Newtons law of motion

1. Sep 26, 2011

apprentice213

1. The problem statement, all variables and given/known data

The problem statement, all variables and given/known data
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.

2. Relevant 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)
3. 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

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 26, 2011

danielakkerma

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

3. Sep 26, 2011

danielakkerma

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

4. Sep 26, 2011

apprentice213

sorry it is kenetic..

Last edited: Sep 26, 2011
5. Sep 26, 2011

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 .. im lack of knowledge ..

Last edited: Sep 26, 2011