In the figure shown, the normal reaction between 2 kg and 1 kg is α newton and normal reaction between 3 kg and 2 kg is β newton.Find (α+β) .(g=10 m / s 2).

Question

In the figure shown, the normal reaction between 2 kg and 1 kg is α newton and normal reaction between 3 kg and 2 kg is β newton.Find (α+β) .(g=10 m / s 2). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

a=(60-(18+30)/6)=2 m / s 2

So, all the three blocks will accelerate upwards with a=2 m / s 2. N1(18+5)= 1 × 2 N1=25 N

N2-(18+15)=3 × 2 N2=39 N

Q. In the figure shown, the normal reaction between $2 k g$ and $1 k g$ is $α$ newton and normal reaction between $3 k g$ and $2 k g$ is $β$ newton.Find $(α + β) . (g = 10 m / s^{2})$ .

$a = \frac{60 - ( 18 + 30 )}{6} = 2 m / s^{2}$

So, all the three blocks will accelerate upwards with $a = 2 m / s^{2}$ .
$N_{1} (18 + 5) = 1 \times 2$
$N_{1} = 25 N$

$N_{2} - (18 + 15) = 3 \times 2$
$N_{2} = 39 N$

Q. In the figure shown, the normal reaction between 2kg and 1kg is α newton and normal reaction between 3kg and 2kg is β newton.Find (α+β).(g=10m/s2).

a=660−(18+30)​=2m/s2 So, all the three blocks will accelerate upwards with a=2m/s2. N1​(18+5)=1×2 N1​=25N N2​−(18+15)=3×2 N2​=39N

Q. In the figure shown, the normal reaction between $2 k g$ and $1 k g$ is $α$ newton and normal reaction between $3 k g$ and $2 k g$ is $β$ newton.Find $(α + β) . (g = 10 m / s^{2})$ .

$a = \frac{60 - ( 18 + 30 )}{6} = 2 m / s^{2}$

So, all the three blocks will accelerate upwards with $a = 2 m / s^{2}$ .
$N_{1} (18 + 5) = 1 \times 2$
$N_{1} = 25 N$

$N_{2} - (18 + 15) = 3 \times 2$
$N_{2} = 39 N$