Calculating tension of objects with a pulley on an incline

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SUMMARY

The discussion focuses on calculating the tension in a cable connecting two masses on a frictionless incline. The user initially attempted the calculation for a 5 kg mass, resulting in a negative acceleration of -2 m/s². The equation used was Ft - (5)(9.8)sin60 = 5a. A suggestion was made to apply a similar approach for the 6 kg mass, but with -a instead of a to account for the direction of acceleration.

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yummallory
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Homework Statement


Calculate the tension in the cable connecting the two masses. Assume all surfaces are frictionless.

http://tinypic.com/r/20jg36s/7

Homework Equations





The Attempt at a Solution


I tried to begin the solution but came out with a negative acceleration (-2m/s^2)
For the 5 kg mass:
Ft-Fg2sin60=ma
Ft-(5)(9.8)sin60=5a
Ft-49sin60=5a
I wasn't sure how to get the equation for the 6 kg mass...
 
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welcome to pf!

hi yummallory! welcome to pf! :smile:

(have a degree: ° and try using the X2 icon just above the Reply box :wink:)
yummallory said:
For the 5 kg mass:
Ft-Fg2sin60=ma
Ft-(5)(9.8)sin60=5a
Ft-49sin60=5a
I wasn't sure how to get the equation for the 6 kg mass...

same way, but with -a instead of a :wink:
 

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