As you are trying to move a heavy box of mass m, you realize that it is too heavy for you to lift by yourself. There is no one around to help, so you attach an ideal pulley tothe box and a massless rope to the ceiling, which you wrap aroundthe pulley. You pull up on the rope to lift the box. Use g or themagnitude of the acceleration due to gravity and neglect frictionforces.
Part A What is the magnitude F of the upward force you must apply to the ropeto start raising the box with constant velocity?

magnitude of applied force is FA = mg + F

Where F is the resultant force downward that the rope moves with

Explanation:

Force downwards F is,

F = FA - T

T is the upwards tension force on the rope

FA is the actual applied force in pulling the rope down.

Therefore, T = FA - F .....equ. (1)

For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)

Therefore, ma = T - mg

T = ma + mg ..... equ. (2)

Equating equ. 1 and 2

T = FA - F = ma + mg

Therefore FA = ma + mg + F

But at constant velocity a = 0

Magnitude of applied force becomes

FA = mg + F

See image below

the answer is a. at the equator

positive reinforcement? ?