The given problem can be solved using the following free-body diagram:
The diagram is the free-body diagram for the pulley that is holding the weight. Where:
[tex]\begin{gathered} T=\text{ tension} \\ m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]
Now we add the forces in the vertical direction:
[tex]\Sigma F_v=T+T-mg[/tex]
Adding like terms:
[tex]\Sigma F_v=2T-mg[/tex]
Now, since the velocity is constant this means that the acceleration is zero and therefore the sum of forces is zero:
[tex]2T-mg=0[/tex]
Now we solve for "T" by adding "mg" from both sides:
[tex]2T=mg[/tex]
Now we divide both sides by 2:
[tex]T=\frac{mg}{2}[/tex]
Now we substitute the values and we get:
[tex]T=\frac{(64\operatorname{kg})(9.8\frac{m}{s^2})}{2}[/tex]
Solving the operations:
[tex]T=313.6N[/tex]
Now we use the free body diagram for the second pulley:
Now we add the forces in the vertical direction:
[tex]\Sigma F_v=T-F[/tex]
The forces add up to zero because the velocity is constant and the acceleration is zero:
[tex]T-F=0[/tex]
Solving for the force:
[tex]T=F[/tex]
Therefore, the pulling force is equal to the tension we determined previously and therefore is:
[tex]F=313.6N[/tex]
