What is the acceleration of each mass with a friction coefficient of .25?

[willingtolearn]

Find the acceleration experienced by each of the two masses shown in the picture below if the coefficient of friction between the 7 kg mass and the plane is .25 ?
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In the picture, is the force of fiction go that way ?
http://img515.imageshack.us/img515/893/img3394dj7.jpg
m1 = 7kg u = .25
m2 = 12kg
F (per) = F (normal)
F (per) = 68.6 cos 37 = 54.8 N
F (parallel) = 68.66 sin 37 = 41.3 N
F net = (12*9.8) - 1.75 = 115.85 N
115.85 / 19 = 6.1 m/s2
So 7 kg is accelerate at 6.1 m/s2 and 12 kg is the same but in the negative direction.
Is this right ?

 

[learningphysics]

can you explain this part:

"F net = (12*9.8) - 1.75 = 115.85 N"

how are you getting 1.75... and which mass are you taking Fnet for...

 

[willingtolearn]

can you explain this part:

"F net = (12*9.8) - 1.75 = 115.85 N"

how are you getting 1.75... and which mass are you taking Fnet for...

in this problem, there are only 2 force make the block able to slid. (and it is force of gravity on 12 kg and force friction on 7 kg.
So F net = F (gravity) - F (fiction)

 

[learningphysics]

but how do you get 1.75N?

Also what about the F (parallel) = 68.66 sin 37 = 41.3 N?

 

[willingtolearn]

F (friction) = .25 time 7 = 1.75

 

[learningphysics]

F (friction) = .25 time 7 = 1.75

But friction = 0.25*Fnormal = 0.25*7*9.8*cos37.

Although it might be faster to do the whole problem at once... I think it's best to divide into 2 sections...

do the freebody diagram of the first mass, get the equations there... then the second mass, get the equations there... use T for tension...

 

[willingtolearn]

ok, I got this
a = 3.3 m/s2

 

[learningphysics]

ok, I got this
a = 3.3 m/s2

looks good to me. that's what I get also.

 

[willingtolearn]

Yes ... thanks learningphysics