Newtons law of motion

[apprentice213]

Homework Statement

Homework Statement
the blocks shown in the figure are connected by flexible, inextensible cords passing over frictionless pulleys. at A, the coefficient fiction is 0.30 while at B, it is 0.40. compute the magnitude and direction of the friction forces and each block.

Homework Equations

for block A=
T-mgsin53.13-Fmax=0
Fmax=0.3(300cos53.13)

for block B=
T-mgsin36.87-Fmax=0
Fmax=0.40(200cos36.87)

m(1)g-T=m(2)

The Attempt at a Solution

for A:
Fmax=(o.30)(300cos53.13)
Fmax=54

T= 300(9.81)sin53.13
T=2354.4

for B :
Fmax=0.4(200cos36.87)
Fmax=54
T=200(9.81)sin36.87
T=1177.20

m1g-T=m2a
a=(2943-T)/200
a=244.3

AM i RiGHT ? please help thnx .. I am too confuse

Homework Statement

Homework Equations

The Attempt at a Solution

 

[danielakkerma]

Hello!
Are the blocks to remain static? Does this question keep them stationary?
Daniel

 

[danielakkerma]

Then this

m1g-T=m2a

is impossible, since a = 0.
So try again, on that point alone.

 

[apprentice213]

sorry it is kenetic..

 

[apprentice213]

for A:

Fmax=μk(m1)gcos53.13
54=μk(m1)gcos53.13
μk=0.03

T-300(9.81)sin53.13-μk(300)9.81cos53.13=(300)a eq(1)
200g-T=200a eq(2)

subtitute (1) and (2) to find a

a=-3.79

for block B:

Fmax=μk(m2)gcos36.87
64=μk(m2)gcos36.87
μk=0.04



T-200(9.81)sin36.87-μk(200)9.81cos36.87=(200)a eq(1)
300g-T=300a eq(2)

subtitute (1) and (2)

to find a

a=0.2


AM i DOiNG RiGHT ?? please comment .. I am lack of knowledge ..