a) To keep the body in static equilibrium, the force exerted by man must be equal to the gravitational
force of the body from the static equilibrium
\[
\begin{gather}
\vec F={\vec F}_g \tag{I}
\end{gather}
\]
In the body acts the gravitational force
\( {\vec F}_g \)
and tension force
\( \vec{T} \)
in the rope
\[
\begin{gather}
{\vec F}_g=\vec{T} \tag{II}
\end{gather}
\]
The tension force is transmitted through the rope to the pulley where the body is attached. For the
system to be in equilibrium, the tension force must be equally divided between both sides of the pulley.
On each side, we have a tension equal to
\( \frac{\vec T}{2} \)
acting on the rope. The lower pulley is attached to another pulley where acts this tension
\( \frac{\vec T}{2} \).
For the equilibrium of the system the tension must be divided between the two sides of the middle
pulley. On each side, we will have a tension equal to
\( \frac{\vec T}{4} \).
This tension is transmitted by the rope to the pulley fixed to the ceiling, this pulley only transmits
the tension from one side to another, where the man holds the rope. Using equation (II), we find
that the gravitational force that will act on the rope at this point is
\( \frac{{\vec F}_g}{4} \),
and a system of two movable pulleys divides the load's weight by 4. Thus, from equation (I) the force
the man exerts will be