Question

Find the acceleration of the $500$g block in figure.

Answer 1

Here, $g$ is acceleration due to gravity

$gm$ unit of mass in CGS unit.

$kg$ is kilogram, unit of mass in SI unit

$T_1$ is the tension on the string connecting A and B

$T_2$ is the tension on the string connecting B and C

$a$ is the acceleration of the system.

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Now write the equation of motion for the three masses.

for mass C,

$\left(0.5kg\right)g-T_2=0.5kg\left(a\right)$

$0.5g-T_2=0.5a$ ------------------------------------(1)

for mass B,

$T_2-\left(0.1kg\right)g\left(\sin \right({30}^{\circ }\left)\right)-T_1=0.1kg\left(a\right)$

$T_2-0.1g\left(\sin \right({30}^{\circ }\left)\right)-T_1=0.1a$ -----------------(2)

for mass A,

$T_1-\left(0.05kg\right)g=0.05kg\left(a\right)$

$T_1-0.05g=0.05a$ --------------------------------(3)

Now, add equation (3) and (2).

$T_2-0.05g-0.1g\left(\sin \right({30}^{\circ }\left)\right)=0.1a+0.05a$

$T_2-0.05g-0.1g\left(\sin \right({30}^{\circ }\left)\right)=0.15a$ ------------(4)

Now add equation (4) and (1).

$0.5g-0.05g-0.1g\left(\sin \right({30}^{\circ }\left)\right)=0.15a+0.5a$

$0.5g-0.05g-0.1g\left(\sin \right({30}^{\circ }\left)\right)=0.65a$

$a=\frac{0.5g-0.05g-0.1g\left(\sin \right({30}^{\circ }\left)\right)}{0.65}$

$a=6.03m/s^2$

so, the acceleration of $500gm$ mass is $a=6.03m/s^2$ in the downward direction.