Find the reading of th spring balance shown in figure. The elevator is going up with an accelertion of g/10, the pulley and the strng are light and the pulley is smooth.

Explanation

When the elevator is accelerated upwards, maximum weight will be recorded
`r-(W+m_a)=0`
`rarr R=W+m_a`
`=m(g+a)` maximum weight
When decelerating upwards, maximum wight will be recorded.
` R+ ma-W=0`
`rarr R=W-ma`
`:. M(g+a)=72 xx9.9`...............i,
` and m(g-a)=60xx9.9`.........2
Now, `mg+ma= 72xx9.9`
and `mg-ma=60xx9.9`
`rarr 2mg=1306.8`
` rarr m= 1306.8/(2xx9.9) = 66 kg`
so, the true weight of he man is 66 kg.
b. Again, to find the acceleration
`mg+ma= 72xx9.9`
`rarr = (72xx9.9-66xx9.9)/66`
` = (9.9xx6)/66= 9.9/11`
`= 0.9 m/s^2`.