|Oct6-12, 06:11 PM||#18|
Newton, Find T
you had T - m1gsin30° - 31.825 = m1a
and m2g - T = m2a
to find T, you need to eliminate a, so multiply the first equation by m2 and the second equation by m1, and then subtract
|Oct6-12, 11:50 PM||#19|
T=m1g sin 30° - 31.825 (m2)
T=(15 kg) (9.80 m/s^2) (sin 30°) - 31.825 (12 kg)
T = m2g (m1)
T=(12 kg) (9.80 m/s^2) (15 kg)
Then Subtract the 2....
1764 - 1382.1 = 381.9 N
T= 381.9 N?
|Oct7-12, 08:17 AM||#20|
Here's a different way, can you tell me if it's right or wrong?
forces on A
u*m*g*Cos30 -T = ma
T - 0.25*15*9.8*cos30 = 15*a
T - 31.82N = 15a .........................................i
Forces on B
m*g-T = ma
12*9.8 -T = 12a
117.6-T = 12a...........................................ii
adding i and ii
-31.82 + 117.6 = 27a
85.78 = 27a
a = 3.17 m/s/s
Now T = 117.6 - 12*9.8 = 0 No motion
|Oct8-12, 04:29 PM||#21|
So any input? I still don't know if I'm doing this right.
|Oct8-12, 04:57 PM||#22|
(alternatively, if you only want T and not a, then you could eliminate a instead of T )
|Oct8-12, 05:26 PM||#23|
I don't like that way, could I also find Acceleration and knowing both masses I could rewrite:
the sum of F=m2g-T=m2a as
T= m2g / m2a = 12(9.8) / 12 (.454 assuming the acceleration i computed is correct) = 21.58 N
T=21.58 N ?
|Oct8-12, 05:36 PM||#24|
Or for box b use sum of F=T m2g so T=m2g.....12(9.8 m/s^2) = 117.6N ?
|Oct8-12, 05:37 PM||#25|
Which T is the same all the way through because it's 1 rope?
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