rigid body 2
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Short working note by Phil, dated 10.24.16, written while preparing a rigid body drawing for Appendix D of the Lagrange multipliers write-up. It analyzes two masses joined by a stick on an inclined ramp with no string, setting up forces with normal and tension terms. It finds an apparent contradiction, then resolves it by showing the stick tension is zero and both masses accelerate at g sin(theta).
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Rigid Body Drawing for Appendix D PhL 10.24.16
I thought this was trivial, but now I see that I really have to solve the whole problem before I can draw the vectors. So I will now solve the problem. This is the problem with no string.
Paradox: I solve the problem below and the result is a contradiction . Watch:
Force on m1 is F1 = m1g + R1 where R1 = N1+ T12 .
Force on m2 is F2 = m2g + R2 where R2 = N2- T12 .
Both F1 and F2 must point along the ramp.
Both masses have the same acceleration so a1 = a2 which means F1/m1 = F2/m2 or m2F1 = m1F2 .
Aside: Does this give some preliminary contradiction?
m2(m1g + R1) = m1(m2g + R2)
or
m2R1 = m1R2
This does seem odd, it says that the constraint forces point in the same direction ! Keep going:
m2(N1+ T12) = m1(N2 - T12)
m2N1 - m1N2 = - (m1+m2)T12
More contradiction! Since both normal forces are perp to the ramp, this says T12 must be perp to the ramp.
So I can stop right here, no need to do algebra below, already have the Paradox.
Resolution: In the case I have described, there is no tension/compression in the stick. Both balls natrually accelerate down the ramp at the same acceleration which is gsinθ. So T12 = 0. Then there is no contradiction, and N1= R1 and N2= R2. So that must be why I added the string!