Question

Three equal weights $A,B$ and $C$ of mass 2 kg each are hanging on a string passing over a fixed frictionless pulley as shown in the figure. The tension in the string connecting weights $B$ and $C$ is

• A. Zero
• B. $13\text{N}$
• C. $3.3\text{N}$
• D. $19.6\text{N}$

Answer 1

The correct option is B $13\text{N}$

FBD of each mass

Let ${}^{\prime }{a}^{\prime }$ be acceleration of each block
then $T-mg=ma$
$T_1+mg-T=ma$
$mg-T_1=ma$
Adding $\overline{mg=3ma}$
$a=\frac{g}{3}$
Now $mg-T_1=ma=\frac{mg}{3}$
$\Rightarrow T_1=\frac{2mg}{3}=\frac{2\times 2\times 9.8}{3}$
$\approx 13\text{N}$

Alternative Method;
$a=\frac{\text{Net unbalanced force}}{\text{total mass}}$
As on right side total mass is $4\text{kg}$ and on left side mass is $2\text{kg}$
$a=\frac{(4-2)g}{6}=\frac{g}{3}m/s^2$

Taking FBD of $C$ block
$T_1=2(g-a)=2(g-\frac{g}{3})=\frac{4g}{3}=13\text{N}$