Explain the motion of blocks connected by a string in Horizontal motion.

When objects are connected by strings and When objects are connected by strings and a force $\mathrm{F}$ is applied either vertically or horizontally or along an inclined plane, it produces a tension $T$ in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.
Horizontal motion:
In this case, mass $\mathrm{m}_2$ is kept on a horizontal table and mass $\mathrm{m}_1$, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface

As both the blocks are connected to the un stretchable string, if $\mathrm{m}_1$ moves with an acceleration a downward then $\mathrm{m}_2$ also moves with the same acceleration a horizontally. The forces acting on mass $\mathrm{m}_2$ are
- Downward gravitational force $\left(\mathrm{m}_2 \mathrm{~g}\right)$
- Upward normal force $(\mathrm{N})$ exerted by the surface
- Horizontal tension (T) exerted by the string
The forces acting on mass $m_1$ are
- Downward gravitational force $\left(\mathrm{m}_1 \mathrm{~g}\right)$
- Tension (T) acting upwards
The free body diagrams for both the masses is shown in figure.

Applying Newton's second law for $\mathrm{m}_1$
$\mathrm{T} \hat{i}-\mathrm{m}_1 \mathrm{~g} \hat{j}=-\mathrm{m}_1 \mathrm{a} \hat{j}$ (alongy direction)
By comparing the components on both sides of the above equation, $\mathrm{T}-\mathrm{m}_1 \mathrm{~g}=-\mathrm{m}_1 \mathrm{a} \ldots \ldots \ldots \ldots .(1)$
Applying Newton's second law for $\mathrm{m}_2$
$\mathrm{Ti}=\mathrm{m}_1$ ai (along $\mathrm{x}$ direction)
By comparing the components on both sides of above equation, $\mathrm{T}=\mathrm{m}_2 \mathrm{a} \ldots \ldots \ldots \ldots . .(2)$
There is no acceleration along $\mathrm{y}$ direction for $\mathrm{m}_2$.
$
\mathrm{N} \hat{j}-\mathrm{m}_2 \mathrm{~g} \hat{j}=0
$
By comparing the components on both sides of the above equation
$
\begin{aligned}
& \mathrm{N}-\mathrm{m}_2 \mathrm{~g}=0 \\
& \mathrm{~N}=\mathrm{m}_2 \mathrm{~g} \ldots \ldots \ldots .(3)
\end{aligned}
$
By substituting equation (2) in equation (1), we can find the tension $T$
$
\begin{aligned}
& \mathrm{m}_2 \mathrm{a}-\mathrm{m}_1 \mathrm{~g}=-\mathrm{m}_1 \mathrm{a} \\
& \mathrm{m}_2 \mathrm{a}+\mathrm{m}_1 \mathrm{a}=\mathrm{m}_1 \mathrm{~g} \\
& \mathrm{a}=\frac{m_1}{m_1+m_2} \mathrm{~g} \ldots \ldots \ldots .
\end{aligned}
$
Tension in the string can be obtained by substituting equation (4) in equation (2)
$
\mathrm{T}=\frac{m_1 m_2}{m_1+m_2} \mathrm{~g}
$
Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an
important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).