What is the force \( \vec F \) that an operator must apply to the rope to keep the 50 kg block in equilibrium?

Problem Data:

• Mass of the suspended block:    m = 50 kg;
• Acceleration due to gravity:    g = 9.8 m/s².

Problem Diagram:

The gravitational force \( \vec F_g \) of the block is transmitted to the support of pulley C. The tension force \( \vec T \) is equally divided between the two sides of pulley C. The rope on the left side of pulley C transmits the tension force \( \vec T \) to the left side of pulley B, therefore the tension force on the right side of pulley B is also equal to \( \vec T \). The rope on the right side of pulley B transmits the tension force \( \vec T \) to pulley C (Figure 1).

Figure 1

The rope on the right side of pulley C transmits the tension force \( \vec T \) to the right side of pulley A, therefore the tension force on the left side of pulley A is also equal to \( \vec T \). Therefore, the force \( \vec F \) applied by the operator is equal to \( \vec T \) (Figure 2).

Figure 2

Solution:

The gravitational force is given by

the gravitational force of the block will be

The three tension forces equal to \( \vec T \) in the ropes of pulley C balance the gravitational force of the block (Figure 1). Applying the equilibrium condition, the summation of the forces must be equal to zero.

substituting the gravitational force found above, in magnitude.

As the force \( \vec F \) exerted by the operator is equal to the tension force \( \vec T \) (Figure 2).