Question

Three masses of $2kg,10kg$ and $5kg$ are connected to one another and held as shown in figure. If coefficient of friction $=0.1$ and $g=10m/s^2$.
Determine:
Value of $T_1$
Value of $T_2$
Value of acceleration

Answer 1

Given that,

Mass $m_1=2kg$

Mass $m_2=10kg$

Mass $m_3=10kg$

Friction coefficient $\mu =0.1$

Let $T_1$ and $T_2$ be the tensions in the strings and a be the acceleration towards right

Then

Balancing condition

$T_2=ma+T_1+\mu mg.....\left(I\right)$

Now, for $T_2$

$T_2=mg-ma....\left(II\right)$

Now, for $T_1$

$T_1=m(g+a)....\left(III\right)$

Now put the value of $T_1$ and $T_2$ in equation (I)

$2(10+a)+0.1\times 10\times 10+10a=50-5a$

$20+2a+10+10a+5a-50=0$

$a=\frac{20}{17}$

$a=1.17m/s^2$

Now, put the value of a in equation (II) and (III)

$T_1=2(10+\frac{20}{17})$

$T_1=23.5N$

Now,

$T_2=5(10-\frac{20}{17})$

$T_2=44.11N$

Hence, the value of acceleration, $T_1$ and $T_2$ is $1.17m/s^2$ ,$23.5N$ and $44.1N$