Physics - Rotation Question with Solution | TestHub

$N_1+N_2=mg=200$.....(i)

$({\vec{\tau }}_{net}{)}_{cm}=0$

Case-1: $F(h-0.9)+N_1(0.3)-N_2(0.3)=0$

$N_2-N_1=\frac{1000}{3}(h-0.9)$.....(ii)

From (i) & (ii):

$2N_1=200-\frac{1000}{3}(h-0.9)$

$N_1=100-\frac{500}{3}(h-0.9)$

For not tipping, $N_1>0$:

$h<0.9+0.6$

$h<1.5m$

Case-2: $({\vec{\tau }}_{net}{)}_{cm}=0$

$F(0.9-h)+N_1(0.3)-N_2(0.3)=0$....(iii)

From (i) & (iii): $N_1=100-\frac{500}{3}(0.9-h)$

For not tipping, $N_1>0$:

$\frac{300}{500}>0.9-h$

$h>0.9-0.6$

$h>0.3m$