rigid body 3
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Short working notes by Phil dated 10.24.16, written to produce a drawing for Appendix D of the Lagrange multipliers write-up. He solves the problem of two masses on a ramp joined by a string, using equal accelerations, the constraint forces N1, N2 and T12, and the string tension T. He finds the normal forces N = -m g cosθ, the common acceleration g sinθ minus |T|, and that the internal forces are equal and opposite. Symbols and vector notation are partly lost in extraction.
AI-written summary; may contain errors. This description is approximate.
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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 the string.
Force on m1 is F1 = m1g + R1+ T where R1 = N1+ T12 . T = force of the string
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+ T) = m1(m2g + R2)
or
m2(R1+ T) = m1R2 .
Now the two Ri vectors are not in the same direction. Write this in more detail
m2(N1+ T12+ T) = m1(N2 - T12)
or
m2N1 - m1N2 = - m2T- m2T12 - m1T12
or
m1N2 - m2N1 = m2T + m2T12 + m1T12
or
m1N2 - m2N1 = m2T + (m1+m2)T12
All terms on the left are in the ' direction, while all terms on the right are in the ' direction. Thus each side is separately zero and we then have
m1N2 = m2N1 N2 = (m2/m1)N1 N2 = N1
m2T = - (m1+m2)T12σ
Now suppose the string is just holding it back a little bit, a very small T to ghe left. Then the first T12 is to the left as well.
T12 = - T
F1 = m1g + N1+ T12 + T
T12 + T = - T + T = - T + T
= T
So
F1 = m1g + N1+ T
F2 = m2g + N2 - T12 = m2g + N2 + T
Summarize:
F1 = m1g + N1+ T
F2 = m2g + N2 + T
Accelerations:
a1 = g + [ N1+ T ]/m1
a2 = g + [ N2 + T]/m2
a1 = g + [ T ]/m1
a2 = g + [ T]/m2
or
a1 = gsinθ + T T = T
a2 = gsinθ + T
a1z = a2z = gsinθ - |T|
Now I know that both these forces are along the ramp. Let be down the ramp and be normal to the ramp. Then
F1 = m1g cosθ + N1
F2 = m2g cosθ + N2
But we know F1 and F2 are down the ramp, so
N1 = -m1g cosθ
N2 = -m2g cosθ
These facts are unaffected by T and the stick and so on. Next,
F1 = m1gsinθ + T
F2 = m2gsinθ + T T < 0 in my picture
So rewrite as
F1 = m1gsinθ + T
F2 = m2gsinθ + T T < 0 in my picture
Note that F1/m1 = F2/m2. Now consider
R1 = N1+ T12
R2 = N2 - T12
R1 = N1
R2 = N2
R1 = T12
R2 = - T12
so these are indeed equal and opposite.