Question

For the situation shown in figure, if the system is released from rest. Determine the value of tension in the string $T_1\text{and}{T}_2$ . $(\text{Take}g=10\frac{m}{s^2})$

• A. $T_1=\frac{256}{13}N;T_2=\frac{136}{13}N$
  
• B. $T_1=\frac{125}{13}N;T_2=\frac{136}{13}N$
• C. $T_1=\frac{256}{13}N;T_2=\frac{272}{13}N$
• D. $T_1=\frac{128}{13}N;T_2=\frac{136}{13}N$

Answer 1

The correct option is A $T_1=\frac{256}{13}N;T_2=\frac{136}{13}N$


Let $5kg$ block is sliding down the inclined plane and $1kg$ block is moving up and pulley is rotation in anticlockwise direction.

FBD of all three components is shown is figure

For $5kg$ block, from Newton's second law
$5g\sin {37}^{\circ }-0.2\times 5g\cos {37}^{\circ }-T_1=5a\Rightarrow 22-T_1=5a.......\left(1\right)$

For pulley, writing net torque
$(T_1-T_2)R=5\alpha \Rightarrow T_1-T_2=10\alpha ....\left(2\right)$
For $1kg$ block, from Newton's second law
$T_2-g=a\Rightarrow T_2-10=a.......\left(3\right)$

From constraint equation, $a=R\alpha$
$\Rightarrow a=\frac{\alpha }{2}$
Adding$\left(1\right),\left(2\right),\left(3\right)$, we get
$26a=12$
$a=\frac{6}{13}\frac{m}{s^2};\alpha =\frac{12}{13}\frac{rad}{s^2}$

Substititing $a,\alpha$ in equations $\left(1\right),\left(3\right)$
$T_1=22-5a=22-5\left(\frac{6}{13}\right)=\frac{256}{13}N;T_2=10+a=10+\left(\frac{6}{13}\right)=\frac{136}{13}N$