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Two masses of $5 \mathrm{~kg}$ and $3 \mathrm{~kg}$ are suspended with help of massless inextensible strings as shown in figure. Calculate $T_1$ and $T_2$ when whole system is going upwards with acceleration $=2 \mathrm{~m} / \mathrm{s}^2$ (use $g=9.8 \mathrm{~ms}^{-2}$ ).


Solution:
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Verified Answer
As the whole system is going up with acceleration $a=$ $2 m s^2$
As given that, $m_1=5 \mathrm{~kg}, m_2=3 \mathrm{~kg}$
Tension, $g=9.8 \mathrm{~m} / \mathrm{s}^2$

Tension a string is equal and opposite in all parts of a string.
For the upper block of mass $5 \mathrm{~kg}$, the forces on mass $m_1$.
$$
\begin{aligned}
&T_1-T_2-m_1 g=m_1 a \\
&T_1-T_2-5 g=5 a \\
&T_1-T_2=5(g+a)
\end{aligned}
$$
For the lower block of mass $3 \mathrm{~kg}$, the force on mass $m_2$
$$
\begin{aligned}
&T_2-m_2 g=m_2 a \\
&T_2-3 g=3 a \\
&T_2=3(g+a)=3(9.8+2)=35.4 \mathrm{~N}
\end{aligned}
$$
From eq. (i),
$$
\begin{aligned}
T_1 &=T_2+5(g+a) \\
&=35.4+5(9.8+2)=94.4 \mathrm{~N}
\end{aligned}
$$
As given that, $m_1=5 \mathrm{~kg}, m_2=3 \mathrm{~kg}$
Tension, $g=9.8 \mathrm{~m} / \mathrm{s}^2$

Tension a string is equal and opposite in all parts of a string.
For the upper block of mass $5 \mathrm{~kg}$, the forces on mass $m_1$.
$$
\begin{aligned}
&T_1-T_2-m_1 g=m_1 a \\
&T_1-T_2-5 g=5 a \\
&T_1-T_2=5(g+a)
\end{aligned}
$$
For the lower block of mass $3 \mathrm{~kg}$, the force on mass $m_2$
$$
\begin{aligned}
&T_2-m_2 g=m_2 a \\
&T_2-3 g=3 a \\
&T_2=3(g+a)=3(9.8+2)=35.4 \mathrm{~N}
\end{aligned}
$$
From eq. (i),
$$
\begin{aligned}
T_1 &=T_2+5(g+a) \\
&=35.4+5(9.8+2)=94.4 \mathrm{~N}
\end{aligned}
$$
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