Three boxes lie on an inclined plane a theta= 30o with respect to the horizontal

Question

Three boxes lie on an inclined plane a theta= 30o with
respect to the horizontal. The boxes are connected by
strings of constant length that lie parallel to the inclined
plane, and an applied force
??????F acts parallel to the inclined plane as illustrated. You are given that
MC= 5.00kg, MB= 2MC , and MA= 3MC and that
the coefficients of kinetic (sliding) friction between the
lower surfaces of the boxes and the slope are the same
and are equal to ? k = 0.200 . The acceleration of the
three boxes up the slope is 1.00 m/s2.
A13. Calculate the magnitude of the applied force.
A14. Calculate the magnitude of the tension in the string joining boxes B and A when the system is moving up the slope
with the acceleration given in part (i)

Three boxes lie on an inclined plane a theta= 30o with respect to the horizontal. The boxes are connected by strings of constant length that lie parallel to the inclined plane, and an applied force ??????F acts parallel to the inclined plane as illustrated. You are given that MC= 5.00kg, MB= 2MC , and MA= 3MC and that the coefficients of kinetic (sliding) friction between the lower surfaces of the boxes and the slope are the same and are equal to ? k = 0.200 . The acceleration of the three boxes up the slope is 1.00 m/s2. A13. Calculate the magnitude of the applied force. A14. Calculate the magnitude of the tension in the string joining boxes B and A when the system is moving up the slope with the acceleration given in part (i)

Explanation / Answer

F = applied system force UP and parallel to plane M = total system mass = 5 + 10 + 15 = 30 kg W = total system weight = 30(9.8) = 294 N Component of W acting parallel & DOWN incline = Wd = 294 sin 30 = 147 N Component of W acting Normal to incline = 294 cos 30 = 255 N µk = 0.200 System friction force acts parallel and DOWN incline = FF = (0.2)(255) = 51 N The SYSTEM acceleration = 1.00 m/s² Fnet = ma = (30)(1) = 30 N Fnet = F + Wd + FF 30 = F - (147 + 51) = F - 198 F = 198 - 30 = 168 N ANS A13 Let T = size of the tension in the string joining boxes B & A: since all boxes are tied together to make up the combined SYSTEM of boxes, the net ACCELERATION of each box UP incline is the SAME = 1.00 m/s². the net FORCE up incline on Box A = ma = (15)(1) = 15 N A Free-Body diagram of Box A shows its weight = mg = (15)(9.8) = 16 N component of A's weight acting parallel and down incline = 16 sin 30 = 8 N component of A's weight acting Nomal to incline = 16 cos 30 = 13.9 N friction force on A acting parallel & down incline = (0.2)(13.9) = 2.8 N tension(T) in string between B & A = 15 = T - (8 + 2.8) = T - 10.8 T = 15 + 10.8 ˜ 26 N ANS A14

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