Question

Calculate tensions $T_1$, $T_2$ and $T_3$ in the given system when whole system is moving upward with an acceleration of $a=2m/s^2$

Answer 1

Hint: The tension in a particular string is affected by the weight of the boxes below it, whether immediate or not. Since the boxes accelerate upward, the force in the upward direction acting on a particular box is greater than the force acting in the downward direction. So in the equation of motion of each body we need to substitute the values and then we will get the answer.

Formula used: In this solution we will be using the following formulae;
$W=mg$ where $W$ is the weight of a body, $m$ is the mass of the body and $g$ is the acceleration due to gravity.
${\overrightarrow{F}}_{NET}=m\overrightarrow{a}$ where $F_{NET}$ is the net force acting on a body, and a is the acceleration of the body.

Complete step by step solution:
In the diagram, the weight of all boxes acts on string $T_1$ , while the weight of mass $m_2$ and mass $m_3$ acts on string $T_2$ , but only the weight of mass $m_3$ acts on string is $T_3$ .
To find the tension in each string we analyse the dynamics of each box. For simplicity, let’s start from the bottom
For the bottom box, the tension $T_3$ acts as the upward force acting on the box to pull it upward (as if to prevent it from falling), while the weight of the box acts downward. Hence, on the bottom box, Newton's second law would be given as
$F_{NET}=T_3-W_3=m_{3}a$ (assuming upward is positive), $a$ is positive (and non-zero) because the whole system accelerates upward, hence, the tension pulling the box must be greater than the weight.
Now, since the weight of a body is given as $W=mg$ then
$\Rightarrow T_3-m_{3}g=m_{3}a$
Making $T_3$ subject, we have
$\Rightarrow T_3=m_{3}a+m_{3}g=m_3(a+g)$
Inserting all known values
$\Rightarrow T_3=3\left(2+9.8\right)=35.4N$
For the box of mass $m_2$ , there are three forces acting on it.
The tension $T_2$ acting upward, the weight $W_2$ acting downwards, and finally the tension $T_3$ acting downwards.
Therefore, we have
$\Rightarrow F_{NET}=T_2-T_3-W_2=m_{2}a$
$\Rightarrow T_2=m_2(a+g)+T_3$
Inserting values, we have
$\Rightarrow T_2=2\left(2+9.8\right)+35.4=59N$
Similarly for top box we have
$\Rightarrow F_{NET}=T_1-T_2-W_1=m_{1}a$
$\Rightarrow T_1=m_1(a+g)+T_2$
Inserting values, we have
$\Rightarrow T_1=5(2+9.8)+59$
$\Rightarrow T_1=82.6N$
Hence, $T_1=82.6N$ , $T_2=59N$ , and $T_3=35.4N$ .

Note:
To avoid confusion, the tension $T_3$ acts downward on the middle box because the action of the tension $T_3$ acting on the bottom box has an equal and opposite tension acting downwards on the middle box. This is analogous to a box suspended by a string attached to a roof. The tension in this string acting upwards keeps the box suspended but at the same time, the string pulls on the roof downwards as though to bring it down.