eptheta
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From HRW 8 Ch 16 Q19
http://img684.imageshack.us/img684/8947/82287578.jpg
What i did:
I resolved the components of tension is each string (T cosX and T sin X) and found X to be 19.47 degrees (sin inverse of [0.25m/0.75 m]). Then i got the value of T as 20.79 N. The Tension in the middle(between A and B) i called T2 and got it as 6.93 N [T sinX= T2]
Then, mass per unit length was found out to be 0.01kg/AB(1.5m) = 6.67x10^-3 kg/m
using the formula v= root of[T/mass per unit length], v= 32.2 m/s
The answer at the back of the book says 32.9 m/s .
Can anyone help me out here ?
Thanks
http://img684.imageshack.us/img684/8947/82287578.jpg
What i did:
I resolved the components of tension is each string (T cosX and T sin X) and found X to be 19.47 degrees (sin inverse of [0.25m/0.75 m]). Then i got the value of T as 20.79 N. The Tension in the middle(between A and B) i called T2 and got it as 6.93 N [T sinX= T2]
Then, mass per unit length was found out to be 0.01kg/AB(1.5m) = 6.67x10^-3 kg/m
using the formula v= root of[T/mass per unit length], v= 32.2 m/s
The answer at the back of the book says 32.9 m/s .
Can anyone help me out here ?
Thanks
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