Answer:
T = (3μmg - Fcosθ)/2
Explanation:
Since the boxes move to the right, the net force on box of mass 2m is
Fcosθ - 2μmg + T = ma (1)where Fcosθ = horizontal component of applied force, 2μmg = frictional force and T = tension in string and a = acceleration of box
The net force on the box with mass m is
μmg - T = ma (2) where μmg = frictional force, T = tension and a = acceleration of boxes.
Equating (1) and (2) we have
Fcosθ - 2μmg + T = μmg - T
collecting like terms
Fcosθ - 2μmg - μmg = -T - T
Fcosθ - 3μmg = -2T
(3μmg - Fcosθ)/2 = T
T = (3μmg - Fcosθ)/2