Answer :
Each worker must exert a force of approximately 165.0 lb to hold up the plank and load.
To solve this problem, we can use the principle of torque equilibrium.
Torque is the product of force and distance from the pivot point.
First, we find the torque exerted by the plank itself.
Since the plank is uniform, we can consider its weight to act at its center, which is 7.5 ft from either worker.
[tex]\[ \text{Torque by plank} = \text{Weight of plank} \times \text{Distance} = 22.0 \, \text{lb} \times 7.5 \, \text{ft} = 165.0 \, \text{lb-ft}\][/tex]
Then, we calculate the torque exerted by the load of blocks. This is given by the product of the load's weight and its distance from the first worker.
[tex]\[ \text{Torque by load} = \text{Weight of load} \times \text{Distance} = 165 \, \text{lb} \times 7.00 \, \text{ft} = 1155.0 \, \text{lb-ft}\][/tex]
For torque equilibrium, the sum of torques must be zero.
[tex]\[ \text{Torque by plank} + \text{Torque by load} = 0\][/tex]
[tex]\[ \text{Torque by plank} = - \text{Torque by load} \][/tex]
[tex]\[ \text{Force exerted by each worker} = \frac{\text{Torque by load}}{\text{Distance from pivot}}\][/tex]
[tex]\[ \text{Force exerted by each worker} = \frac{1155.0 \, \text{lb-ft}}{7.00 \, \text{ft}}\][/tex]
Force exerted by each worker ≈ 165.0 lb
Therefore, each worker must exert a force of approximately 165.0 lb to hold up the plank and load.
