Question

Tension in the rope at the rigid support is $\left(g=10m/s^2\right)$

Answer 1

Step 1. Given data

The mass of the man $A$ is $M_A=60kg$

The acceleration of the man $A$ is $a_A=2m/s^2$

The mass of the man $B$ is $M_B=50kg$

The mass of the man $C$ is $M_C=40kg$

The acceleration of the man $C$ is $a_C=-1m/s^2$

Step 2. Formula used

1. $F=ma$
2. $W=mg$
3. $T-W=F$

Where $m$ is the mass , $W$ is the weight, $g$ acceleration due top gravity, $a$ is the acceleration, $T$ is the tension

Step 3. Solution

Let us draw the free body diagram of man $A$.

There are two forces acting on man $A$, the upward tension due to the rope and his weight downward.

The mass of the man $A$ is $M_A=60kg$

The acceleration of the man $A$ is $a_A=2m/s^2$

$$\begin{gathered} T_A-W_A=F_A \\ \Rightarrow T_A-M_{A}g=M_A{a}_A \\ \Rightarrow T_A=M_A{a}_A+M_{A}g \\ \Rightarrow T_A=M_A\left(a_A+g\right) \\ \Rightarrow T_A=60\left(10+2\right) \\ \Rightarrow T_A=720N..............\left(i\right) \end{gathered}$$

Let us draw the free body diagram of man $B$.

There are two forces acting on man $B$ , the upward tension due to the rope and his weight downward.

The mass of the man $B$ is $M_B=50kg$

Since, the man is moving down with constant velocity equal to $2m/s$, he has zero acceleration

The acceleration of the man $B$ is $a_B=0m/s^2$

$$\begin{gathered} T_B-W_B=F_B \\ \Rightarrow T_B-M_{B}g=M_B{a}_B \\ \Rightarrow T_B=M_B{a}_B+M_{B}g \\ \Rightarrow T_B=M_B\left(a_B+g\right) \\ \Rightarrow T_B=50\left(0+10\right) \\ \Rightarrow T_B=500N...............\left(ii\right) \end{gathered}$$

Let us draw the free body diagram of man$C$.

There are two forces on him, the upward tension due to the rope and his weight downward.

The mass of the man $C$ is $M_C=40kg$

The acceleration of the man $C$ is $a_A=-1m/s^2$

$$\begin{gathered} T_C-W_C=F_C \\ \Rightarrow T_C-M_{C}g=M_C{a}_C \\ \Rightarrow T_C=M_C{a}_C+M_{C}g \\ \Rightarrow T_C=M_C\left(a_C+g\right) \\ \Rightarrow T_C=50\left(-1+10\right) \\ \Rightarrow T_C=450N\left(iii\right) \end{gathered}$$

The total tension in the rigid support has to balance the tensions in the rope due to individual actions of the three men.
Let the total tension at the rigid support be $T$

$\therefore T=T_A+T_B+T_C=720+500+360=1580N$

Hence, the total tension on the rigid support is $1580N$