Question

The friction coefficient between the table and the block shown in the figure, is $0.2$. Find the tensions in the two strings:

Answer 1

From the $FBD$ of blocks (15 kg block assumed to move downwards) and writing Newton's 2nd law of motion ,$F=Ma$

for 15 kg block : $15g-T_1=15a$ ...(1)

for 5 kg hanging block : $T_2-5g=5a$ ...(2)

for 5 kg block on the horizontal surface, we use Newton's second law in two direction (i.e$X-axis$ and $Y-axis$)

Balancing force along $Y$-direction : $N=5g$, and using concept of friction $f=\mu N$, we get $f=0.2\times 5\times g=g$

Now along $X$-direction : $T_1-T_2-f=5a$ ....(3)

using value of $f$ and adding eq 1, eq 2 and eq 3 we get

$\Rightarrow$ $15g-5g-f=(15+5+5)\times a$

$\Rightarrow$ $9g=25a$

$\Rightarrow$ $a=\frac{9g}{25}=\frac{9\times 10}{25}=\frac{18}{5}\frac{m}{s^2}$ (since $g=10$)

now using eq 1 : $T_1=150-\frac{15\times 18}{5}=150-54=96N$

using eq 2 : $T_2=50+\frac{5\times 18}{5}=68N$

Hence values of $T_1=96N$ and $T_2=68N$