Asked by Anonymous
A pulley is suspended by a Cord (C). on one end of the pulley there is a 1.2 kg block and on the other is a 3.2 kg block.Determine the tension in this cord after the masses are released and before one hits the ground. Ignore the mass of the pulley and cords
Answers
Answered by
drwls
Solve these two simultaneous equations with unknowns a (acceleration) and T (tension).
M1*g - T = M1*a
T - M2*g = M2*a
This leads to
a = [(M1 -M2)/(M1 +M2)]*g
and
T = [2*M1*M2/(M1+M2)]*g
M1*g - T = M1*a
T - M2*g = M2*a
This leads to
a = [(M1 -M2)/(M1 +M2)]*g
and
T = [2*M1*M2/(M1+M2)]*g
Answered by
Hunter
34N
Answered by
bjk
17.1
Answered by
I’m not a physics god, but
I did the Ft=g[(2Mm)/(M+m)] equation and ended up with 17.1N
Which plugged in was
Ft=9.8[((2)(1.2)(3.2))/(1.2+3.2)]
Which plugged in was
Ft=9.8[((2)(1.2)(3.2))/(1.2+3.2)]
Answered by
l
if you use the formula as such as
ft=(2*m1*m2/m1+m2)2g
you will get the right answer which is 34.210
ft=(2*m1*m2/m1+m2)2g
you will get the right answer which is 34.210
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