Two 5 kg. masses connected by a light rope act upon a force, F, which is parallel to the incline of 37 degrees

Question

The coefficents of friction between the masses incline are mew sub s (u s)= 0.4 and mew sub k (u k)= 0.2
a. find the minimum force necessary to prevent the masses from sliding down the incline
b. find the tension in the rope reuslting from this force
*The masses are then given a brief push up the incline. Using the force found in part a), find

c. the acceleration of the system
d. the tension in the rope

*The masses are stopped and then given a brief push down the incline. Using the force found in part a) which continues to point up the incline, find
e. the acceleration of the system
f. the tension in the rope.

I know the answers, I just cannot figure out the work:
ANSWERS
a. 28 N
b. 14 N
c. 4.8 m/s2 down the incline
d. 14 N
e. 1.6 m/s 2 down the incluine
f. 14 N

Answer 1

The force balance on the lower mass
T = mgsinθ - μmgcosθ = mg(sinθ - μcosθ)
F = T + mg(sinθ - μcosθ) = 2mg(sinθ - μcosθ)
F = 2(5)(9.80665)(sin37 - 0.4cos37)
a. F ≈ 27.69014 N ≈ 27.690 N
b. T = F/2 ≈ 13.845 N
a = (F - 2mg(sinθ + μcosθ))/(2m)
a = (F/(2m) - (sinθ + μcosθ))
a = (27.690/10 - g(sin37 + 0.2cos37))
c. a ≈ - 4.699163 m/s^2 ≈ 4.70 m/s^2 down
T = F/2
d. T ≈ 27.690 N
a = (27.690/10 - g(sin37 - 0.2cos37))
e. a ≈ - 1.566388 m/s^2 ≈ 1.6 m/s^2 down
f. T = F/2 ≈ 27.690 N

Answer 2

Draw this out: the 30N rigidity hits the 25kg merchandise from the left and the 40N rigidity from the front. the 30N rigidity hitting the section will reason an equation that appears like this: acceleration=30N/25kg=30kg*mps^2/ 25kg=a million.2m/s^2 (mps=m/s^2) the 40Nforce hitting the front will reason this equation: acceleration=40N/25kg=40kg*mps^2/25kg= a million.6m/s^2 contained interior the precise the object will flow at an innovations-set decrease back left yet extra beneficial decrease back as a results of extra constructive foce from the front.