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|    Message 17,486 of 17,516    |
|    Jonathan Thornburg [remove -color t to Luigi Fortunati    |
|    Re: Acting forces (2/2)    |
|    27 Nov 25 16:33:34    |
      [continued from previous message]              because the whole system is accelerating. In fact, because we're taking       m1 = m2 = m3 and and all three masses have the same common acceleration,       it must be the case that F_net_on_m1 = F_net_on_m2 = F_net_on_m3.                     On the other hand, suppose that m1=10, m2=80, and m3=10. Then (11) gives        F3 = - ((m2+m3)/(m1+m2+m3)) F1 + (m1/(m1+m2+m3)) F2        = -0.9 F1 + 0.1 F2        = -95       and (19) gives        F4 = (m3/(m1+m2+m3)) F1 - (m1+m2)/(m1+m2+m3) F2        = 0.1 F1 - 0.9 F2        = +55       This means the net forces on the three bodies are now        F_net_on_m1 = F1 + F3 = 100-95 = +5        F_net_on_m2 = -F3 - F4 = -(-95) - (55) = 95-55 = +40        F_net_on_m3 = F2 + F4 = -50 +55 = +5       The forces are now different from the equal-mass case (which is not       surprising, since we've changed the mass of each individual body).       Notice that the forces are in the ratio        F_net_on_m1 = F_net_on_m3 = (1/8) F_net_on_m2       which just match the ratios of the masses,        m1 = m3 = (1/8) m2       This is just what we would expect from Newton's 2nd law, since all       all three masses have the same common acceleration.                            So, the simple answer to Luigi's questions is "it depends on the body's       internal mass distribution".              --       -- "Jonathan Thornburg [remove -color to reply]" |
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