Newton, Find T

tiny-tim

i'm confused

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

SherBear

Sorry Tim =/

T=m1g sin 30° - 31.825 (m2)

T=(15 kg) (9.80 m/s^2) (sin 30°) - 31.825 (12 kg)

T=1382.1 N?

Then...2nd Equation

T = m2g (m1)

T=(12 kg) (9.80 m/s^2) (15 kg)

T=1764 N?

Then Subtract the 2....

1764 - 1382.1 = 381.9 N

T= 381.9 N?

SherBear

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

SherBear

So any input? I still don't know if I'm doing this right.

tiny-tim

yes, that's the right method, but in A you've missed out the m₁gsin30°

(alternatively, if you only want T and not a, then you could eliminate a instead of T )

SherBear

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 ?

SherBear

Or for box b use sum of F=T m2g so T=m2g.....12(9.8 m/s^2) = 117.6N ?

SherBear

Which T is the same all the way through because it's 1 rope?