Newtons law of motion

  • #1

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.

2.jpg


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 .. im too confuse

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Hello!
Are the blocks to remain static? Does this question keep them stationary?
Daniel
 
  • #3
Then this
m1g-T=m2a
is impossible, since a = 0.
So try again, on that point alone.
 
  • #4
sorry it is kenetic..
 
Last edited:
  • #5
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:

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