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Unformatted text preview: x locomotive 1 2 1 2 2 F 1 F 2 44. The two buckets and the cords must have the same acceleration. We write ? F = m a from the force diagram for the lower bucket: F Tbottomlow m 2 g = m 2 a , which gives F Tbottomlow = m 2 ( g + a ) = (3.5 kg)(9.80 m/s 2 + 1.60 m/s 2 ) = 40 N. The tensions shown in the diagram are exerted by the cords. If we consider the lower cord, which now has weight w bottom , as the system, the tensions will be in the opposite directions. F Tbottomhigh F Tbottomlow w bottom = ( w bottom / g ) a , which gives F Tbottomhigh = F Tbottomlow + w bottom + ( w bottom / g ) a , = 40 N + 2.0 N + [(2.0 N)/(9.80 m/s 2 )](1.60 m/s 2 ) = 42.3 N. We write ? F = m a from the force diagram for the upper bucket: F Ttoplow F Tbottomhigh m 1 g = m 1 a , which gives F Ttoplow = F Tbottomhigh...
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This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at University of California, Berkeley.
 Fall '08
 all
 Acceleration, Force

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