rigid body
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Working note by Phil, dated 10.24.16, written to prepare a drawing for Appendix D of the Lagrange multipliers write-up. He sets up two masses on a ramp with normal forces and an internal force T12, and requires equal accelerations along the ramp. Algebra in rotated and unrotated axes gives two incompatible equations, which would require -cotθ = tanθ, so he concludes the setup is paradoxical.
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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.
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.
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So let and be the usual axes to the right and up, and let ' and ' be rotated axes so ' is down the ramp.
m1g = - m1g F1 = F1' T12 = T12 '
m2g = - m2g F2 = F2'
Then equations become
Force on m1 is F1 = m1g + R1 = - m1g + R1 N1 = N1' N1 > 0
R1 = N1+ T12 R1 = N1'+ T12 '
so then
F1 = - m1g + N1'+ T12 '
Force on m2 is F2 = m2g + R2 = - m2g + R2 N2 = N2' N2 > 0
R2 = N2 - T12 R2 = N2' - T12 '
so then
F2 = - m2g + N2' - T12 '
Now write out F1/m1 = F2/m2
[ - m1g + N1'+ T12 ' ] / m1 = [ - m2g + N2' - T12 ' ] / m2
I now need to write
' = down slope = (cosθ - sinθ )
' = perp to slope = (cosθ + sinθ )
Now the acceleration equation tells us
[ - m1g + N1 (cosθ + sinθ )+ T12 (cosθ - sinθ ) ] / m1
= [ - m2g + N2(cosθ + sinθ ) - T12 (cosθ - sinθ ) ] / m2
We can balance the x and y components separately to get two equations:
First here is the equation followed by the y equation
[ N1 ( sinθ ) + T12 (cosθ ) ] / m1 = [ N2( sinθ ) - T12 (cosθ ) ] / m2
[ - m1g + N1 (cosθ )+ T12 ( - sinθ ) ] / m1 = [ - m2g + N2(cosθ ) - T12 (- sinθ ) ] / m2
Rewrite these as
[ N1 sinθ + T12cosθ ] / m1 = [ N2 sinθ - T12cosθ ] / m2
[ - m1g + N1cosθ - T12sinθ ] / m1 = [ - m2g + N2cosθ + T12 sinθ ] / m2
Known quantities are m1, m2, g, sinθ, cosθ. Simplify
m2[ N1 sinθ + T12cosθ ] = m1 [ N2 sinθ - T12cosθ ]
m2[ - m1g + N1cosθ - T12sinθ ] = m1 [ - m2g + N2cosθ + T12 sinθ ]
[ N1 m2sinθ + m2T12cosθ ] = [ N2 m1sinθ - m1T12cosθ ]
[ - m1m2g + N1m2cosθ - T12m2sinθ ] = [ - m1m2g + N2m1cosθ + T12 m1sinθ ]
[ m2sinθN1 + m2cosθT12 ] = [ m1sinθN2 - m1cosθT12 ]
[ m2cosθN1 - m2sinθT12 ] = [ m1cosθN2 + m1sinθT12 ]
m2sinθN1 - m1sinθN2 = - m2cosθT12 - m1cosθT12
m2cosθN1 - m1cosθN2 = m1sinθT12 + m2sinθT12
(m2N1 - m1N2)sinθ = (- m2 - m1)T12cosθ
(m2N1 - m1N2)cosθ = (m1 + m2)T12sinθ
(m2N1 - m1N2) = - (m2 + m1)T12cotθ
(m2N1 - m1N2) = (m1 + m2)T12 tanθ
Something is wrong because the two left sides are the same and the right sides are not the same. To make the right sides the same you would need to have
- (m2 + m1)T12cotθ = (m1 + m2)T12 tanθ
or
- cotθ = tanθ
and this is a contradiction, confirming the Paradox.