Explain the motion of blocks connected by a string in (i) vertical motion (ii) horizontal motion .

Explanation

Vertical motion: Consider two blocks of masses `m_(1)` and `m_(2) (m_(2) gt m_(1))` connected by a light and inextensible string that passes over a pulley as shown in Figure. Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving,`m_(2)` vertically upward and `m_(1)` downward with same acceleration a. The gravitational force `m_(1)g` on mass `m_(1)` is used in lifting the mass `m_(2)`
The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.


Applying Newton.s second law for mass `m_(2)`
`T hat j- m_(2)g hat j= m_(2)a hat j`
The left hand side of the above equation is the total force that acts on `m_(2)` and the right hand side is the product of mass and acceleration of `m_(2)` in y direction. By comparing the components on both sides, we get
`T- m_(2)g= m_(2)a`
Similarly, applying Newton.s second law for mass `m_(1)`
`T hat j- m_(1)g hat j= -m_(1)a hat j`
As mass `m_(1)` moves downward (`-hat j`), its acceleration is along (`-hat j`)
By comparing the components on both sides, we get
`T- m_(1)g= -m_(1)a`
`m_(1)g-T= m_(1)a`
Adding equations (1) and (2), we get
`m_(1)g-m_(2)g = m_(1)a+ m_(2)a`
`(m_(1)-m_(2))g= (m_(1)+m_(2))a`
From equation (3), the acceleration of both the masses is
`a= ((m_(1)-m_(2))/(m_(1)+m_(2)))g`
If both the masses are equal (`m_(1) = m_(2)`), from equation (4)
a=0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest. To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
`T-m_(2)g= m_(2)((m_(1)-m_(2))/(m_(1)+m_(2)))g`
`T= m_(2)g+ m_(2)((m_(1)-m_(2))/(m_(1)+m_(2)))g`
By taking m,g common in the RHS of equation (5)
`T= m_(2)g(1+ (m_(1)-m_(2))/(m_(1)+m_(2)))`
`T= m_(2)g((m_(1)+m_(2)+ m_(1)-m_(2))/(m_(1)+m_(2)))`
`T= (2m_(1)m_(2))/(m_(1)+m_(2))g`
Equation (4) gives only magnitude of acceleration.
For mass `m_(1)` the acceleration vector is given by `vec a= -((m_(1)-m_(2))/(m_(1)+m_(2)))g hatj`
For mass m, the acceleration vector is given by `vec a= -((m_(1)-m_(2))/(m_(1)+m_(2)))g hatj`
Case 2: Horizontal motion: In this case, mass `m_2` is kept on a horizontal table and mass `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 unstretchable string, if `m_1` moves with an acceleration a downward then `m_2` also moves with the same acceleration a horizontally.
The forces acting on mass `m_2` are
(i) Downward gravitational force `(m_2g)`
(ii) Upward normal force (N) exerted by the surface
(iii) Horizontal tension (T) exerted by the string Free body diagram The forces acting on mass `m_1` are
(i) Downward gravitational force `(m_1g)`
(ii) Tension (T) acting upwards
The free body diagrams for both the masses is shown in figure 2.
Applying Newton.s second law for `m_1`
` T hat j - m_1 g hatj - m_1 a hatj ` (along y direction )
By comparing the components on both sides of the above equation,
` T - m_1g = - m_1a` ...(1)
Applying Newton.s second law for `m_2`
`T hati = m_2 a hati ` (along x direction)
By comparing the components on both sides of above equation,
`T = m_2 a` ....(2)
There is no acceleration along y direction for `m_2`
By comparing the components on both sides of the above equation
`N - m_2 g = 0`
By substituting equation (2) in equation (1), we can find the tension T
`m_2 a - m_1 g = - m_1 a`
`m_2 a + m_1 a = m_1 g `
`m_2a + m_1a = m_1 g `
`a = (m_1)/(m_1+m_2) g`
Tension in the string can be obtained by substituting equation (4) in equation (2)
`T = (m_1 m_2)/(m_1 +m_2)` ....(5)
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).