Question

In the system shown above, calculate $a,T_1,T_2,T_1^{\prime }$ and $T_2^{\prime }$.

Answer 1

Let the acceleration be 'a' to left side

Free body diagram

Now,

Applying Newton's II of motion and valency forces

For FBD is

$60-T_1=6a$ .........$\left(1\right)$

For FBD (ii)

$N-30=0\Rightarrow N=30$(No acceleration in vertical direction)

$T_1-T_2=3a$ ........$\left(2\right)$

For FDB(iii)

$T_2-10=a$ .......$\left(3\right)$

For FBD (iv)

Component of $T_1$ in direction of $T_1^1$ will be $T_1\cos 4$

and the net force will be zero

$\Rightarrow T_1^1+2T_1\cos 45=0$ ..........$\left(4\right)$

Similarly, for FBD(v)

$T_2^1+2T_2\cos {45}^o=0$ ........$\left(5\right)$

now adding equation $\left(1\right),\left(2\right)$ and $\left(3\right)$

$60-10=10a$

$50=10a$

$\Rightarrow a=5m{s}^{-2}$

from equation (i)

$60-T_1=6\times 5$

$\Rightarrow T_1=60-3a$

$\Rightarrow T_1=30N$

From equation (iii)

$T_2-10=5$

$\Rightarrow T_2=15N$

From equation$\left(4\right)$

$T_1^1=-2\times 30\cos {41}^o\Rightarrow -20\sqrt{2}N$

From equation $\left(5\right)$

$T_2^1=-2\times 15\cos {45}^o\Rightarrow -15\sqrt{2}N$.