How Does an Elevator's Acceleration Affect Spring Balance Readings?

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SUMMARY

The discussion focuses on calculating the reading of a spring balance in an elevator accelerating upwards at g/10 (0.98 m/s²). The effective acceleration acting on the loads in the elevator is determined to be 10.78 m/s². The tension in the string for a 1.5 kg load is expressed as 16.7 - 3a, while for a 3 kg load, it is 32.34 + 3a. The participant is seeking assistance to identify potential errors in their calculations and to solve the problem accurately.

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Homework Statement



Find the reading of the spring balance shown in the figure (see attachment). The elevator is going up with an acceleration of g/10 , the pulley and the string are light weight and the pulley is smooth.


Homework Equations


F= ma


The Attempt at a Solution


Since the elevator is accelerating upwards with a magnitude of 0.98 m/s^2 (g/10) , the acceleration acting on the items in the elevator should be g+g/10 = 10.78 m/s^2.
By assuming that the 3k load has a relative acceleration of "a", the tension on the string of the 1.5 kg load will be
16.7 - 3a
While on the 3kg load it will be
32.34 + 3a

I'm sure i have made a blunder somewhere up here, yet i can't seem to find it. How do i further solve the question?
 
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