hint: Look for the instant center of link BC to determine the velocity of point B at position 2.
- Solve: Use the work-energy and kinematics equations to find vA.
6 comments:
I'm having some trouble figuring out where I have gone wrong. I have the following...
v2=v1*sqrt(3)
v1 = sqrt( (g((ma*(h_a1-h_a2) + mb*(h_b1-h_b2)))/(2(ma+mb)) )
h_a1=0.15
h_b1=-0.3
h_a2=0.3
h_b2=-0.6
I end up with something around 1.02. I feel like my velocity relation must be wrong, but I don't see how. If anybody could point me in the right direction, I'd appreciate it!
I followed the page problem guide and am running into some trouble with how to come up with the correct V_1 and V_2 equations. I broke up the work-energy equation, and worked on the kinematics but am having trouble figuring out the instant centers and how they play into finding these values. I know it should be fairly straight forward, but I think fall-break-it is keeping me from seeing it. I summed some of the forces and solved, but my value's were no where near what they should. If anyone could help me out, I would appreciate it.
--Sam--
I understand most of the terms that you have written down. Let me see if I can reconstruct your work-energy equation from what you have here:
T2 = V1 - V2, OR
(1/2)*ma*va2^2 + (1/2)*mb*vb2^2 = (g((ma*(h_a1-h_a2) + mb*(h_b1-h_b2)))
Is the above your work-energy equation?
If so, then it appears that you then used va2 = vb2 (that is, the speed of A is the same as the speed of B at position 2) to get your final answer. This is not correct.
If you locate the IC of link BC, you will find that this IC lies at point B, making vb2 = 0. With this, you can solve the WE equation for its one unknown: va2.
Does this help?
Yes that is the work-energy equation I used.
I've gotten the right numbers now that I'm using vb=0 but I don't understand how the IC is at point B. I would think that vc would be in the downward direction, forcing the IC to be somewhere in the horizontal plane, not along the BC axis.
* At Position 2, bar AC is vertical. Since AC is rotating about point O, then vC is horizontal, pointing to the right.
* Block B always moves in the vertical direction, so vB is vertical.
* Constructing perpendiculars to vB and vC shows that their intersection is at B. Hence B is the IC, and vB = 0.
Does this help?
I just realized that I had points C and B switched on my FBD. Everything makes sense now, sorry!
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