What is the reading of the spring balance in the following device? (Answer in kg).

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Solution

Let T to be the reading of the spring balance, then

20g - T = 20 a.......(i)
Similarly,

$T - 10 g = 10 a . . . . . . . (i i)$
From (i) and (ii)
$a = \frac{g}{3} \frac{m}{s^{2}} a n d T = \frac{40 g}{3} N$
So, the reading is $\frac{40}{3} k g$

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What is the reading of the spring balance in the following device? (Answer in kg).

Let T to be the reading of the spring balance, then 20g - T = 20 a.......(i) Similarly, T−10g=10 a.......(ii) From (i) and (ii) a=g3ms2 and T=40g3N So, the reading is 403kg

Let T to be the reading of the spring balance, then

20g - T = 20 a.......(i)
Similarly,

$T - 10 g = 10 a . . . . . . . (i i)$
From (i) and (ii)
$a = \frac{g}{3} \frac{m}{s^{2}} a n d T = \frac{40 g}{3} N$
So, the reading is $\frac{40}{3} k g$