Acceleration coefficient

  • #1
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 ?
------------
In the picture, is the force of fiction go that way ?
http://img515.imageshack.us/img515/893/img3394dj7.jpg [Broken]
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 ?
 
Last edited by a moderator:

Answers and Replies

  • #2
learningphysics
Homework Helper
4,099
6
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...
 
  • #3
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)
 
  • #4
learningphysics
Homework Helper
4,099
6
but how do you get 1.75N?

Also what about the F (parallel) = 68.66 sin 37 = 41.3 N?
 
  • #5
F (friction) = .25 time 7 = 1.75
 
  • #6
learningphysics
Homework Helper
4,099
6
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...
 
  • #9
Yes ... thanks learningphysics
 

Related Threads on Acceleration coefficient

Replies
14
Views
2K
Replies
3
Views
6K
Replies
2
Views
3K
Replies
3
Views
10K
Replies
15
Views
2K
Replies
10
Views
17K
Replies
2
Views
5K
Replies
4
Views
2K
Replies
8
Views
13K
Replies
10
Views
8K
Top